用户: Solution/ 习题: 实分析

If you find any typesetting problems feel free to correct it since I am busy.

Also, there are some troublesome questions I left blank, you can complete them if you are interested.

Sets

Problem 1. Given a linear space $X$, and its linear subspace $V$, then there is a linear subspace $X$ of $W$, and $X$ is the direct sum of $V$ and $W$.

Proof. First we recall Zorn’s Lemma:

Lemma. Let $X$ be a partially ordered non-empty set for which every totally ordered subset has an upper bound. Then $X$ has a maximal element.

Let $\Gamma$ denote the collection of all linear subspaces of $X$, all of which share with $V$ only the $0$ vector. It is trivial that zero space is in $\Gamma$, so it is not empty. Define a partial order $\prec$ as follows: $Y_1\prec Y_2$ if $Y_1\subset Y_2$.

Now we verify that any totally ordered subset ${\Gamma }^{\prime }$ of $\Gamma$ has an upper bound. It suffices to show that $Y^{\prime }={\bigcup }_{Y\in {\Gamma }^{\prime }}Y$ is in $\Gamma$.

First, it is obvious that $Y^{\prime }\cap V=\varnothing$.

Second, verify that $Y^{\prime }$ is a linear subspace of $X$. For any $y_1,y_2\in Y^{\prime }$ and scalar $k$, there is $Y_1,Y_2\in {\Gamma }^{\prime }$, and $y_i\in Y_i(i=1,2)$. Since ${\Gamma }^{\prime }$ is totally ordered, we could assume $Y_1\subset Y_2$. Then $y_i\in Y_2$, so $y_1+y_2,k{y}_1\in Y_2\subset Y^{\prime }$, since $Y_2$ is a linear subspace of $X$.

Thus we could use Zorn’s Lemma to get a maximal element $W$ of $\Gamma$. We will show that $X=V\oplus W$. If it is not the case, there exists $x$ in $X$, $x$ cannot be decomposed as sum of vectors in $V$ and $W$. Then let $W^{\prime }$ denote the subspace generated by $x,W$, and $W^{\prime }\in \Gamma$, contradicting to the maximality of $W$.

$\square$

Problem 2. If $A$ is countable, $f:A\to B$ is surjective, then $B$ is countable.

Proof. Since $f$ is surjective, then for any $b\in B$, the pre-image of $b$ is non-empty. Choose $g(b)\in f^{-1}(b)$ arbitrarily for all $b\in B$, then $g$ is injective. Thus $B,g(B)$ are equipotent, but $g(B)\subset A$ implies that $g(B)$ is countable, then $B$ is countable.

$\square$

Problem 3. Suppose that $A\subset \mathbb{R}$, for any $x\in A$, there is $\delta >0$, such that$$(x-\delta ,x+\delta )\cap A$$is countable. Then $A$ is countable.

Proof. $\forall x\in A,\exists a_x<x<b_x\in \mathbb{Q}$, with $(a_x,b_x)\cap A$ countable. Then: $$A=\bigcup _{x\in A}\left(\left(a_x,b_x\right)\cap A\right)\subset \bigcup _{\begin{array}{c}a<b\in \mathbb{Q},\\ (a,b)\cap A\text{ countable}\end{array}}\left(a,b\right)\cap A$$is countable since it is a subset of a countable union of countable sets.

$\square$

Problem 4. Construct a bijection from ${\mathbb{R}}^2$ to $\mathbb{R}$.

Proof. The readers are invited to surf the internet for answers.(Or can someone complete this?)

Hint: Consider the numbers in $[0,1)$, and write them in decimals. Assume that we write the finite decimals NOT in the form of $0.XXX...XXX\mathrm{999...}$, but $0.XXX...XXX\mathrm{000...}$

Then we can "split" a number into two in the following way:

Example: $\mathrm{0.23989479837099981...}$

Write a bar after each number other than $9$: $0.2|3|98|94|7|98|3|7|0|9998|1|...$

We have infinite many bars since there is no $0.XXX...XXX\mathrm{999...}$

Then we get two numbers: $0.2|98|7|3|0|1|...$ and $0.3|94|98|7|9998|...$. We can prove that this gives a bijection between $[0,1)$ and $[0,1)\times [0,1)$ by writing its inverse (which is obvious).

$\square$

Measure Space

Problem 5. Given a measure space $(X,\mathcal{X})$, and an arbitrary function $f:X\to \mathbb{R}$. We have$$\mathcal{A}=\{A\subset \mathbb{R}:f^{-1}(A)\in \mathcal{X}\}$$is a $\sigma$-algebra.

Proof. By definition, we will check the following conditions:

1.	$\varnothing ,\mathbb{R}\in \mathcal{A}$: $f^{-1}(\varnothing )=\varnothing \in \mathcal{X},f^{-1}(\mathbb{R})=X\in \mathcal{X}.$
2.	if $A\in \mathcal{A}$, then $A^c\in \mathcal{A}$: $f^{-1}(A^c)={\left(f^{-1}(A)\right)}^c\in \mathcal{X}.$
3.	if ${\left\{A_n\right\}}_{n=1}^{\infty }$ is a sequence in $\mathcal{A}$, then ${\bigcup }_{n=1}^{\infty }{A}_n\in \mathcal{A}$: $f^{-1}({\bigcup }_{n=1}^{\infty }{A}_n)={\bigcup }_{n=1}^{\infty }{f}^{-1}(A_n)\in \mathcal{X}.$

$\square$

Problem 6. Determine if it is true and give a proof: The collection of irrationals in $[0,1]$, the decimal expansion of which contain no digit “$7$”, is Borel.

Proof. The following discussion is within $[0,1]$ and using decimal expansion. Let $A$ denote those numbers whose expansion contain digit “$7$”, then it suffices to check $A$ is Borel. (Why?)

Let $B_n$ denote numbers with the $n$-th position of expansion “$7$”, then $B_n$ could be decomposed of countable disjoint intervals, thus it is Borel. Since $A=\bigcup B_n$, $A$ is also Borel.

$\square$

Problem 7. The $\sigma$-algebra generated by intervals of the form $(-\infty ,a)$ is exactly $\mathcal{B}(\mathbb{R})$.

Proof. With the closure of complement and countable union operations in $\sigma$-algebra, it is not hard to construct all open intervals, then the $\mathcal{B}(\mathbb{R})$.

$\square$

Problem 8. Determine if it is true and give a proof: Let $\mathcal{C}$ denote the $\sigma$-algebra of the collection of all subsets of $\mathcal{R}$, then the following set functions are measures on $(\mathbb{R},\mathcal{C})$:$${\mu }_1\left(A\right)=\{\begin{array}{ll}0,& \text{if }A\text{ is finite}\\ \infty ,& \text{if }A\text{ is infinite.}\end{array},{\mu }_2\left(A\right)=\{\begin{array}{ll}0,& \text{if }A\text{ is countable}\\ \infty ,& \text{if }A\text{ is uncountable.}\end{array}$$

Proof. Since ${\mu }_1\left(\mathbb{Z}\right)=\infty \ne 0={\sum }_{n\in \mathbb{Z}}^{}{\mu }_1\left(\{n\}\right)$, we know ${\mu }_1$ is not a measure.

It is trivial to check by definition that ${\mu }_2$ is a measure.

$\square$

Problem 9. Let $\mathcal{C}$ denote the collection of subsets of $\mathbb{N}$ that make this limit exist:$$\underset{n\to \infty }{\lim }\frac{|A\cap \left\{1,\cdots ,n\right\}|}{n}$$

And in case it exists, let $p(A)$ denote it. Determine if $(\mathbb{N},\mathcal{C},p)$ is a measure space.

Proof. In fact, the countable additivity is not satisfied.

First we construct a set $A\in \mathbb{N}$, and the limit does not exist: $$A=\left\{j\in \mathbb{N},\exists n\in {\mathbb{Z}}_{+},2^n\le j<3\times 2^{n-1}\right\}$$

Then denote $A_n=|A\cap \left\{1,\cdots ,n\right\}|$, it is easy to observe: $$\begin{array}{c}{A}_{3\times 2^{n-1}}=\sum _{k=1}^n{2}^{k-1}=2^n-1,A_{2^n-1}=2^{n-1}-1\\ \underset{n\to \infty }{\lim }\frac{A_{3\times 2^{n-1}}}{3\times 2^{n-1}}=\frac{2}{3},\underset{n\to \infty }{\lim }\frac{A_{2^n-1}}{2^n-1}=\frac{1}{2}\end{array}$$

Which implies what we want. Then we seperate $A$ into countable disjoint union of unit subsets. Each of them are in $\mathcal{C}$, but $A$ is not. A contradiction!

$\square$

Problem 10. Assume $(X,\mathcal{X},\mu )$ to be a measure space, for any $E_1,E_2\in \mathcal{X}$, we have$$\mu \left(E_1\cup E_2\right)+\mu \left(E_1\cap E_2\right)=\mu \left(E_1\right)+\mu \left(E_2\right).$$

Proof. If the right hand side is infinity, then it is trivial.

Otherwise, due to countable additivity, we obtain $$\begin{array}{rl}& \mu (E_1\cup E_2)=\mu (E_1\cap E_2)+\mu (E_1-E_2)+\mu (E_2-E_1)\\ & \mu (E_1)=\mu (E_1\cap E_2)+\mu (E_1-E_2)\\ & \mu (E_2)=\mu (E_1\cap E_2)+\mu (E_2-E_1)\end{array}$$

Which implies what we want.

$\square$

Problem 11. Let $E_1,E_2,\cdots ,E_n$ be Borel subsets of $[0,1],{\sum }_{k=1}^{n}m(E_k)>n-1$, then ${\bigcap }_{k=1}^n{E}_k$ is not empty.

Proof. Let $F_k=[0,1]\backslash E_k$, then ${\bigcap }_{k=1}^n{E}_k=\left[0,1\right]\backslash {\bigcup }_{k=1}^n{F}_k$, which implies: $$\begin{array}{c}m\left(\bigcap _{k=1}^n{E}_k\right)=1-m\left(\bigcup _{k=1}^n{F}_k\right)\ge 1-\sum _{k=1}^{n}m(F_k)=1-\sum _{k=1}^n(1-m(E_k))>0\end{array}$$

and it is obvious that ${\bigcap }_{k=1}^n{E}_k$ is not empty.

$\square$

Problem 12. (Poincaré Recurrence Theorem) Let $(X,\mathcal{X},\mu )$ be a probability space (a measure space with $\mu (X)=1$), $f:X\to X$ is a measure-preserving transformation, which means: for any $E$ measurable, $f^{-1}(E)$ is also measurable, and their measure coincide. Then for any set $A$ of positive measure, there exists $x\in A,n\in {\mathbb{N}}_{+}$, with $f^{\circ n}(x)\in A$.

Proof. We prove it by contradiction. If $m(A)>0$, but $\forall x\in A,n\in {\mathbb{N}}_{+},f^{\circ n}(x)\notin A$, then $f^{\circ n}(A)\cap A=\varnothing$, and we can easily prove by contradiction $f^{\circ (-n)}(A)\cap A=\varnothing$. Then again, $A,f^{-1}(A),\cdots ,f^{\circ (-n)}(A),\cdots$ are disjoint. Finally we obtain $$\mu \left(\bigcup _{n=1}^{\infty }{f}^{\circ (-n)}(A)\right)=\sum _{n=1}^{\infty }\mu \left(f^{\circ (-n)}(A)\right)=\sum _{n=1}^{\infty }\mu (A)=\infty$$

Contradicting to $\mu (X)=1$!

$\square$

Problem 13. $\forall E\subset \mathbb{R},m^{\ast }(E)=\inf \{m^{\ast }(U),U\text{open},E\subset U\}$

Proof. For any $\epsilon >0$, let $\{I_k\}$ be a countable open cover of $E$, $E\subset \bigcup I_k,{\sum }_{}^{}m(I_k)\le m^{\ast }(E)+\epsilon$.

Since $U=\bigcup I_k$ is open, and $E\subset U$, we get: $$m^{\ast }(U)\le \sum m^{\ast }(I_k)=\sum m(I_k)\le m^{\ast }(E)+\epsilon .$$

Using the definition of infimum, we are done.

$\square$

Problem 14. If $E,F\subset \mathbb{R}$ with $m^{\ast }(E△F)=0$, where $E△F=(E\backslash F)\cup (F\backslash E)$. $E$ is Lebesgue measurable if and only if $F$ is Lebesgue measurable.

Proof. Due to the symmetry between $E,F$, we only deal with the case where $E$ is measurable. Then for any $T\subset \mathbb{R}$, we have: $$m^{\ast }(T)=m^{\ast }(T\cap E)+m^{\ast }(T\cap E^c).$$To prove $F$ is measurable, it suffices to show: $$m^{\ast }(T)\ge m^{\ast }(T\cap F)+m^{\ast }(T\cap F^c).$$

Then we observe and make use of the assumption: $$T\cap F\subset (T\cap E)\cup (E△F),m^{\ast }(T\cap F)\le m^{\ast }(T\cap E)$$$$T\cap F^c\subset (T\cap E^c)\cup (E△F),m^{\ast }(T\cap F^c)\le m^{\ast }(T\cap E^c)$$

Which implies what we want.

$\square$

Problem 15. $\forall A,B\subset \mathbb{R},m^{\ast }(A\cup B)+m^{\ast }(A\cap B)\le m^{\ast }(A)+m^{\ast }(B).$

Proof. Using 13, we have $\forall \epsilon >0,\exists U,V$ open, with: $$A\subset U,B\subset V,m^{\ast }(U)+m^{\ast }(V)<m^{\ast }(A)+m^{\ast }(B)+\epsilon .$$

Since 10 holds for $U,V$, we obtain $$\begin{array}{rl}& m^{\ast }(A\cap B)+m^{\ast }(A\cup B)\le m^{\ast }(U\cap V)+m^{\ast }(U\cup V)\\ =& m^{\ast }(U)+m^{\ast }(V)<m^{\ast }(A)+m^{\ast }(B)+\epsilon .\end{array}$$

Then we are done since $\epsilon$ is arbitrary.

$\square$

Problem 16. Let $d(A,B)>0$, then $m^{\ast }(A\cup B)=m^{\ast }(A)+m^{\ast }(B)$.

Proof. If the right hand side is infinity, then it is trivial.

Otherwise, for any $\epsilon >0$, let $\{I_k\}$ be a countable open cover of $A\cup B$, ${\sum }_{}^{}m(I_k)\le m^{\ast }(A\cup B)+\epsilon$.

Since $d(A,B)>0$, we assume $m(I_k)<d(A,B)$, then $\{I_k\}$ cound be divided into two groups: $\{I_{1k}\}$ covering $A$ and $\{I_{2k}\}$ covering $B$. So we have $$\begin{array}{c}{m}^{\ast }(A)+m^{\ast }(B)\le \sum m(I_{1k})+\sum m(I_{2k})\\ =\sum m(I_k)\le m^{\ast }(A\cup B)+\epsilon .\end{array}$$

Then we are done since $\epsilon$ was arbitrary.

$\square$

Problem 17. Assume $m^{\ast }(E)<\infty$, and for any reals $a<b$,$$b-a=m^{\ast }((a,b)\cap E)+m^{\ast }((a,b)\backslash E).$$

Then $E$ satisfies Caratheodory condition.

Proof. $\forall T\subset \mathbb{R}$, our goal is to prove $m^{\ast }(T)\ge m^{\ast }(T\cap E)+m^{\ast }(T\cap E^c)$.

For any $\epsilon >0$, let $\{I_k\}$ be a countable set of open intervals covering $T$, $T\subset \bigcup I_k,\sum m(I_k)\le m^{\ast }(T)+\epsilon$.

We obtain: $$\begin{array}{c}T\cap E\subset \bigcup (I_k\cap E),T\cap E^c\subset \bigcup (I_k\cap E^c)\\ m^{\ast }(T\cap E)+m^{\ast }(T\cap E^c)\le \sum (m^{\ast }(I_k\cap E)+m^{\ast }(I_k\cap E^c))\\ =\sum m^{\ast }(I_k)<m^{\ast }(T)+\epsilon .\end{array}$$

Since $\epsilon$ was arbitrary, we are done.

$\square$

Construct the following:

Problem 18. $A\subset \mathbb{R}$ with $0<4m(A^{\circ })=2m(A)=m(\bar{A})<\infty$.

Proof. Observing the properties of rational numbers, we define $$\begin{array}{rlrlrl}& A_1=(0,2)\cap \mathbb{Q},& & A_1^{\circ }=\varnothing ,& & \overline{A_1}=[0,2]\\ & A_2=(2,3)\backslash \mathbb{Q},& & A_2^{\circ }=\varnothing ,& & \overline{A_2}=[2,3]\\ & A_3=(3,4),& & A_3^{\circ }=(3,4),& & \overline{A_3}=[3,4]\end{array}$$

Then it is easy to check $A_1\cup A_2\cup A_3$ is what we want.

$\square$

Problem 19. Construct a sequence of disjoint subsets of $\mathbb{R}$, such that the measure of the closure of their union is greater than the sum of the measure of their closure.

Proof. Let $\{r_n\}$ denote the sequence of all rational numbers. Then $A_n=\{r_n\}$ is a unit set. We have: $$m\left(\overline{\bigcup _{n=1}^{\infty }{A}_n}\right)=\infty >0=\sum _{n=1}^{\infty }m(\overline{A_n}).$$

$\square$

Measurable Functions

Problem 20. Given a measure space $(X,\mathcal{X})$, then $f:X\to \mathbb{R}$ is $\mathcal{X}$-measurable if and only if for any integer $k,n$, we have:$$\left\{x:f(x)>\frac{k}{2^n}\right\}\in \mathcal{X}$$

Proof. For any $c\in \mathbb{R}$, let a sequence of the form $k/2^n$ approximate it.

$\square$

Problem 21. Given a measure space $(X,\mathcal{X})$, and a sequence of measuable functions $f_n:X\to \mathbb{R}$.Denote$$E=\left\{x\in X:\left\{f_n(x)\right\}\text{ converges in }\mathbb{R}\right\}.$$

Then $E$ is measurable.

Proof. In fact, $\mathrm{limsup}{f}_n(x),\mathrm{liminf}{f}_n(x)$ are measurable. We have $E=\left\{x\in X:\mathrm{limsup}{f}_n(x)=\mathrm{liminf}{f}_n(x)\right\}$ is also measurable.

$\square$

Problem 22. Let $f:\mathbb{R}\to \mathbb{R}$ be pointwise differentiable, $E\subset \mathbb{R},m(E)=0$. Then $m(f(E))=0$.

Proof. We set $E_n=\{x\in E:|f^{\prime }(x)|<n\}$, then $f(E_n)$ increases to $f(E)$. Since $m(E)=0$, for any $\epsilon >0$, there exists an open set $U$ such that $E\subset U$ and $m(U)<\epsilon /n$.

For each $x\in E_n$, there exsits a binary interval $J_x$ such that $x\in J_x\subset U$ and $m(f(J_x))<nm(J_x)$.

Then $\mathcal{J}=\{J_x,x\in E_n\}$ is a cover of $E_n$. By removing the duplicate parts we obtain ${\mathcal{J}}_1\subset \mathcal{J}$, a disjoint cover of $E_n$. We have the estimate $$m(f(E_n))\le \sum m(f(J_x))\le n\sum m(J_x)=nm\left(\bigcup J_x\right)\le nm(U)<\epsilon$$

Thus $m(f(E_n))=0$ since $\epsilon$ was arbitrary. And finally $$m(f(E))=m(\lim f(E_n))=\lim m(f(E_n))=0.$$

$\square$

Problem 23. Let $f:{\mathbb{R}}^2\to \mathbb{R}$ be a continuous function. For any $x\in \mathbb{R}$, let $p(x)$ denote the number of y that satisfies $f(x,y)=0$. Then $p(x)$ is Borel measurable.

Proof. Since $p(x)$ is a positive integer-valued function, it suffices to prove $\{x:p(x)\ge n\}$ is borel for any positive integer $n$. Observe that $$\begin{array}{c}\left\{x:p(x)\ge n\right\}=\left\{x:\exists \text{ distinct }{y}_1,\cdots ,y_n,f(x,y_k)=0\right\}\\ =\left\{x:\exists N,m\in {\mathbb{N}}_{+},\exists \text{ distinct }{y}_1,\cdots ,y_n,|y_k|\le N,|y_j-y_k|\ge \frac{1}{m},f(x,y_k)=0\right\}\\ =\bigcup _{N=1}^{\infty }\bigcup _{m=1}^{\infty }{E}_n\end{array}$$

Where we denote $$E_n=\left\{x:\exists \text{ distinct }{y}_1,\cdots ,y_n,|y_k|\le N,|y_j-y_k|\ge \frac{1}{m},f(x,y_k)=0\right\}$$

Thus our problem is reduced to prove $E_n$ is closed. If so, it is Borel, then countable union of Borel sets is still Borel.

$\forall x_0\in E_n^{\prime },\exists \left\{x_r\right\}\subset E_n,\lim x_r=x_0$. since $x_r\in E_n$, $$\exists y_{r1},\cdots ,y_{rn}\text{ pairwise disjoint, }|y_{rk}|\le N,|y_{rj}-y_{rk}|\ge \frac{1}{m},f(x_r,y_{rk})=0.$$

Then for any $r,k$, $y_{rk}$ is bounded by $N$. Thus $\{(y_{r1},\cdots ,y_{rn})\}\subset {\mathbb{R}}^n$ is also bounded, and contains a subsequence $\{(y_{r_{t}1},\cdots ,y_{r_{t}n})\}$ which converges to $(y_{01},\cdots ,y_{0n})$. Still we have $|y_{0k}|\le N,|y_{0j}-y_{0k}|\ge \frac{1}{m}$, and since $f$ is continuous, we obtain $$f(x_0,y_{0k})=f(\lim \left(x_{r_t},y_{r_{t}k}\right))=\lim f(x_{r_t},y_{r_{t}k})=0.$$

Finally $x_0$ is in $E_n$. Thus $E_n$ is closed.

$\square$

Problem 24. Let $\mu$ be a Borel measure on $\mathbb{R}$. Then real-valued function$$f(x)=\underset{r\to 0}{\mathrm{liminf}}\frac{\mu ((x-r,x+r))}{r}$$

is Borel measuable.

Proof. Since $f(x)={\lim }_{r_0\to 0^{+}}{f}_{r_0}(x)$, where we denote $$f_{r_0}(x)=\underset{0<r<r_0}{\inf }\frac{\mu ((x-r,x+r))}{r}.$$

It is obvious that for a fixed $x$, the function $f_{r_0}(x)$ of $r_0$ is decreasing. Thus we obtain $f(x)=\sup f_{1/n}(x)$. Next we shall prove this $f_{1/n}(x)$ is Borel measurable, then the supremum of it is still Borel measurable.

We claim that for any $n\in {\mathbb{N}}_{+},c\in \mathbb{R}$, the set $\{f_{1/n}(x)<c\}$ is open.

For any $x_0\in \{f_{1/n}(x)<c\}$, we have $\{f_{1/n}(x_0)<c\}$. Using definition, there exists $0<r<1/n$, such that $\mu ((x_0-r,x_0+r))<cr$. Thus take one more $0<s\ll r$, such that $\mu ((x_0-r,x_0+r))<c(r-s)$.

For any $x\in U(x_0,s)$, we have $U(x,r-s)\subset U(x_0,r)$. Then $$\mu ((x-(r-s),x+(r-s)))\le \mu ((x_0-r,x_0+r))<c(r-s)$$$$\underset{0<r<\frac{1}{n}}{\inf }\frac{\mu ((x-r,x+r))}{r}\le \frac{\mu ((x-(r-s),x+(r-s)))}{r-s}<c$$Thus $f_{1/n}(x)<c$, namely $U(x_0,s)\subset \{f_{1/n}(x)<c\}$, so it is open, then Borel measurable.

$\square$

Lebesgue Integral

Problem 25. Let $(X,\mathcal{X},\mu )$ be a measure space with $\mu (X)<\infty$. $f:X\to \mathbb{R}$ is integrable if and only if:$$\sum _{n=1}^{\infty }{2}^n\mu (\left\{x\in X:|f(x)|>2^n\right\})<\infty .$$

Proof. Let $A_n=\left\{x\in X:2^n<|f(x)|\le 2^{n+1}\right\}$, then we denote: $$B_n=\bigcup _{m=n}^{\infty }{A}_m$$

this union is disjoint.Thus $$\begin{array}{c}I=\sum _{n=1}^{\infty }{2}^n\mu (B_n)=\sum _{n=1}^{\infty }{2}^n\sum _{m=n}^{\infty }\mu (A_m)=\sum _{m=1}^{\infty }\mu (A_m)\sum _{n=1}^m{2}^n\\ =\sum _{m=1}^{\infty }{\int }_{}^{}(2^{m+1}-1){\chi }_{A_m}\mathrm{d}\mu ={\int }_{}^{}\sum _{m=1}^{\infty }(2^{m+1}-1){\chi }_{A_m}\mathrm{d}\mu \\ {\int }_{}^{}|f(x)|-1\mathrm{d}\mu \le I<{\int }_{}^{}2|f(x)|-1\mathrm{d}\mu \end{array}$$

Since $\mu (X)<\infty$ and $|f|$ is integrable if and only if $f$ is integrable, we are done.

$\square$

Problem 26. Suppose ${\left\{f_n\right\}}_1^{\infty }\subset L^{+},f_n\to f$ pointwise, and $\int f=\lim \int f_n<\infty$. Then ${\int }_{E}f=\lim {\int }_E{f}_n$ $\forall E\in \mathcal{M}$. However, this need not be true if $\int f=\lim f_n=\infty$.

Proof. *:Applying Fatou Lemma to the complement of $E$.

Let $E\in \mathcal{M}$ and $\int f<\infty$, we have: $$\int f=\int \left({\chi }_{E}f+{\chi }_{E^c}f\right)=\int {\chi }_{E}f+\int {\chi }_{E^c}f$$And similarly for substituting $f_n$ for $f$ above. Now, since $f_n\to f\Rightarrow {\chi }_F{f}_n\to {\chi }_{F}f,\forall F\in \mathcal{M}$, we may apply Fatou’s Lemma as follows: $${\int }_{E}f=\int \underset{n\to \infty }{\mathrm{liminf}}{\chi }_E{f}_n\le \underset{n\to \infty }{\mathrm{liminf}}{\int }_E{f}_n\stackrel{\ast }{=}\underset{n\to \infty }{\mathrm{liminf}}\left(\int f_n-{\int }_{E^c}{f}_n\right)\stackrel{\ast \ast }{=}\int f-\underset{n\to \infty }{\mathrm{limsup}}{\int }_{E^c}{f}_n$$Where we have $\stackrel{\ast }{=}$ since ${\int }_{E}f=\int {\chi }_{E}f+\int {\chi }_{E^c}f$, and $\stackrel{\ast \ast }{=}$ since $\mathrm{liminf}\int f_n=\lim \int f_n=\int f$ and $\mathrm{liminf}-\int g=-\lim \sup \int g$. However, since all terms above are finite, we may gain by apply Fatou’s Lemma (and in noticing the similarity to the steps made above) to see that: $$\underset{n\to \infty }{\mathrm{limsup}}{\int }_{E^c}{f}_n=\underset{n\to \infty }{\mathrm{limsup}}\left(\int f_n-{\int }_E{f}_n\right)=\int f-\underset{n\to \infty }{\mathrm{liminf}}{\int }_E{f}_n\le \int f-{\int }_{E}f.$$And thus by substituting this in, we have: $${\int }_{E}f\le \underset{n\to \infty }{\mathrm{liminf}}{\int }_E{f}_n\le \underset{n\to \infty }{\mathrm{limsup}}{\int }_E{f}_n\le {\int }_{E}f.$$And therefore all the inequalities in the equation(s) above are actually equalities, and so we have: $$\underset{n\to \infty }{\mathrm{liminf}}{\int }_E{f}_n=\underset{n\to \infty }{\mathrm{limsup}}{\int }_E{f}_n=\underset{n\to \infty }{\lim }{\int }_{E}f={\int }_{E}f.$$We now turn our attention showing the above result need not hold if $\int f=\lim \int f=\infty$ by means of a counter-example. Let $E=(0,1],f={\chi }_{[2,\infty )}$, and $f_n={\chi }_{[2,\infty )}+n{\chi }_{(0,1/n]}$. Then $f_n\to f$ p.w., and: $${\int }_{(0,1]}{f}_n=n\mu ((0,1/n])=1\forall n\in \mathbb{N}\Rightarrow \underset{n\to \infty }{\lim }{\int }_{(0,1]}{f}_n=1.$$However, ${\int }_{(0,1]}f=0$, thus ${\int }_{E}f=\lim {\int }_E{f}_n$ need not be true if $\lim \int f_n=\int f=\infty$.

$\square$

Problem 27. Suppose that $\left\{f_j\right\}$ is a sequence in $L^1$ such that ${\sum }_1^{\infty }{\int }_j\left|f_j\right|<\infty$. Then ${\sum }_1^{\infty }{f}_j$ converges a.e. to a function in $L^1$, and $\int {\sum }_1^{\infty }{f}_j={\sum }_1^{\infty }\int f_j$.

Proof. $\int {\sum }_1^{\infty }\left|f_j\right|={\sum }_1^{\infty }\int \left|f_j\right|<\infty$, so the function $g=$ ${\sum }_1^{\infty }\left|f_j\right|$ is in $L^1$. In particular, ${\sum }_1^{\infty }\left|f_j(x)\right|$ is finite for a.e. $x$, and for each such $x$ the series ${\sum }_1^{\infty }{f}_j(x)$ converges. Moreover, $\left|{\sum }_1^n{f}_j\right|\le g$ for all $n$, so we can apply the dominated convergence theorem to the sequence of partial sums to obtain $\int {\sum }_1^{\infty }{f}_j={\sum }_1^{\infty }\int f_j$.

$\square$

Problem 28. Let $f$ be a non-negative measurable function, $\{E_k{\}}_{k=1}^{\infty }$ is a sequence of measurable sets, such that$$\underset{k\to \infty }{\lim }\mu (X\backslash E_k)=0.$$

Then$$\lim {\int }_{E_k}^{}f\mathrm{d}\mu ={\int }_{}^{}f\mathrm{d}\mu .$$

Proof.

Hint: Let $An$ denote the $x$ that $f(x)>n$. Use theories of summation of series to solve the problem when the equality given takes real values.

$\square$

Problem 29. Let $f$ be a positive-valued measurable function. For any $\alpha >0$,$$\inf \left\{{\int }_E^{}f\mathrm{d}\mu :\mu (E)\ge \alpha \right\}>0.$$

Proof. We prove it by contradiction. If there exists $\alpha >0$, such that $$\inf \left\{{\int }_E^{}f\mathrm{d}\mu :\mu (E)\ge \alpha \right\}=0,$$

then we can choose a sequence $\{E_k{\}}_{k=1}^{\infty }$, each with $\mu (E_k)\ge \alpha$, and ${\int }_{E_k}^{}f\mathrm{d}\mu \le 1/2^k$.

Let $F=\mathrm{limsup}{E}_k$, it is not hard to observe $$\mu (F)=\lim \mu \left(\bigcup _{k=n}^{\infty }{E}_k\right)\ge \alpha$$$${\int }_F^{}f\mathrm{d}\mu \le {\int }_{{\bigcup }_{k=n}^{\infty }{E}_k}^{}f\mathrm{d}\mu \le \sum _{k=n}^{\infty }{\int }_{E_k}^{}f\mathrm{d}\mu \le \sum _{k=n}^{\infty }\frac{1}{2^k}=\frac{1}{2^{n-1}}$$

Let $n\to \infty$, we obtain ${\int }_F^{}f\mathrm{d}\mu =0$, then $f=0$ holds a.e. on $F$, contradicting to being positive-valued.

$\square$

Problem 30. Let $f$ be a non-negative measurable function, and there is a constant $c$, such that ${\int }_{}^{}{f}^n\mathrm{d}\mu =c$ holds for any $n\in {\mathbb{N}}^{+}$. Then there exists a measurable set $E$, $f(x)={\chi }_E(x)$ a.e.

Proof. First we compute $F(x)=\mathrm{liminf}f(x{)}^n$. For $f(x)<1,=1,>1$, $F(x)=0,1,\infty$ respectively.

By Fatou’s Lemma, $${\int }_{}^{}F\le \mathrm{liminf}{\int }_{}^{}{f}^n=\mathrm{liminf}{\int }_{}^{}f=c$$

We obtain for a.e. $F<\infty \Rightarrow f\le 1\Rightarrow f^2\le f$. But on the other hand we have ${\int }_{}^{}{f}^2={\int }_{}^{}f$, then for a.e. $f^2=f\Rightarrow f=0\text{ or }f=1$.

Thus let $E=\{f(x)=1\}$, it is exactly what we want.

$\square$

Problem 31. For any $x\in [0,1)$, define by its binary decomposition a sequence of functions $x=0.a_1(x)a_2(x)\cdots$. For those without an unique decomposition, we choose the one with the ending of $000\cdots$.

Let $f:[0,1)\to \mathbb{R}$ be Lebesgue measurable, then$$\underset{n\to \infty }{\lim }{\int }_0^{1}f(x)a_n(x)\mathrm{d}x=\frac{1}{2}{\int }_0^{1}f(x)\mathrm{d}x.$$

Proof.

1.	If $f$ is continuous on $[0,1]$, then it is also uniformly continuous. For any $\epsilon >0$, there exists $N>0$, such that $|f(x)-f(y)|\le \epsilon$ for any $|x-y|<1/2^{N+1}$. Thus for any $n>N$, we have $$|{\int }_{\frac{k}{2^n}}^{\frac{k+\frac{1}{2}}{2^n}}f(x)\mathrm{d}x-{\int }_{\frac{k+\frac{1}{2}}{2^n}}^{\frac{k+1}{2^n}}f(x)\mathrm{d}x|\le {\int }_{\frac{k}{2^n}}^{\frac{k+\frac{1}{2}}{2^n}}\epsilon \mathrm{d}x=\frac{\epsilon }{2^{n+1}}$$$$|{\int }_0^{1}f(x)a_n(x)\mathrm{d}x-{\int }_0^{1}f(x)(1-a_n(x))\mathrm{d}x|\le 2^n\frac{\epsilon }{2^{n+1}}=\frac{\epsilon }{2}$$ Which implies the limit. Now we approximate any measurable function by a continuous one.
2.	If $f$ is a indicator function: $f={\chi }_E$, $E\subset [0,1)$, by Lusin’s theorem, for each $\delta >0$ there exists a continuous $h:\mathbb{R}\to [0,1]$ such that $m(\{f\ne h\})<\delta$, thus ${\int }_0^1|f-h|\mathrm{d}x\le \delta$. We already know ${\int }_0^{1}h{a}_n\mathrm{d}x\to \frac{1}{2}{\int }_0^{1}h\mathrm{d}x$, and we have $$|{\int }_0^{1}f{a}_n\mathrm{d}x-{\int }_0^{1}h{a}_n\mathrm{d}x|\le {\int }_0^1|f-h|\mathrm{d}x\le \delta$$ Since $\delta$ was arbitrary, we are done.
3.	If $f$ is a simple function, it is trivial by the above.
4.	Finally for any measurable $f$, WLOG we assume $f\ge 0$. Let an increasing sequence of simple functions approximate it and...

$\square$

Problem 32. Let $\mu (X)<\infty ,f:X\to \left[0,\infty \right)$ be bounded. Then$${\int }_{}^{}f\mathrm{d}\mu =\inf \left\{{\int }_{}^{}\phi \mathrm{d}\mu :\phi \ge f,\phi \text{ simple}\right\}$$Moreover, prove by examples that the condition $\mu (X)<\infty$ is necessary.

Proof. Let $|f|<M$. We obtain by $\mu (X)<\infty$ the following: $$M\mu \left(X\right)-{\int }_{}^{}f\mathrm{d}\mu ={\int }_{}^{}M-f\mathrm{d}\mu =\text{sup}\left\{{\int }_{}^{}\psi \mathrm{d}\mu ,\psi \text{ simple},0\le \psi \le M-f\right\}$$$$=\text{sup}\left\{{\int }_{}^{}M-\phi \mathrm{d}\mu ,\phi \text{ simple},f\le \phi \le M\right\}=M\mu \left(X\right)-\text{inf}\left\{{\int }_{}^{}\phi \mathrm{d}\mu ,\phi \text{ simple},\phi \ge f\right\}$$

Which implies what we want.

For an example, let $\mu$ be a counting measure on $(\mathbb{N},2^{\mathbb{N}},\mu )$ and $f:\mathbb{N}\to \left[0,\infty \right)$ be $f(n)=1/n^2$, then it is bounded, and we have $${\int }_{}^{}f\mathrm{d}\mu =\text{sup}\left\{{\int }_{}^{}\phi \mathrm{d}\mu ,\phi \text{ simple}\le f\right\}=\sum _1^{\infty }\frac{1}{n^2}=\frac{{\pi }^2}{6}.$$

But for any simple $\phi \ge f$, assume $\phi ={\sum }_{k=1}^n{a}_k{\chi }_{E_k}$, then each $a_k>0$, and ${⨆}_{k=1}^n{E}_k=\mathbb{N}$. Thus there exists a $E_k$, which is infinite, so $\int \phi \mathrm{d}\mu =\infty$: $$\inf \left\{{\int }_{}^{}\phi \mathrm{d}\mu :\phi \ge f,\phi \text{ simple}\right\}=\infty .$$

$\square$

Problem 33. Let $f$ be integrable. For any $\epsilon >0$, there exists $E$ measurable, $\mu (E)<\infty$, such that$$|{\int }_{}^{}f\mathrm{d}\mu -{\int }_E^{}f\mathrm{d}\mu |<\epsilon$$

Proof. WLOG we assume $f\ge 0$. Let $F_n=\{f>\frac{1}{n}\}$ and $F=\{f>0\}$, then $F_n$ increases to $F$, ${\chi }_{F_n}$ increases to ${\chi }_F$. By MCT we have $$\int f=\int {\chi }_{F}f=\lim \int {\chi }_{F_n}f=\lim {\int }_{F_n}f.$$

Now it suffices to check $\mu (F_n)<\infty$. It is guaranteed by $$\mu \left(F_n\right)={\int }_{F_n}^{}1<{\int }_{F_n}^{}nf<n{\int }_{}^{}f<\infty .$$

$\square$

Problem 34. Let ${\left\{f_n\right\}}_{n=1}^{\infty }$ be a sequence of integrable functions that converge to $0$ a.e. Constant $M>0$ such that$${\int }_{}^{}\max \left\{|f_1(x)|,|f_2(x)|,\cdots ,|f_n(x)|\right\}\mathrm{d}\mu <M$$

for any positive integer $n$. Then$$\underset{n\to \infty }{\lim }{\int }_{}^{}{f}_n\mathrm{d}\mu =0.$$

Proof. Let $g_n(x)=\max \left\{|f_1(x)|,\cdots ,|f_n(x)|\right\}$ be a monotone increasing sequence of positive measurable functions and let $g=\lim g_n$, we obtain by MCT: $$\underset{n\to \infty }{\lim }{\int }_{}^{}{g}_n\mathrm{d}\mu ={\int }_{}^{}g\mathrm{d}\mu$$

Since ${\int }_{}^{}{g}_n\mathrm{d}\mu <M$, ${\int }_{}^{}g\mathrm{d}\mu \le M$. Thus $g$ is integrable.

On the other hand, $|f_n|\le g_n\le g$, then by DCT and $f_n\to 0$ a.e.: $$\underset{n\to \infty }{\lim }{\int }_{}^{}{f}_n\mathrm{d}\mu ={\int }_{}^{}0\mathrm{d}\mu =0.$$

$\square$

Problem 35. Compute$$\underset{n\to \infty }{\lim }{\int }_0^{\infty }{\left(1+\frac{x}{n}\right)}^n{\text{e}}^{-2x}\mathrm{d}x.$$

Proof. 1.

$\square$

Problem 36. Let $f:X\to \mathbb{R}$ be measurable, $\mu (X)<\infty$. Then$$\underset{n\to \infty }{\lim }{\int }_{}^{}|\cos (\pi f(x)){|}^n\mathrm{d}\mu =\mu (f^{-1}(\mathbb{Z})).$$

Proof. Let $g_n(x)=|\cos (\pi f(x)){|}^n,\tilde{X}=f^{-1}(\mathbb{Z})$, then $\tilde{X}$ is measurable since $f$ is measurable.

For $x\in \tilde{X}$, $g_n(x)=1$. Otherwise, $g_n(x)\to 0$. Thus $g_n(x)\searrow {\chi }_{\tilde{X}}$.

By MCT and $\mu (X)<\infty$, we obtain $$\underset{n\to \infty }{\lim }{\int }_{}^{}{g}_n\mathrm{d}\mu ={\int }_{}^{}{\chi }_{\tilde{X}}\mathrm{d}\mu =\mu (\tilde{X}).$$

$\square$

Problem 37. Compute$$\underset{n\to \infty }{\lim }{\int }_0^{n^2}\frac{x^{\frac{1}{n}}}{1+x^{2n}}\mathrm{d}x$$

Proof. 1.

$\square$

Problem 38. Let $f_n$ be measurable. For any $x\in X$, $|f_n(x)|<1/n^2$, for any positive integer $n$. Let $g$ be integrable. Then for any $E$ measurable, the following holds:$${\int }_E\sum f_{n}g\mathrm{d}\mu =\sum {\int }_E{f}_{n}g\mathrm{d}\mu .$$

Proof. WLOG we assume $f_n,g\ge 0$, then prove it by lemmas:

Lemma. Let $f_n\ge 0$ be measurable, $f_n\to f$ pointwise, $\int f=\lim \int f_n<\infty$, then for any $E$ measurable, we have $${\int }_{E}f=\lim {\int }_E{f}_n.$$

Lemma. Let $f_n\ge 0$ be measurable, $f=\sum f_n$, then $\int f=\sum \int f_n$.

Back to our proof, we check two things at first: ${\sum }_{k=1}^n{f}_{k}g\to {\sum }_{k=1}^{\infty }{f}_{k}g$, which is obvious; second: $${\int }_{}^{}\sum _{k=1}^n{f}_{k}g\to {\int }_{}^{}\sum _{k=1}^{\infty }{f}_{k}g<\infty$$

This convergence is by the second lemma, and finity by $f_n<1/n^2$.

By the first lemma, for any $E$ measurable, we have $${\int }_E^{}\sum _{k=1}^{\infty }{f}_{k}g=\lim {\int }_E^{}\sum _{k=1}^n{f}_{k}g=\lim \sum _{k=1}^n{\int }_E^{}{f}_{k}g,$$

where the latter equality is by the second lemma.

$\square$

Problem 39. Let $f:[0,1)\to \mathbb{R}$ be Lebesgue measurable. For each $n\in {\mathbb{N}}_{+}$, define$$f_n(x)=n{\int }_{\frac{k}{n}}^{\frac{k+1}{n}}f(t)\mathrm{d}t,x\in \left[\frac{k}{n},\frac{k+1}{n}\right),k=0,1,\cdots ,n-1.$$

Then$$\underset{n\to \infty }{\lim }{\int }_0^1|f_n(x)-f(x)|\mathrm{d}x=0.$$

Proof.

1.	If $f$ is continuous on $[0,1]$, then it is also uniformly continuous. For any $\epsilon >0$, there exists $N>0$, such that $|f(x)-f(y)|\le \epsilon$ for any $|x-y|<1/N$. Thus for any $n>N$, we have $${\int }_{\frac{k}{n}}^{\frac{k+1}{n}}|f_n(x)-f(x)|\mathrm{d}x\le n{\int }_{\frac{k}{n}}^{\frac{k+1}{n}}{\int }_{\frac{k}{n}}^{\frac{k+1}{n}}|f(t)-f(x)|\mathrm{d}t\mathrm{d}x\le n{\int }_{\frac{k}{n}}^{\frac{k+1}{n}}\epsilon \frac{1}{n}\mathrm{d}x=\frac{\epsilon }{n}.$$ Add them, by the arbitrary of $\epsilon$, we obtain the limit. Now we approximate any measurable function by a continuous one.
2.	If $f$ is a indicator function: $f={\chi }_E$, $E\subset \left[0,1\right)$, by Lusin’s theorem, for each $\delta >0$ there exists a continuous $h:\mathbb{R}\to \mathbb{R}$ such that $m(\{f\ne h\})<\delta$, thus ${\int }_0^1|f-h|\mathrm{d}x\le \delta$. We already know ${\int }_0^1|h_n(x)-h(x)|\mathrm{d}x\to 0$, and we have $${\int }_0^1|f_n(x)-h_n(x)|\mathrm{d}x=\sum _{k=0}^{n-1}{\int }_{\frac{k}{n}}^{\frac{k+1}{n}}|f_n(x)-h_n(x)|\mathrm{d}x$$$$\le \sum _{k=0}^{n-1}\frac{1}{n}n{\int }_{\frac{k}{n}}^{\frac{k+1}{n}}|f(t)-h(t)|\mathrm{d}t={\int }_0^1|f(t)-h(t)|\mathrm{d}t.$$ Then we estimate $${\int }_0^1|f_n(x)-f(x)|\mathrm{d}x\le {\int }_0^1|f_n(x)-h_n(x)|+|h_n(x)-h(x)|+|h(x)-f(x)|\mathrm{d}x$$$$\le 2\delta +{\int }_0^1|h_n(x)-h(x)|\mathrm{d}x.$$ Since $\delta$ is arbitrary, we are done.
3.	If $f$ is a simple function, it is trivial by the above.
4.	Finally for any measurable $f$, WLOG we assume $f\ge 0$. Let an increasing sequence of simple functions approximate it and...

$\square$

Problem 40. Let $f_n:[0,1]\to \mathbb{R}$, $f_n\ge 0$ Lebesgue measurable, and converges to $f$ a.e.

Let $F(x)={\int }_0^{x}f(y)\mathrm{d}y,F_n(x)={\int }_0^x{f}_n(y)\mathrm{d}y$. Then:$${\int }_0^{1}f+F\mathrm{d}x\le \mathrm{liminf}{\int }_0^1{f}_n+F_n\mathrm{d}x$$

Proof. By Fatou’s Lemma, $$\mathrm{liminf}{\int }_0^1{f}_n(x)+F_n(x)\mathrm{d}x\ge {\int }_0^1\mathrm{liminf}(f_n(x)+F_n(x))\mathrm{d}x$$$$={\int }_0^{1}f(x)+\mathrm{liminf}{\int }_0^{x}f(y)\mathrm{d}y\mathrm{d}x$$$$\ge {\int }_0^{1}f(x)+{\int }_0^x\mathrm{liminf}f(y)\mathrm{d}y\mathrm{d}x={\int }_0^{1}f(x)+F(x)\mathrm{d}x$$

$\square$

Problem 41. Let $f,f_n,g\in {\mathcal{L}}^2(\mathbb{R}),n=1,2,\cdots$, $f_n$ converges to $f$ a.e. and$${\int }_{\mathbb{R}}^{}|f_n(x){|}^2\mathrm{d}x\le 1.$$

Then$$\underset{n\to \infty }{\lim }{\int }_{\mathbb{R}}^{}|f_n(x)-f(x)||g(x)|\mathrm{d}x=0.$$

Proof. For any $\epsilon >0$, there exists a $\delta >0$ such that for any $m(A)<\delta$ we have $${\int }_A^{}{g}^2<\epsilon$$

At the same time, denote $E_n=\left\{|x|>n\right\}$, then there exists a $N$: $${\int }_{E_N}^{}{g}^2<\epsilon$$

By Erogoff’s theorem, there exists a measurable $F$ such that $m(\mathbb{R}\backslash (E_N\cup F))<\delta$, and $f_n(x){|}_F\rightrightarrows f(x){|}_F$. Then there exits a $N_1$ such that for any $k\ge N_1,x\in F$, we have $|f_k-f{|}^2<\epsilon /m(F)$. Thus $${\int }_F^{}|f_k-f{|}^2<\frac{\epsilon }{m(F)}m(F)=\epsilon .$$

Again we also have $${\int }_{\mathbb{R}\backslash (E_N\cup F)}^{}{g}^2<\epsilon \Rightarrow {\int }_{\mathbb{R}\backslash F}^{}{g}^2<2\epsilon .$$

Thus for any $n\ge N+N_1$, we have $${\int }_{\mathbb{R}}^{}|f_n-f|g\le {\int }_{\mathbb{R}-F}^{}|f_n-f|g+{\int }_F^{}|f_n-f|g$$$$\le \sqrt{{\int }_{\mathbb{R}}^{}|f_n-f{|}^2{\int }_{\mathbb{R}-F}^{}{g}^2}+\sqrt{{\int }_F^{}|f_n-f{|}^2{\int }_{\mathbb{R}}^{}{g}^2}$$$$\le 2\sqrt{2\epsilon }+\Vert g{\Vert }_2\sqrt{\epsilon }.$$

Then we are done.

$\square$

Problem 42. Let $f_n$ be a sequence of non-negative Borel measurable functions on $[0,1]$ that converges to $f$ a.e. Moreover, we have$${\int }_0^1{f}_n(x)\log (f_n(x)+2)\mathrm{d}x\le 1.$$

Then $f$ is Lebesgue measurable, and $\lim {\int }_0^1|f_n(x)-f(x)|\mathrm{d}x=0$.

Proof. We prove it by Vitali Convergence Theorem:

Lemma. Let $\mathcal{F}$ be a uniformly integrable and tight family of integrable functions, and a subfamily $f_n$ converges to $f$ pointwise, then $f$ is integrable, and $\lim {\int }_0^1|f_n(x)-f(x)|\mathrm{d}x=0$.

Since $m([0,1])=1<\infty$, the tightness is trivial. We only need to show that $f_n$ is uniformly integrable.

For any $\epsilon >0$, we choose a $M>0$ such that $\epsilon \log \left(M+2\right)>1$, then denote $F_n=\{f_n\ge M\}$. Thus for $\delta =\epsilon /M$, we have for any $m(E)<\delta$: $${\int }_E^{}{f}_n={\int }_{E\cap F_n}^{}{f}_n+{\int }_{E\backslash F_n}^{}{f}_n\le \frac{1}{\log \left(M+2\right)}+Mm(E)<2\epsilon$$

Thus we are done.

$\square$

Problem 43. Let $f:\mathbb{R}\to \mathbb{R}$ be Lebesgue integrable, such that$$\underset{\epsilon \searrow 0}{\mathrm{liminf}}{\int }_{{\mathbb{R}}^2}^{}\frac{|f(x)f(y)|}{|x-y{|}^2+\epsilon }\mathrm{d}x\mathrm{d}y<\infty .$$Prove that $f=0$ a.e.

Proof. Without loss of generality, assume that $f(x)\ge 0.$
By Fatou’s lemma we have$${\int }_{{\mathbb{R}}^2}\frac{f(x)f(y)}{|x-y{|}^2}\mathrm{d}x\mathrm{d}y\le M<\infty ,$$where $M$ denotes the limit given above. Suppose the otherwise, there exists an $n\in {\mathbb{N}}^{\ast }$ such that $\mu (E:=\{x\mid f(x)\ge \frac{1}{n}\})>0$.
Then we have$${\int }_{E\times E}\frac{1}{|x-y{|}^2}\mathrm{d}x\mathrm{d}y\le n^{2}M.$$It indicates that $\frac{1}{|x-y{|}^2}{\chi }_E(x){\chi }_E(y)$is integrable, and that Fubini’s theorem works.
Since $\mu (E)>0$, we can take a Lebesgue density point $x\in E$ and an $r$ such that $\mu (E\cap (x-\delta ,x+\delta ))>\delta ,\forall \delta \in (0,r).$
Thus ${\int }_E\frac{1}{|x-y{|}^2}\mathrm{d}y\ge \frac{1}{\delta }\to \infty$ as $\delta \to 0,$a contradiction to Fubini’s theorem since Lebesgue density points also have positive measure.

$\square$

Problem 44. Let $f:\mathbb{R}\to \mathbb{R}$ be Lebesgue integrable. Then for $x\in \mathbb{R}$ a.e. the series converges:$$\sum _{n=1}^{\infty }\frac{1}{\sqrt{n}}f(x-\sqrt{n}).$$

Proof. We prove that for any $a<b$, the series is integrable on $[a,b]$. Observe that $${\int }_a^b\sum _{n=1}^{\infty }\frac{1}{\sqrt{n}}f(x-\sqrt{n})\mathrm{d}x=\sum _{n=1}^{\infty }\frac{1}{\sqrt{n}}{\int }_a^{b}f(x-\sqrt{n})\mathrm{d}x$$$$=\sum _{n=1}^{\infty }\frac{1}{\sqrt{n}}{\int }_{\mathbb{R}}^{}f(x){\chi }_{[a-\sqrt{n},b+\sqrt{n}]}\mathrm{d}x={\int }_{\mathbb{R}}^{}f(x)S(x)\mathrm{d}x$$

Where we denote $S(x)={\sum }_{n=1}^{\infty }\frac{1}{\sqrt{n}}{\chi }_{[a-\sqrt{n},b-\sqrt{n}]}$. We claim that $S(x)$ is bounded by $C=C(a,b)$ independent of $x$:

For any $x\in \mathbb{R}$, the sum is taken on those $n$ with $a-\sqrt{n}\le x\le b-\sqrt{n}$, namely:$$S\left(x\right)=\sum _{a-x\le \sqrt{n}\le b-x}\frac{1}{\sqrt{n}}$$

If $a-x\le 1$, we have $$S(x)=\sum _{1\le n\le (b-x{)}^2}\frac{1}{\sqrt{n}}\le {\int }_0^{{\left(b-x\right)}^2}\frac{\mathrm{d}t}{\sqrt{t}}=2(b-x)\le 2(b-a+1).$$

If $a-x>1$, we have $$S(x)=\sum _{(a-x{)}^2\le n\le (b-x{)}^2}\frac{1}{\sqrt{n}}\le {\int }_{(a-x{)}^2-1}^{(b-x{)}^2}\frac{\mathrm{d}t}{\sqrt{t}}$$$$=2((b-x)-\sqrt{(a-x{)}^2-1})\le 2(b-a+1).$$

The last inequality by the monotonicity.

$\square$

Problem 45. Define $f(0)=0$, and $f(x)=x^2\sin (1/x)$ for $x\ne 0$, then $f$ is a function of bounded variation on $[-1,1]$.

Proof. Trivial if you make use of Lagrange’s mean value theorem.

$\square$

Problem 46. Construct a monotone increasing function $f:\mathbb{R}\to \mathbb{R}$, such that the set of continuous points of $f$ is the complement of $\mathbb{Q}$.

Proof. It suffices to consider $[0,1]\to \mathbb{R}$ and extend by $f(x+1)=f(x)+f(1)$. Let $R(x)$ denote the Riemann function, and define $f(x):[0,1]\to \mathbb{R}$: $$R(x)=\{\begin{array}{ll}\frac{1}{q},& x=\frac{p}{q}\in [0,1],p,q\in \mathbb{N},\left(p,q\right)=1\\ 0,& x\in [0,1]\backslash \mathbb{Q}\end{array},f(x)=\sum _{0\le r\le x}{R}^3(r)$$

Since $f(x)\le f(1)=\sum \varphi (n)/n^3<\sum 1/n^2$, $f(x)$ is well defined. The monotone property is obvious.

For any $x_0=p/q$, the discontinuity follows from the observation: $$\forall \delta >0,\exists x_0-\delta <x<x_0,f(x_0)-f(x)\ge \frac{1}{q^3}$$

Now for any $x_0\in [0,1]\backslash \mathbb{Q}$, we will show that $f(x)$ is continuous at $x_0$. For each $\epsilon >0$, denote $K=[1/\epsilon ]+1$, then those $x=p/q$ with $q\le K$ is finite. Let $\delta$ be half of the minimum distance between the points and $x_0$. Then for any $x\in B(x_0,\delta )$, we have $$|f(x)-f(x_0)|\le \sum _{r\in I}{R}^3(r)\le \sum _{n\ge K+1}\frac{\varphi (n)}{n^3}\le \frac{1}{K}\le \epsilon ,$$

where $I$ is a close interval with end points $x,x_0$.

$\square$

Problem 47. Let $f:\mathbb{R}\to \mathbb{R}$ be monotone increasing, $E\subset \mathbb{R}$ is Lebesgue measurable, For any $\epsilon >0$, there exists a sequence of open intervals $(a_i,b_i)$, such that$$E\subset \bigcup (a_i,b_i),\sum |f(b_i)-f(a_i)|<\epsilon$$

Then $f^{\prime }(x)=0$ a.e. $x\in E$.

Proof. The monotone property implies that $f$ is differentiable a.e. and $f^{\prime }$ is integrable. For any $\epsilon >0$, $${\int }_E^{}{f}^{\prime }(x)\mathrm{d}x\le \sum {\int }_{a_i}^{b_i}{f}^{\prime }(x)\mathrm{d}x\le \sum f(b_i)-f(a_i)<\epsilon$$

Thus ${\int }_E{f}^{\prime }(x)\mathrm{d}x=0$. We have $f^{\prime }(x)=0$ a.e. $x\in E$.

$\square$

Problem 48. Let $f:[0,1]\to [0,1]$ be a continuous bijection, but not absolutely continuous. Then there exists $E\subset [0,1]$, such that $m(E)=0,m(f(E))>0$.

Proof. First we prove two lemmas.

Lemma. Let $f$ be differentiable on $E\subset \mathbb{R}$, and $|f^{\prime }|\le M$ on $E$, then $m^{\ast }(f(E))\le M{m}^{\ast }(E)$.

Proof. It is a more general case of 22, thus can be proved in a similar way.

$\square$

Lemma. Let $f$ be measurable on $[a,b]$, $E\subset [a,b]$ is measurable, and $f$ is differentiable everywhere on $E$, then $m^{\ast }(f(E))\le {\int }_E|f^{\prime }|$.

Proof. For each $\epsilon >0$, denote $E_k=\{x\in E:(k-1)\epsilon \le |f^{\prime }|<k\epsilon \}$, by the first lemma we have $$m^{\ast }(f(E_k))\le k\epsilon m(E_k)=(k-1)\epsilon m(E_k)+\epsilon m(E_k)\le {\int }_{E_k}^{}|f^{\prime }|+\epsilon m(E_k)$$

Add them together, we obtain $m^{\ast }(f(E))\le {\int }_E^{}|f^{\prime }|+\epsilon m(E)$, then we are done since $\epsilon$ was arbitrary.

$\square$

Back to our proof, we will first show $f\in \mathrm{BV}$ from continuous and bijective property. Then by assuming the conclusion is false, we will prove $f\in \mathrm{AC}$, a contradiction.

We have two approaches to prove $f\in \mathrm{BV}$. One is that $f$ is monotone as a continuous bijection. The other one can deal with more general cases.

$\forall 0=x_0<x_1<\cdots <x_n=1$, we denote $J_k$ as the close interval with endpoints $f(x_{k-1}),f(x_k)$. Then by continuity, we have $J_k\subset f([x_{k-1},x_k])=I_k$. Thus $$\sum _{k=1}^n|f(x_k)-f(x_{k-1})|={\int }_0^1\sum _{k=1}^n{\chi }_{J_k}\mathrm{d}y\le {\int }_0^1\sum _{k=1}^n{\chi }_{I_k}\mathrm{d}y\le 1$$

Where we used the bijective property. In fact, even not being bijective our conclusion $f\in \mathrm{BV}$ still holds as long as the number of every point’s preimage is bounded by a same number.

Now assume $f$ preserves zero measure. By $f\in \mathrm{BV}$ we know $f^{\prime }$ exists a.e. and it is integrable. Hence there exists $E\subset [0,1]$ such that $m(E)=0$ and $f^{\prime }$ exists everywhere in $[0,1]\backslash E$. By the assumption, $m(f(E))=0$. We observe that $\forall (x,y)\subset [0,1],|f(y)-f(x)|\le m(f([x,y]))\le m^{\ast }(f([x,y]\backslash E))+m(f(E))\le {\int }_x^y|f^{\prime }(t)|\mathrm{d}t$, the last inequality by the second lemma.

Now we make use of the integrablility of $f^{\prime }$ and the absolute continuity of Lebesgue integral, for any $\epsilon >0$, there exists $\delta >0$, ${\int }_A|f^{\prime }|<\epsilon$ as long as $m(A)<\delta$.

Finally we check the definition of absolute continuity of $f$. For each $\epsilon >0$, for any $0\le a_1<b_1<\cdots <a_n<b_n\le 1$ with $\sum (b_k-a_k)<\delta$, we have $$\sum _{k=1}^n|f(b_k)-f(a_k)|\le \sum _{k=1}^n{\int }_{a_k}^{b_k}|f^{\prime }|={\int }_{{\bigcup }_{k=1}^n(a_k,b_k)}^{}|f^{\prime }|<\epsilon .$$

$\square$

Problem 49. Let $f:[0,1]\to \mathbb{R}$, prove the following are equivalent:

1.	$f\in \mathrm{AC}$, and there exists $h$ bounded and measurable, such that $f^{\prime }=h$ a.e.
2.	There exists $L>0$, such that for any $0\le x,y\le 1$, $|f(x)-f(y)|\le L|x-y|$.

Proof. $(2)\Rightarrow (1)$: It follows easily from checking definition that $f\in \mathrm{AC}$. By Lebesgue differentiation theorem, $\exists h$ measurable, $h:[0,1]\to \mathbb{R}$, such that $h=f^{\prime }$ a.e. For those differentiable points, $|f^{\prime }|$ is bounded by Lipschitz condition. For the rest, just assign $h=0$, then $h$ is bounded.

$(1)\Rightarrow (2)$: Since Lebesgue differentiation theorem, $f(x)=f(0)+{\int }_0^{x}h(t)\mathrm{d}t$. Thus $$|f(y)-f(x)|=|{\int }_x^{y}h(t)\mathrm{d}t|\le {\int }_x^y|h(t)|\mathrm{d}t\le L|y-x|.$$

$\square$

Problem 50. Let $f:{\mathbb{R}}^2\to \mathbb{R}$, $f\in C^1$ and for any $(x_0,y_0)\in {\mathbb{R}}^2$, we have$$\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}{|}_{\left(x_0,y_0\right)}\ne 0$$

Then for $E=\{(x,y):f(x,y)=0\}$ is a set of measure zero.

Proof. On any straight line $y=x+t$, the directional derivative never vanishes, and also it is continuous, then $f$ is absolutely monotone, with no more than one zero. Thus $m(E)\le m(\{0\}\times \mathbb{R})=0$.

$\square$

Problem 51. Let $E\subset \mathbb{R}$ be measurable and $m(E)>0$, and $f:\mathbb{R}\to \mathbb{R}$ be non-negative Lebesgue measurable function. If $F(x)={\int }_{E}f(x-t)\mathrm{d}t$ is Lebesgue integrable on $\mathbb{R}$, then $f$ is also Lebesgue integrable.

Proof. Observe by Tonelli’s theorem, $f(x-t)$ is measurable and $f\ge 0$ that $$\infty >{\int }_{\mathbb{R}}^{}F(x)\mathrm{d}x={\int }_{\mathbb{R}}^{}{\int }_{\mathbb{R}}^{}f(x-t){\chi }_E(t)\mathrm{d}t\mathrm{d}x$$$$={\int }_{\mathbb{R}}^{}{\chi }_E(t){\int }_{\mathbb{R}}^{}f(x-t)\mathrm{d}x\mathrm{d}t=m(E){\int }_{\mathbb{R}}^{}f(x)\mathrm{d}x$$

Thus $f$ is also Lebesgue integrable.

$\square$

$L^p$ spaces

Problem 52. Let $f_n:[0,1]\to \mathbb{R}$ be a sequence of Lebesgue integrable functions, and converges in measure to $f:[0,1]\to \mathbb{R}$, which is Lebesgue measurable. Moreover we have $\Vert f_n{\Vert }_3\le 1$, then $f_n\to f$ in $L^2([0,1])$.

Proof. For any $\epsilon >0$, there exists $N$ such that for $E_n=\{|f_n-f|\ge \epsilon \}$, we have $m(E_n)<\epsilon$ for any $n>N$.

$${\int }_0^1|f_n-f{|}^2\le {\int }_{E_n}^{}|f_n-f{|}^2+{\int }_{E_n^c}^{}|f_n-f{|}^2$$$$\le {\epsilon }^{2}m(E_n^c)+{\left({\int }_0^1|f_n-f{|}^3\right)}^{\frac{2}{3}}(m(E_n){)}^{\frac{1}{3}}\le {\epsilon }^2+{\epsilon }^{\frac{1}{3}}\Vert f_n-f{\Vert }_3^2$$

Since $f_n\to f$ in measure, there is a subsequence $f_{n_k}\to f$ a.e. Thus we have $${\int }_0^1|f{|}^3={\int }_0^1\mathrm{liminf}|f_{n_k}{|}^3\le \mathrm{liminf}{\int }_0^1|f_{n_k}{|}^3\le 1,$$

which implies $\Vert f_n-f{\Vert }_3^2\le (\Vert f_n{\Vert }_3+\Vert f{\Vert }_3{)}^2\le 4$. Thus we are done.

$\square$

Problem 53. Let $1\le p<\infty$, $f_n,f\in L^p$, and $f_n\to f$ a.e. Then $\Vert f_n-f{\Vert }_p\to 0\iff \Vert f_n{\Vert }_p\to \Vert f{\Vert }_p$.

Proof. If $\Vert f_n-f{\Vert }_p\to 0$, then $|\Vert f_n{\Vert }_p-\Vert f{\Vert }_p|\le \Vert f_n-f{\Vert }_p\to 0$, thus $\Vert f_n{\Vert }_p\to \Vert f{\Vert }_p$.

If $\Vert f_n{\Vert }_p\to \Vert f{\Vert }_p$, we notice that $|x{|}^p$ is lower convex, thus by Jensen’s inequality we have $$|\frac{f_n-f}{2}{|}^p\le \frac{|f_n{|}^p+|f{|}^p}{2}.$$

The right-hand side converges to $|f{|}^p$ a.e. and its integral also converges to the integral of $|f{|}^p$, thus by DCT we have $${\int }_{}^{}|\frac{f_n-f}{2}{|}^p\to {\int }_{}^{}0=0,$$

which means $\Vert f_n-f{\Vert }_p\to 0$.

$\square$

Problem 54. Let $E,F\subset \mathbb{R}$ be with finite Lebesgue measure.

1.	The convolution ${\chi }_E\ast {\chi }_F$ is continuous.
2.	${\chi }_E\ast {\chi }_F\in L^p(\mathbb{R})$ for $1\le p<\infty$.

Proof. (1) A more general case: let $f$ be integrable, $g$ be bounded and measurable, then $f\ast g$ is continuous.

We denote $\Psi :\mathbb{R}\to L^1(\mathbb{R}),x\mapsto f(x-y)$ and $\Phi :L^1(\mathbb{R})\to \mathbb{R},h\mapsto {\int }_{\mathbb{R}}^{}hg$, then $f\ast g=\Phi \circ \Psi$.

Now it suffices to show both $\Phi$ and $\Psi$ are continuous.

For $\Phi$, it is trivial by that $g$ is bounded and measurable.

For $\Psi$, we observe that as $h\to 0$, we have $${\int }_{\mathbb{R}}^{}|f(x-y)-f(x+h-y)|\mathrm{d}y={\int }_{\mathbb{R}}^{}|f(y-h)-f(y)|\mathrm{d}y\to 0$$

Which is a common proposition used very common, for instance in the proof of the density of $C_c$ in $L^p$.

(2) Since ${\chi }_E\in L^1,{\chi }_F\in L^p$, we have ${\chi }_E\ast {\chi }_F\in L^p(\mathbb{R})$.

$\square$

Problem 55. Let $f:[0,1]\to \mathbb{R}$ be Lebesgue measurable, and for any $g\in L^2[0,1]$, we have $fg\in L^2([0,1])$, then $f\in L^{\infty }([0,1])$.

Proof. If $f\notin L^{\infty }([0,1])$, then for $E_n=\{n\le |f|<n+1\}$, we have $m(E_n)>0$ frequently. Take a subsequence while not changing the notion, we have $f\ge {\sum }_{n=1}^{\infty }n{\chi }_{E_n}$, and then $f^2\ge {\sum }_{n=1}^{\infty }{n}^2{\chi }_{E_n}$.

Let $g={\sum }_{n=1}^{\infty }\frac{{\chi }_{E_n}}{n\sqrt{m\left(E_n\right)}}$, then ${\int }_0^1{g}^2={\sum }_{n=1}^{\infty }\frac{1}{n^2}<\infty$, $g\in L^2([0,1])$. But ${\int }_0^1{f}^2{g}^2\ge {\sum }_{n=1}^{\infty }1=\infty$.

$\square$

Problem 56. Let $f:\mathbb{R}\to \mathbb{R}$ be Lebesgue integrable, then$$\underset{h\to 0}{\lim }{\int }_{\mathbb{R}}^{}|f(x+h)-f(x)|\mathrm{d}x=0.$$

Proof.

Since in our system the density of ${\mathrm{C}}_c^{\infty }(\mathbb{R})$ in $L^1(\mathbb{R})$ was proved through this proposition, we have to prove it in another way, otherwise we will fall into circular argument. So you are invited to consult related books for answers.

$\square$

Problem 57. Let $f_n,f:[0,1]\to \mathbb{R}$ be Lebesgue integrable, and $f_n\to f$ in $L^1([0,1])$.

1.	There exists a subsequence $f_{n_k}$ that converges to $f$ a.e.
2.	Show by a example that ‘a.e.’ can not be replaced with ‘uniformly’.

Proof. (1) It suffices to show $f_n\to f$ in measure. For any $\eta >0$, we denote $E=\{|f_n-f|\ge \eta \}$, then observe $${\int }_0^1|f_n-f|\mathrm{d}x\ge {\int }_E^{}|f_n-f|\mathrm{d}x\ge \eta m(E),$$

which forces $m(E)$ to be null.

(2) Let $f_n=n{\chi }_{[0,\frac{1}{n}]},f=0$, then for any subsequence $f_{n_k}$, we have $|f_{n_k}(0)-f(0)|=n_k$.

$\square$

Problem 58. Let $f:\mathbb{R}\to \mathbb{R}$ be a Lebesgue measurable bijection, and $f(x+y)=f(x)+f(y)$, then $f$ is continuous.

Proof. Using Lusin’s theorem, there exists a compact $K\subset [0,1]$ such that $m(K)>2/3$ and $f$ continuous on $K$. Let $\epsilon >0$. In fact, $f$ is uniformly continuous on $K$, so there exists $\delta >0$ such that $|x-y|<\delta$ implies $|f(x)-f(y)|<\epsilon$ ; without loss of generality, suppose $\delta <1/3$.

Let $0<h<\delta$. Notice that the intersection between $K$ and $K-h$ is nonempty; otherwise $$1+h=\mu ([-h,1])\ge \mu (K\cup K-h)=\mu (K)+\mu (K-h)=2\mu (K)>4/3$$

So $h>1/3$ whereas $h<\delta <1/3$ by assumption.

Let $x_0\in K\cap (K-h)$. We have $|f(x_0+h)-f(x_0)|<\epsilon$, hence $|f(h)|<\epsilon$ because $f$ is additive. You deduce that $f$ is continuous at $0$.

Finally, it is straightforward to conclude that $f$ is continuous on $\mathbb{R}$.

$\square$

Problem 59. $f:[0,1]\to \mathbb{R}$ be absolutely continuous, and $f^{\prime }\in L^2([0,1])$, then there exists a constant $C>0$ such that for any $0\le x\le y\le 1$, we have $|f(x)-f(y)|\le C\sqrt{y-x}$.

Proof. $$|f(x)-f(y)|=|{\int }_x^y{f}^{\prime }(t)\mathrm{d}t|\le {\int }_x^y|f^{\prime }(t)|\mathrm{d}t\le {\int }_0^1|f^{\prime }(t){|}^2\mathrm{d}t\sqrt{y-x}$$

Where we used the Hölder’s inequality.

$\square$

Problem 60. Let $H$ be the set of every absolutely continuous functions $f:[0,1]\to \mathbb{R}$ that satisfies$$\Vert f{\Vert }_H:=\Vert f{\Vert }_2+\Vert f^{\prime }{\Vert }_2<\infty .$$

Then $(H,\Vert \cdot {\Vert }_H)$ is Banach space.

Proof. It is obvious that $\Vert \cdot {\Vert }_H$ is a norm. Then it suffices to check any Cauchy sequence converges to some point in $H$.

If $f_n$ is Cauchy in $H$, then $f_n$ and $f_n^{\prime }$ are both Cauchy in $L^2([0,1])$. Thus we assume $f_n\to f,f_n^{\prime }\to g$ in $L^2([0,1])$. To show $f_n\to f$ in $H$, we must check $f^{\prime }=g$ a.e.

Since $f_n\in AC$, we have $$f_n(x)=f_n(0)+{\int }_0^x{f}_n^{\prime }(t)\mathrm{d}t$$

In the following we will deduce $f(x)=f(0)+{\int }_0^{x}g(t)\mathrm{d}t$ from it, which implies $f^{\prime }=g$ a.e.

Since $f_n\to f$ in $L^2([0,1])$, we have a subsequence $f_{n_k}\to f$ a.e.

By Holder’s inequality we have $$|{\int }_0^x{f}_n^{\prime }(t)\mathrm{d}t-{\int }_0^{x}g(t)\mathrm{d}t|\le \sqrt{x}{\int }_0^x(f_n^{\prime }(t)-g(t){)}^2\mathrm{d}t\to 0$$

Thus we are done by taking the limit in the subsequence.

$\square$

Problem 61. Let $f:\mathbb{R}\to \mathbb{R}$ be a Lebesgue integrable, and for any $\phi :\mathbb{R}\to \mathbb{R}$ such that $\phi \ge 0$ is bounded and continuous, we have $\int \phi f\ge 0$. Then $f\ge 0$ a.e.

Proof. If $f<0$ on $E$ with $m(E)>0$, then ${\int }_{E}f=-A<0$, where $A>0$.

By the absolutely continuity of the integral of $|f|$, $\exists \delta >0$, and ${\int }_T|f|<A/2$ as long as $m(T)<\delta$.

Using Lusin’s theorem, there exists $h:\mathbb{R}\to [0,1]$ continuous, such that for $F=\{h\ne {\chi }_E\}$, we have $m(F)<\delta$. Then $|\int (h-{\chi }_E)f|\le {\int }_F|f|<A/2$. Hence $\int hf=\int (h-{\chi }_E)f+\int {\chi }_E<A/2+(-A)=-A/2<0$.

$\square$

Problem 62. Let $f_n\in L^2([0,1])$, and constants $A,B$ such that for each $n$, the following holds:$$0<A<\Vert f_n{\Vert }_1\le \Vert f_n{\Vert }_2<B.$$

Let a sequence of non-negative reals $c_n$ such that for $F(x)={\sum }_1^{\infty }|c_n{f}_n(x)|$, we have $F(x)<\infty$ a.e. Denote $E_n=\{F(x)>n\}$, prove the following:

1.	${\lim }_{n\to \infty }m(E_n)=0$.
2.	${\int }_{E_n}^{}|f_k(x)|\mathrm{d}x\le B\sqrt{m(E_n)}$.
3.	${\sum }_1^{\infty }{c}_k<\infty$.

Proof. (1) ${\lim }_{n\to \infty }m(E_n)=m({\bigcap }_{n=1}^{\infty }{E}_n)=m(\{F(x)=\infty \})=0.$

(2) ${\int }_{E_n}^{}|f_k|\le {\left({\int }_{E_n}^{}|f_k{|}^2\right)}^{\frac{1}{2}}{\left({\int }_{E_n}^{}{1}^2\right)}^{\frac{1}{2}}\le \Vert f{\Vert }_2\sqrt{m(E_n)}\le B\sqrt{m(E_n)}.$

(3) From (2) we know that ${\int }_{E_n^c}|f_k(x)|\mathrm{d}x\ge A-B\sqrt{m(E_n)}$.

By (1), there exists a sufficiently large $n$ such that ${\int }_{E_n^c}|f_k(x)|\mathrm{d}x\ge A/2>0$.

Now sum the above inequalities over all $k\ge 1$ with weights $c_k$. This leads to the following:$${\int }_{E_n^c}|F(x)|\mathrm{d}x\ge \frac{A}{2}\sum _{k\ge 1}{c}_k.$$But the left hand side is finite according to the choice of $E_n$, so $\sum _{k\ge 1}{c}_k<\infty .$

$\square$

Problem 63. Let $\mathcal{F}$ be a nonempty set of some Lebesgue integrable functions on $\mathbb{R}$ that satisfies the following:

•	If $f,g\in \mathcal{F}$, set $h(x)=\max (f(x),g(x))$, then $h\in \mathcal{F}$.
•	$M=\sup \left\{{\int }_{\mathbb{R}}^{}f(x)\mathrm{d}x\right\}<\infty$.

Then there exists $\phi$ such that the following holds:

•	For any $f\in \mathcal{F}$, $f\le \phi$ holds a.e.
•	${\int }_{\mathbb{R}}\phi =M$.

Proof. Select a sequence $\{f_n\}$ in $\mathcal{F}$ such that $\int f_n\to M$. Then let $g_n=\max (f_1,\cdots ,f_n)$, and $\phi =\sup f_n$, we have for any fixed $x$, $g_n(x)$ increases to $\phi (x)$. Thus $\lim \int g_n=\int \phi$. But on one hand $g_n\in \mathcal{F}\Rightarrow \int g_n\le M\Rightarrow \lim \int g_n\le M$, on the other hand $g_n\ge f_n\Rightarrow \lim \int g_n\ge \lim \int f_n=M$. Thus $\int \phi =M$.

If there is some $f\in \mathcal{F}$ such that $f>\phi$ on a positive measure, consider $h_n=\max (g_n,f)\in \mathcal{F}$.

By the constructions above, $(h_n{)}_{n\ge 1}$ is an increasing sequence of functions and thus converges pointwise to a function $h$.

Since by the given condition that $\int h_n\le M$ and the fact that $h_n\ge h_1$ where $h_1$ is integrable, the monotone convergence theorem holds for $(h_n{)}_{n\ge 1}$ and thus $\int h\le M$.

However, $h>\phi$ on a positive measure and $h\ge \phi$ everywhere. These imply that $$M\ge \int h>\int \phi =M,$$which is absurd.

(Proof by contradiction is not necessary, but it is shorter.)

$\square$

Problem 64. Let $A=(a_{ij}{)}_{1\le i,j\le n}$ be an $n-$order real square symmetric matrix, and $f(x_1,x_2,\cdots ,x_n)=\exp \left(-{\sum }_{j,k=1}^n{x}_{jk}{a}_j{a}_k\right)$.

1.	Prove that $f\in L^1({\mathbb{R}}^n)$ iff $A$ is positive definite.
2.	Assuming $A$ is positive definite, compute ${\int }_{{\mathbb{R}}^n}f(x)dx$.

Proof.

1.

Since $A$ is real and symmetric, it can be orthogonal diagonalized. Assume $Q$ is orthogonal with $\det Q=1$, such that $QA{Q}^{-1}=D=\mathrm{diag}\{{\lambda }_1,{\lambda }_2,\cdots ,{\lambda }_n\}$, where each ${\lambda }_i$ is an eigenvalue of $A$. Then $f(x)=\exp (-x^{T}Ax)=\exp (-x^T{Q}^{T}DQx)=\exp (-y^{T}Dy)$, where we denote $y=Qx$. Thus $${\int }_{{\mathbb{R}}^n}^{}f(x)\mathrm{d}x={\int }_{{\mathbb{R}}^n}^{}\exp \left(-y^{T}Dy\right)\mathrm{d}x={\int }_{{\mathbb{R}}^n}^{}\exp \left(-y^{T}Dy\right)\mathrm{d}y={\int }_{{\mathbb{R}}^n}^{}\exp \left(-\sum _{k=1}^n{\lambda }_k{y}_k^2\right)\mathrm{d}{y}_1\cdots \mathrm{d}{y}_n.$$ Therefore, $f\in L^1({\mathbb{R}}^n)$ iff $\forall i=1,2,\cdots n,{\lambda }_i>0$ iff $A$ is positive definite.

2.

By the above discussions, using Tonelli’s theorem we have $${\int }_{{\mathbb{R}}^n}^{}f(x)\mathrm{d}x={\int }_{{\mathbb{R}}^n}^{}\exp \left(-\sum _{k=1}^n{\lambda }_k{y}_k^2\right)\mathrm{d}{y}_1\cdots \mathrm{d}{y}_n=\prod _{k=1}^n{\int }_{\mathbb{R}}^{}\exp \left(-{\lambda }_k{t}^2\right)\mathrm{d}t=\sqrt{\frac{{\pi }^n}{\det A}}.$$

$\square$