1. Introduction

Although we often deal with length, area and volume, have you ever wondered what they are?
Grothendieck once wrote in his autobiography: What I found most unsatisfactory in my mathematics textbooks was the absence of any serious attempt to tackle the meaning of the idea of the arc-length of a curve, or the area of a surface or the volume of a solid. I resolved therefore to make up for this defect once I found time to do so.
We often talk about the length or the area and volume are those geometric objects in the Euclidean space abstracted from the mind, you can feel very strongly that length, area and volume should actually be the same concept. Let’s name these three words measure. For example, we might say, "the length of this rope is $1$ meter" (note that you have imagined the rope as a straight line), and "the area of this square is $1$ square meter." But how do we talk about the measure of a strange geometric object if it cannot be considered in Euclidean space? Even, can we discuss the measure of a set? In order to strictly define the so-called measure, we should at least look at the characteristics of the measure in the Euclidean space, and we should extract the essence of the measure from the Euclidean space. In an interval $(a,b]$ on the real axis $\mathbb{R}$, we naturally want to define its measure as the length of the interval $b-a$. In addition, the most basic thing we should think of is that we give an interval whose measure is non-negative. Then, when the interval moves on $\mathbb{R}$, its measure should remain the same. If there are several such disjoint intervals, then when we measure the union of these intervals, it should be equals to measure these intervals separately and then add them up. So let’s make a tentative definition and see if it works.

Definition 1.0.1.
A measure $\lambda$ on $\mathbb{R}$ consist of some data

•	$\lambda :\Im (\mathbb{R})⟶{\mathbb{R}}_{+}\cup \left\{+\infty \right\}$
•	$\lambda (]a,b])=b-a$
•	$\lambda (A+x)=\lambda (A)$ for all $A\subset \mathbb{R}$ and $x\in \mathbb{R}$ where $A+x$$:=$$\left\{x+y|y\in A\right\}$
•	If$A={\bigcup }_{j=1}{A}_j$and$A_j\bigcap A_k=\varnothing$, then$\lambda (A)={\sum }_{k⩾1}$$\lambda$$\left(A_k\right)$

Remark 1.0.2. We denote all subsets of $\mathbb{R}$ by $\Im (\mathbb{R})$.

We will construct a weird subset of $\mathbb{R}$, which shows that if we really define the measure on $\mathbb{R}$ like this, then some sets will be excluded from $\Im (\mathbb{R})$.
We define an equivalent relation on $\mathbb{R}$, $x\sim y$ if and only if $y-x\in \mathbb{Q}$, then we construct $\Lambda =\mathbb{R}/\sim$. It is clear that $\Lambda$ is uncountable. $\Lambda$ consists of a family of the equivalent classes $[x]$. We select a representative element of $[x]$ for all equivalent classes, then it form a subset $\Omega$ of $\mathbb{R}$ and we may assume that $\Omega \subset (0,1)$.
Claim 1

$$\begin{array}{r}\{\begin{array}{l}\Omega +q=\Omega +p,p=qp,q\in \mathbb{Q}\\ \left(\Omega +q\right)\bigcap \left(\Omega +p\right)=\varnothing ,p\ne qp,q\in \mathbb{Q}\end{array}\end{array}$$

Assume that $(\Omega +p)\bigcap (\Omega +q)\ne \varnothing$, then we can find a $x$ such that $x=\alpha +p=$$\beta +q$ where $\alpha ,\beta \in \Omega$, it implies that $\alpha -\beta =q-p$, hence $\alpha \sim \beta$ (note that $q,p\in \mathbb{Q}$, hence $q-p\in \mathbb{Q}$). Since the construction of $\Omega$, it force $\alpha =\beta$, and hence $q=p$.

Claim 2

If $E\subset F\subset \mathbb{R}$, then $\lambda (E)⩽\lambda (F)$.

Since $F=E\bigcup (F\backslash E)$, we set $E=A_1$ and $F\backslash E=A_2$ and $A_j=\varnothing$ for $j⩾3$, then $$\begin{array}{rl}\lambda (F)& =\lambda (E\cup (F\backslash E)\cup \varnothing \cup \cdots )\\ & =\lambda (E)+\lambda (F\backslash E)+\lambda (\varnothing )+\cdots \\ \text{(since }\lambda (\varnothing )=0)& =\lambda (E)+\lambda (F\backslash E)\end{array}$$By the definition of $\lambda$, the inequality $\lambda (E)⩽\lambda (F)$ hold.

Claim 3

$$\lambda \left(\sum _{\begin{array}{c}q\in \mathbb{Q}\\ -1<q<1\end{array}}\left(\Omega +q\right)\right)⩽\lambda \left((-1,2)\right)=3$$

We know that $(\Omega +q)\subset (-1,2)$ since $\Omega \subset (0,1)$ and $-1<q<1$. This implies that $\left({\sum }_{\begin{array}{c}q\in \mathbb{Q}\\ -1<q<1\end{array}}\left(\Omega +q\right)\right)\subset$$(-1,2)$, then this claim is clear by using claim 2.

Remark 1.0.3. Since $(\Omega +q)\bigcap (\Omega +p)=\varnothing$ while $p\ne q$, then we can denote ${⨆}_{p\in \mathbb{Q}}$$\left(\Omega +p\right)$ by ${\sum }_{p\in \mathbb{Q}}\left(\Omega +p\right)$.

Claim 4

$$\lambda \left(\sum _{\begin{array}{c}q\in \mathbb{Q}\\ -1<q<1\end{array}}\left(\Omega +q\right)\right)=0$$

Since $\lambda \left({\sum }_{\begin{array}{c}q\in \mathbb{Q}\\ -1<q<1\end{array}}\left(\Omega +q\right)\right)={\sum }_{\begin{array}{c}q\in \mathbb{Q}\\ -1<q<1\end{array}}\lambda \left(\left(\Omega +q\right)\right)⩽3$, $\lambda (\Omega +q)=\lambda (\Omega )$$=0$ (note that if $\lambda (\Omega +q)$ is strictly positive, then the inequality can not be hold) and therefore $\lambda \left({\sum }_{\begin{array}{c}q\in \mathbb{Q}\\ -1<q<1\end{array}}\left(\Omega +q\right)\right)=0$.

Claim 5

$$(0,1)\subset \sum _{\begin{array}{c}q\in \mathbb{Q}\\ -1<q<1\end{array}}\left(\Omega +q\right)$$

Choose $x\in (0,1)$, then there exist $\alpha \in [x]\bigcap \Omega$ and $\alpha \in (0,1)$. We now know that $x-\alpha =q\in \mathbb{Q}$ which bounded by $(-1,1)$. Therefore, $x=\alpha +q\in (\Omega +q)\subset$${\sum }_{\begin{array}{c}q\in \mathbb{Q}\\ -1<q<1\end{array}}\left(\Omega +q\right)$.

In summary, there is a contradiction here, it conclude that ${\sum }_{\begin{array}{c}q\in \mathbb{Q}\\ -1<q<1\end{array}}\left(\Omega +q\right)$ is non-measurable. $\Im (\mathbb{R})$ is not all subsets of $\mathbb{R}$, there are some non-measurable sets with respect to these four properties of $\lambda$.
Therefore, the above definition is not appropriate, and it is impossible to generalize it to any set. We should reduce the conditions that appear in the definition.