(待修改! ) Given a set Ω and C⊂ℑ(Ω), μ is a function μ:C⟶R+∪{+∞}
Definition 3.0.1. 1. For E∈C, we say that μ is continuous from below at E if [ ∀(En∈C)n⩾1 and {En}n⩾1 an increasing sequence and {En}n⩾1⟶E (i.e. En⊂En+1 and ⋃n⩾1En=E) ] ⟹μ(En)⟶μ(E).
2. Similarly, for E∈C, we say that μ is continuous from above at E if [ ∀(En∈C)n⩾1 and {En}n⩾1 an decreasing sequence and {En}n⩾1⟶E (i.e. En+1⊂En and ⋂n⩾1En=E) and ∃n0 such that μ(En0)<+∞ for some n0 ] ⟹μ(En)⟶μ(E). We say that μ is continuous at E if it continuous both from below and above at E.
Remark 3.0.2. The symbol ⟶ means that convergence up to the category of sets or convergence up to the category of topological spaces.
Remark 3.0.3. The condition ∃n0 such that μ(En0)<+∞ can avoid that the following situation
Assume that R=Ω and C=ℑ(R) and En:={[n,+∞)∣n⩾1}, then there is a decreasing sequence {En}n⩾1 and {En}n⩾1⟶∅. We know that μ(∅)=0. If there is not exists n0 such that μ(En0)<+∞, then μ(En0)↛μ(∅)=0.
Lemma 3.0.4. Assume that A⊂ℑ(Ω) an algebra and μ:A⟶R+∪{+∞}is additive. Then
1. μ is σ-additive ⟹μ is continuous at E for ∀E∈A.
2. μ is continuous from below ⟹μ is σ-additive.
3. μ is continuous from above at ∅ and <a id="muisfinite">μ is finite</a> ⟹μ is σ-additive.
Proof.1⟩ In one hand, let E∈A and consider an increasing converges sequence {En∈A}n⩾1⟶E. Set F1F2Fn=E1=E2\E1⋮=En\En−1
Then Fj∩Fk=∅ for any j and k. It is clear that {Fn}n⩾1⟶E. So E=∑k⩾1Fk and Fk∈A since Fk⊂Ek∈A. Then μ(E)=k⩾1∑μ(Fk)=n→+∞limk=1∑nμ(Fk)=n→+∞limμ(k=1∑nFk)=n→+∞limμ(En)Consequently, μ is continuous from below at E∈A.
In the other hand, let E∈A and an decreasing sequence {En∈A}n⩾1 and μ(En0)<+∞ for some n0. Set G1Gk=En0\En0+1⋮=En0\En0+k⋮
Then there is an increasing converges sequence {Gk}k⩾1⟶En0\E, Gk∈A since Gk⊂E1∈A. Then μ(Gk)⟶μ(En0\E). But μ(En0)−μ(E)=μ(En0\E)=k→+∞limμ(En0\En0+k)=k→+∞lim(μ(En0)−μ(En0+k))It conclude that μ(E)=limk→+∞μ(En0+k).
2⟩ Assume that μ is continuous from below, let E=∑k⩾1Ek where Ek, E∈A.
Observation ∑k=1nEk⊂E since ∑k⩾1Ek=E.
Therefore, μ(∑k=1nEk)⩽μ(E), moreover, ∑k=1nμ(Ek)⩽μ(E) since μ is additive.
We claim that k⩾1∑μ(Ek)=μ(E)Let Fn=∑k=1nEk∈A, {Fn}n⩾1⟶E is an increasing converges sequence. Then μ(Fn)⟶μ(E). But μ(Fn)=μ(∑k=1nEk)=∑k=1nμ(Ek) which implies that ∑k=1nμ(Ek)⟶μ(E). It is easy to see that limn→+∞∑k=1nμ(Ek)=∑k⩾1μ(Ek) and therefore ∑k⩾1μ(Ek)=μ(E).
3⟩ Assume that μ is continuous from above at ∅ and μ(Ω)<+∞ (i.e. μ is finite). Let Ek, E∈A and E=∑k⩾1Ek. We construct Fn=k⩾n∑Ek=E\(j=1∑n−1Ek)∈AThen {Fn}n⩾1⟶∅ is a decreasing converges sequence. Since μ is finite, μ(F1)<+∞. By the the condition of continuity of μ at ∅, it conclude that μ(Fn)⟶∅ immediately. μ(E)=μ(k=1∑nEk⋃k>n∑Ek)=k=1∑nμ(Ek)+μ(k>n∑Ek)=k=1∑nμ(Ek)+μ(Fn+1)Then μ(E)=n→+∞limμ(E)=n→+∞lim(k=1∑nμ(Ek)+μ(Fn+1))=n→+∞limk=1∑nμ(Ek)=k⩾1∑μ(Ek)
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Example 3.0.5. Consider Ω=(0,1) and A={(a,b]∣0⩽a<b<1} is an algebra over (0,1), we define μ:A(a,b]⟶R+∪{+∞}↦{+∞,a=0b−a,a=0Then μ is additive but not σ-additive. μ is not finite and we claim that μ is continuous from above at ∅.
Let {En∈A}n⩾1⟶∅ be a decreasing converges sequence and we can write En=(an,1,bn,1]⋃⋯⋃(an,kn,bn,kn]We assume that μ(En0)<+∞, hence an0,j=0 for all j∈[1,kn0]. Since {En}n⩾1⟶∅, n→+∞limμ(En)=(n⩾n0),n→+∞limμ(En)=μ(∅)=0So the condition μ is finite is necessary.
Theorem 3.0.6.F⊂ℑ(Ω) is a semi-algebra over Ω and μ:F⟶R+∪{+∞} a function which is additive, then
1. ∃ a function
ν:A(F)⟶R+∪{+∞}
which is additive.
2. ν is an extension of μ.
3. The extension of μ is unique.
Proof.1⟩ Let A∈A(F), then we have A=∑j=1nEj where Ej∈F. We define ν(A):=j=1∑nν(Ej)extensionj=1∑nμ(Ej)We claim that ν is well-defined first.
Assume that A=∑j=1nEj=∑k=1mFk where Ej,Fk∈F, what we need to show is ∑j=1nμ(Ej)=∑k=1nμ(Fk). Since Ej⊂A=∑k=1nFk, Ej=Ej⋂∑k=1mFk=∑k=1mcontain inF(Ej⋂Fk), the additivity of μ guarantee that μ(Ej)=μ(k=1∑m(Ej⋂Fk))=k=1∑mμ(Ej⋂Fk)Then ν(A)=j=1∑nμ(Ej)=j=1∑nk=1∑mμ(Ej⋂Fk)By the symmetry of Ej and Fk, we can write ν(A)=k=1∑mμ(Fk)=k=1∑mj=1∑nμ(Fk⋂Ej)Consequently, ν is well-defined. Next we claim that ν is additive.
Given A=∑j=1nEj and B=∑k=1mFk where Ej,Fk∈F and A∩B=∅. We write A⋃B=(j=1∑nEj+k=1∑mFk)∈FThen ν(A⋃B)(by the definition of ν)=j=1∑nμ(Ej)+k=1∑mμ(Fk)=ν(A)+ν(B)2⟩ It is clear that ν is an extension of μ by the definition of ν.
3⟩ At the last step, we need to proof that the extension of μ is unique. In other words, if there are two functions ν1,ν2:A(F)⟶R+∪{+∞}Then {μ1(A)=μ2(A),∀A∈Aν1andν2are both additive⟹ν1(B)=ν2(B)for allB∈A(F)Let B∈A(F), B=∑j=1nEj where Ej∈F, μ1(B)(ν1 agree with ν2 at ∀A∈F)=μ1(j=1∑nEj)=j=1∑nμ1(Ej)=j=1∑nμ2(Ej)=μ2(j=1∑nEj)=μ2(B)
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Remark 3.0.7. If the condition μ is additive replaced to be μ is σ-additive, then the extension ν of μ is also σ-additive in the theorem. Assume that A=∑j⩾1Aj where A, Aj∈A(F) and ν:F⟶R+∪{+∞} and μ:F⟶R+∪{+∞} is σ-additive. It sufficient to show that ν(A)=∑j⩾1ν(Aj).
A=∑j=1nEj where Ej∈F since A∈A(F), Ak=∑l=1mkEk,l where Ek,l∈F. Then ν(A)=j=1∑nμ(Ej)=j=1∑nμ(Ej⋂A)=j=1∑nμ(Ej⋂k⩾1∑Ak)=j=1∑nμ(Ej⋂k⩾1∑l=1∑mkEk,l)=j=1∑nμ(k⩾1∑l=1∑mk(Ej⋂Ek,l))=j=1∑nk⩾1∑l=1∑mkμ(Ej⋂Ek,l)=k⩾1∑l=1∑mk(j=1∑nμ(Ek,l⋂Ej))=k⩾1∑l=1∑mkμ(j=1∑n(Ek,l⋂Ej))=k⩾1∑l=1∑mkμ(Ek,l⋂A)=k⩾1∑l=1∑mkμ(Ek,l)=k⩾1∑ν(Ak)