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Vitali–Hahn–Saks theorem

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Inmathematics, theVitali–Hahn–Saks theorem, introduced byVitali (1907),Hahn (1922), andSaks (1933), proves that under some conditions a sequence ofmeasures converging point-wise does so uniformly and the limit is also a measure.

Statement of the theorem

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If(S,B,m){\displaystyle (S,{\mathcal {B}},m)} is ameasure space withm(S)<,{\displaystyle m(S)<\infty ,} and a sequenceλn{\displaystyle \lambda _{n}} ofcomplex measures. Assuming that eachλn{\displaystyle \lambda _{n}} isabsolutely continuous with respect tom,{\displaystyle m,} and that for allBB{\displaystyle B\in {\mathcal {B}}} the finite limits existlimnλn(B)=λ(B){\displaystyle \lim _{n\to \infty }\lambda _{n}(B)=\lambda (B)}. Then the absolute continuity of theλn{\displaystyle \lambda _{n}} with respect tom{\displaystyle m} is uniform inn{\displaystyle n}, that is,limBm(B)=0{\displaystyle \lim _{B}m(B)=0} implies thatlimBλn(B)=0{\displaystyle \lim _{B}\lambda _{n}(B)=0} uniformly inn{\displaystyle n}. Alsoλ{\displaystyle \lambda } is countably additive onB{\displaystyle {\mathcal {B}}}.

Preliminaries

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Given a measure space(S,B,m),{\displaystyle (S,{\mathcal {B}},m),} a distance can be constructed onB0,{\displaystyle {\mathcal {B}}_{0},} the set of measurable setsBB{\displaystyle B\in {\mathcal {B}}} withm(B)<.{\displaystyle m(B)<\infty .} This is done by defining

d(B1,B2)=m(B1ΔB2),{\displaystyle d(B_{1},B_{2})=m(B_{1}\Delta B_{2}),} whereB1ΔB2=(B1B2)(B2B1){\displaystyle B_{1}\Delta B_{2}=(B_{1}\setminus B_{2})\cup (B_{2}\setminus B_{1})} is thesymmetric difference of the setsB1,B2B0.{\displaystyle B_{1},B_{2}\in {\mathcal {B}}_{0}.}

This gives rise to a metric spaceB0~{\displaystyle {\tilde {{\mathcal {B}}_{0}}}} by identifying two setsB1,B2B0{\displaystyle B_{1},B_{2}\in {\mathcal {B}}_{0}} whenm(B1ΔB2)=0.{\displaystyle m(B_{1}\Delta B_{2})=0.} Thus a pointB¯B0~{\displaystyle {\overline {B}}\in {\tilde {{\mathcal {B}}_{0}}}} with representativeBB0{\displaystyle B\in {\mathcal {B}}_{0}} is the set of allB1B0{\displaystyle B_{1}\in {\mathcal {B}}_{0}} such thatm(BΔB1)=0.{\displaystyle m(B\Delta B_{1})=0.}

Proposition:B0~{\displaystyle {\tilde {{\mathcal {B}}_{0}}}} with the metric defined above is acomplete metric space.

Proof: LetχB(x)={1,xB0,xB{\displaystyle \chi _{B}(x)={\begin{cases}1,&x\in B\\0,&x\notin B\end{cases}}}Thend(B1,B2)=S|χB1(s)χB2(x)|dm{\displaystyle d(B_{1},B_{2})=\int _{S}|\chi _{B_{1}}(s)-\chi _{B_{2}}(x)|dm}This means that the metric spaceB0~{\displaystyle {\tilde {{\mathcal {B}}_{0}}}} can be identified with a subset of theBanach spaceL1(S,B,m){\displaystyle L^{1}(S,{\mathcal {B}},m)}.

LetBnB0{\displaystyle B_{n}\in {\mathcal {B}}_{0}}, withlimn,kd(Bn,Bk)=limn,kS|χBn(x)χBk(x)|dm=0{\displaystyle \lim _{n,k\to \infty }d(B_{n},B_{k})=\lim _{n,k\to \infty }\int _{S}|\chi _{B_{n}}(x)-\chi _{B_{k}}(x)|dm=0}Then we can choose a sub-sequenceχBn{\displaystyle \chi _{B_{n'}}} such thatlimnχBn(x)=χ(x){\displaystyle \lim _{n'\to \infty }\chi _{B_{n'}}(x)=\chi (x)} existsalmost everywhere andlimnS|χ(x)χBn(x)|dm=0{\displaystyle \lim _{n'\to \infty }\int _{S}|\chi (x)-\chi _{B_{n'}(x)}|dm=0}. It follows thatχ=χB{\displaystyle \chi =\chi _{B_{\infty }}} for someBB0{\displaystyle B_{\infty }\in {\mathcal {B}}_{0}} (furthermoreχ(x)=1{\displaystyle \chi (x)=1}if and only ifχBn(x)=1{\displaystyle \chi _{B_{n'}}(x)=1} forn{\displaystyle n'} large enough, then we have thatB=lim infnBn=n=1(m=nBm){\displaystyle B_{\infty }=\liminf _{n'\to \infty }B_{n'}={\bigcup _{n'=1}^{\infty }}\left({\bigcap _{m=n'}^{\infty }}B_{m}\right)} thelimit inferior of the sequence) and hencelimnd(B,Bn)=0.{\displaystyle \lim _{n\to \infty }d(B_{\infty },B_{n})=0.} Therefore,B0~{\displaystyle {\tilde {{\mathcal {B}}_{0}}}} is complete.

Proof of Vitali-Hahn-Saks theorem

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Eachλn{\displaystyle \lambda _{n}} defines a functionλ¯n(B¯){\displaystyle {\overline {\lambda }}_{n}({\overline {B}})} onB~{\displaystyle {\tilde {\mathcal {B}}}} by takingλ¯n(B¯)=λn(B){\displaystyle {\overline {\lambda }}_{n}({\overline {B}})=\lambda _{n}(B)}. This function is well defined, this is it is independent on the representativeB{\displaystyle B} of the classB¯{\displaystyle {\overline {B}}} due to the absolute continuity ofλn{\displaystyle \lambda _{n}} with respect tom{\displaystyle m}. Moreoverλ¯n{\displaystyle {\overline {\lambda }}_{n}} is continuous.

For everyϵ>0{\displaystyle \epsilon >0} the setFk,ϵ={B¯B~: supn1|λ¯k(B¯)λ¯k+n(B¯)|ϵ}{\displaystyle F_{k,\epsilon }=\{{\overline {B}}\in {\tilde {\mathcal {B}}}:\ \sup _{n\geq 1}|{\overline {\lambda }}_{k}({\overline {B}})-{\overline {\lambda }}_{k+n}({\overline {B}})|\leq \epsilon \}}is closed inB~{\displaystyle {\tilde {\mathcal {B}}}}, and by the hypothesislimnλn(B)=λ(B){\displaystyle \lim _{n\to \infty }\lambda _{n}(B)=\lambda (B)} we have thatB~=k=1Fk,ϵ{\displaystyle {\tilde {\mathcal {B}}}=\bigcup _{k=1}^{\infty }F_{k,\epsilon }}ByBaire category theorem at least oneFk0,ϵ{\displaystyle F_{k_{0},\epsilon }} must contain a non-empty open set ofB~{\displaystyle {\tilde {\mathcal {B}}}}. This means that there isB0B{\displaystyle B_{0}\in {\mathcal {B}}} and aδ>0{\displaystyle \delta >0} such thatd(B,B0)<δsupn1|λ¯k0(B¯)λ¯k0+n(B¯)|ϵ.{\displaystyle d(B,B_{0})<\delta \Rightarrow \sup _{n\geq 1}|{\overline {\lambda }}_{k_{0}}({\overline {B}})-{\overline {\lambda }}_{k_{0}+n}({\overline {B}})|\leq \epsilon .}On the other hand, anyBB{\displaystyle B\in {\mathcal {B}}} withm(B)δ{\displaystyle m(B)\leq \delta } can be represented asB=B1B2{\displaystyle B=B_{1}\setminus B_{2}} withd(B1,B0)δ{\displaystyle d(B_{1},B_{0})\leq \delta } andd(B2,B0)δ{\displaystyle d(B_{2},B_{0})\leq \delta }. This can be done, for example by takingB1=BB0{\displaystyle B_{1}=B\cup B_{0}} andB2=B0(BB0){\displaystyle B_{2}=B_{0}\setminus (B\cap B_{0})}. Thus, ifm(B)δ{\displaystyle m(B)\leq \delta } andkk0{\displaystyle k\geq k_{0}} then|λk(B)||λk0(B)|+|λk0(B)λk(B)||λk0(B)|+|λk0(B1)λk(B1)|+|λk0(B2)λk(B2)||λk0(B)|+2ϵ{\displaystyle {\begin{aligned}|\lambda _{k}(B)|&\leq |\lambda _{k_{0}}(B)|+|\lambda _{k_{0}}(B)-\lambda _{k}(B)|\\&\leq |\lambda _{k_{0}}(B)|+|\lambda _{k_{0}}(B_{1})-\lambda _{k}(B_{1})|+|\lambda _{k_{0}}(B_{2})-\lambda _{k}(B_{2})|\\&\leq |\lambda _{k_{0}}(B)|+2\epsilon \end{aligned}}}Therefore, by the absolute continuity ofλk0{\displaystyle \lambda _{k_{0}}} with respect tom{\displaystyle m}, and sinceϵ{\displaystyle \epsilon } is arbitrary, we get thatm(B)0{\displaystyle m(B)\to 0} impliesλn(B)0{\displaystyle \lambda _{n}(B)\to 0} uniformly inn.{\displaystyle n.} In particular,m(B)0{\displaystyle m(B)\to 0} impliesλ(B)0.{\displaystyle \lambda (B)\to 0.}

By the additivity of the limit it follows thatλ{\displaystyle \lambda } isfinitely-additive. Then, sincelimm(B)0λ(B)=0{\displaystyle \lim _{m(B)\to 0}\lambda (B)=0} it follows thatλ{\displaystyle \lambda } is actually countably additive.

References

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Basic concepts
Sets
Types ofmeasures
Particular measures
Maps
Main results
Other results
ForLebesgue measure
Applications & related
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