Vitali–Hahn–Saks theorem explained

In mathematics, the Vitali–Hahn–Saks theorem, introduced by,, and, proves that under some conditions a sequence of measures converging point-wise does so uniformly and the limit is also a measure.

Statement of the theorem

(S,l{B},m)

is a measure space with

m(S)<infty,

and a sequence

λ_n

of complex measures. Assuming that each

λ_n

is absolutely continuous with respect to

and that a for all

B\inl{B}

the finite limits exist

\lim_n\toinftyλ_n(B)=λ(B).

Then the absolute continuity of the

λ_n

with respect to

is uniform in

that is,

\lim_Bm(B)=0

implies that

\lim_Bλ_n(B)=0

uniformly in

Also

is countably additive on

l{B}.

Preliminaries

Given a measure space

(S,l{B},m),

a distance can be constructed on

l{B}_0,

the set of measurable sets

B\inl{B}

with

m(B)<infty.

This is done by defining

d(B_1,B₂₎=m(B_1\DeltaB_2),

where

B_1\DeltaB₂=(B_1\setminusB₂₎\cup(B_2\setminusB₁₎

is the symmetric difference of the sets

B_1,B_2\inl{B}_0.

This gives rise to a metric space

\tilde{l{B}_0}

by identifying two sets

B_1,B_2\inl{B}₀

when

m(B_1\DeltaB_2)=0.

Thus a point

\overline{B}\in\tilde{l{B}_0}

with representative

B\inl{B}₀

is the set of all

B_1\inl{B}₀

such that

m(B\DeltaB₁₎=0.

Proposition:

\tilde{l{B}_0}

with the metric defined above is a complete metric space.

Proof: Let $\chi_B(x)=\begin1,&x\in B\\0,&x\notin B\end$ Then $d(B_1,B_2)=\int_S|\chi_(s)-\chi_(x)|dm$ This means that the metric space

\tilde{l{B}_0}

can be identified with a subset of the Banach space

L^1(S,l{B},m)

Let

B_n\inl{B}₀

, with

\lim_d(B_n,B_k)=\lim_\int_S|\chi_(x)-\chi_(x)|dm=0

Then we can choose a sub-sequence

\chi
	B_n'

such that

\lim_n'\toinfty

\chi
	B_n'

(x)=\chi(x)

exists almost everywhere and

\lim_n'\toinfty\int_{S|\chi(x)-\chi}


	B_n'(x)

|dm=0

. It follows that

\chi=\chi
	B_infty

for some

B_infty\inl{B}₀

(furthermore

\chi(x)=1

if and only if

\chi
	B_n'

(x)=1

for

large enough, then we have that

B_infty=\liminf_n'\toinftyB_n'=

	infty}B
{cup
	m\right)

the limit inferior of the sequence) and hence

\lim_n\toinftyd(B_infty,B_n)=0.

Therefore,

\tilde{l{B}_0}

is complete.

Proof of Vitali-Hahn-Saks theorem

Each

λ_n

defines a function

\overline{λ}_{n(\overline{B})}

\tilde{l{B}}

by taking

\overline{λ}_{n(\overline{B})=λ}_n(B)

. This function is well defined, this is it is independent on the representative

of the class

\overline{B}

due to the absolute continuity of

λ_n

with respect to

. Moreover

\overline{λ}_n

is continuous.

For every

\epsilon>0

the set

F_=\

is closed in

\tilde{l{B}}

, and by the hypothesis

\lim_n\toinftyλ_n(B)=λ(B)

we have that

\tilde=\bigcup_^F_

By Baire category theorem at least one

F
	k_0,\epsilon

must contain a non-empty open set of

\tilde{l{B}}

. This means that there is

\overline{B_{0}\in\tilde{l{B}}}

and a

\delta>0

such that

d(B,B_0)<\delta

implies

\sup_n\geq1

\|\overline{λ}
	k₀

(\overline{B})-\overline{λ}
	k_0+n

(\overline{B})|\leq\epsilon

On the other hand, any

B\inl{B}

with

m(B)\leq\delta

can be represented as

B=B_1\setminusB₂

with

d(B_1,B_0)\leq\delta

and

d(B_2,B_0)\leq\delta

. This can be done, for example by taking

B_1=B\cupB₀

and

B_2=B_{0\setminus(B\cap}B₀₎

. Thus, if

m(B)\leq\delta

and

k\geqk₀

then

\begin|\lambda_k(B)|&\leq|\lambda_(B)|+|\lambda_(B)-\lambda_k(B)|\\&\leq|\lambda_(B)|+|\lambda_(B_1)-\lambda_k(B_1)|+|\lambda_(B_2)-\lambda_k(B_2)|\\&\leq|\lambda_(B)|+2\epsilon\end

Therefore, by the absolute continuity of

λ
	k₀

with respect to

, and since

\epsilon

is arbitrary, we get that

m(B)\to0

implies

λ_n(B)\to0

uniformly in

In particular,

m(B)\to0

implies

λ(B)\to0.

By the additivity of the limit it follows that

is finitely-additive. Then, since

\lim_m(B)λ(B)=0

it follows that

is actually countably additive