THC Science

In mathematics, the Vitali–Hahn–Saks theorem, introduced by Vitali (1907), Hahn (1922), and Saks (1933), 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[edit]

If $(S,{\mathcal {B}},m)$ is a measure space with $m(S)<\infty ,$ and a sequence $\lambda _{n}$ of complex measures. Assuming that each $\lambda _{n}$ is absolutely continuous with respect to $m,$ and that a for all $B\in {\mathcal {B}}$ the finite limits exist $\lim _{n\to \infty }\lambda _{n}(B)=\lambda (B).$ Then the absolute continuity of the $\lambda _{n}$ with respect to $m$ is uniform in $n,$ that is, $\lim _{B}m(B)=0$ implies that $\lim _{B}\lambda _{n}(B)=0$ uniformly in $n.$ Also $\lambda$ is countably additive on ${\mathcal {B}}.$

Preliminaries[edit]

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

d(B_{1},B_{2})=m(B_{1}\Delta B_{2}),

where

B_{1}\Delta B_{2}=(B_{1}\setminus B_{2})\cup (B_{2}\setminus B_{1})

is the symmetric difference of the sets

B_{1},B_{2}\in {\mathcal {B}}_{0}.

This gives rise to a metric space ${\tilde {{\mathcal {B}}_{0}}}$ by identifying two sets $B_{1},B_{2}\in {\mathcal {B}}_{0}$ when $m(B_{1}\Delta B_{2})=0.$ Thus a point ${\overline {B}}\in {\tilde {{\mathcal {B}}_{0}}}$ with representative $B\in {\mathcal {B}}_{0}$ is the set of all $B_{1}\in {\mathcal {B}}_{0}$ such that $m(B\Delta B_{1})=0.$

Proposition: ${\tilde {{\mathcal {B}}_{0}}}$ with the metric defined above is a complete metric space.

Proof: Let

\chi _{B}(x)={\begin{cases}1,&x\in B\\0,&x\notin B\end{cases}}

Then

d(B_{1},B_{2})=\int _{S}|\chi _{B_{1}}(s)-\chi _{B_{2}}(x)|dm

This means that the metric space

{\tilde {{\mathcal {B}}_{0}}}

can be identified with a subset of the Banach space

L^{1}(S,{\mathcal {B}},m)

.

Let $B_{n}\in {\mathcal {B}}_{0}$ , with

\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-sequence

\chi _{B_{n'}}

such that

\lim _{n'\to \infty }\chi _{B_{n'}}(x)=\chi (x)

exists almost everywhere and

\lim _{n'\to \infty }\int _{S}|\chi (x)-\chi _{B_{n'}(x)}|dm=0

. It follows that

\chi =\chi _{B_{\infty }}

for some

B_{\infty }\in {\mathcal {B}}_{0}

(furthermore

\chi (x)=1

if and only if

\chi _{B_{n'}}(x)=1

for

n'

large enough, then we have that

B_{\infty }=\liminf _{n'\to \infty }B_{n'}={\bigcup _{n'=1}^{\infty }}\left({\bigcap _{m=n'}^{\infty }}B_{m}\right)

the limit inferior of the sequence) and hence

\lim _{n\to \infty }d(B_{\infty },B_{n})=0.

Therefore,

{\tilde {{\mathcal {B}}_{0}}}

is complete.

Proof of Vitali-Hahn-Saks theorem[edit]

Each $\lambda _{n}$ defines a function ${\overline {\lambda }}_{n}({\overline {B}})$ on ${\tilde {\mathcal {B}}}$ by taking ${\overline {\lambda }}_{n}({\overline {B}})=\lambda _{n}(B)$ . This function is well defined, this is it is independent on the representative $B$ of the class ${\overline {B}}$ due to the absolute continuity of $\lambda _{n}$ with respect to $m$ . Moreover ${\overline {\lambda }}_{n}$ is continuous.

For every $\epsilon >0$ the set

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 in

{\tilde {\mathcal {B}}}

, and by the hypothesis

\lim _{n\to \infty }\lambda _{n}(B)=\lambda (B)

we have that

{\tilde {\mathcal {B}}}=\bigcup _{k=1}^{\infty }F_{k,\epsilon }

By Baire category theorem at least one

F_{k_{0},\epsilon }

must contain a non-empty open set of

{\tilde {\mathcal {B}}}

. This means that there is

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

and a

\delta >0

such that

d(B,B_{0})<\delta

implies

\sup _{n\geq 1}|{\overline {\lambda }}_{k_{0}}({\overline {B}})-{\overline {\lambda }}_{k_{0}+n}({\overline {B}})|\leq \epsilon

On the other hand, any

B\in {\mathcal {B}}

with

m(B)\leq \delta

can be represented as

B=B_{1}\setminus B_{2}

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\cup B_{0}

and

B_{2}=B_{0}\setminus (B\cap B_{0})

. Thus, if

m(B)\leq \delta

and

k\geq k_{0}

then

{\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 of

\lambda _{k_{0}}

with respect to

m

, and since

\epsilon

is arbitrary, we get that

m(B)\to 0

implies

\lambda _{n}(B)\to 0

uniformly in

n.

In particular,

m(B)\to 0

implies

\lambda (B)\to 0.

By the additivity of the limit it follows that $\lambda$ is finitely-additive. Then, since $\lim _{m(B)\to 0}\lambda (B)=0$ it follows that $\lambda$ is actually countably additive.

References[edit]

Hahn, H. (1922), "Über Folgen linearer Operationen", Monatsh. Math. (in German), 32: 3–88, doi:10.1007/bf01696876
Saks, Stanislaw (1933), "Addition to the Note on Some Functionals", Transactions of the American Mathematical Society, 35 (4): 965–970, doi:10.2307/1989603, JSTOR 1989603
Vitali, G. (1907), "Sull' integrazione per serie", Rendiconti del Circolo Matematico di Palermo (in Italian), 23: 137–155, doi:10.1007/BF03013514
Yosida, K. (1971), Functional Analysis, Springer, pp. 70–71, ISBN 0-387-05506-1

THC Science

Bringing Science to the Cannabis Conversation!

Cannabis

Statement of the theorem[edit]

Preliminaries[edit]

Proof of Vitali-Hahn-Saks theorem[edit]

References[edit]

Leave a Reply