Suppose $(X,\mathscr{F},\mu)$, and $\{f_n,f:n\in\mathbb{N}\}$ be a a collection of measurable functions.
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It is a well known fact that if $f_n$ converges to $f$ point-wise $\mu$-a.s., then $f_n$ converges to $f$ weakly in measure, that is, for any $A\in\mathscr{F}$ with $\mu(A)<\infty$ and any $\delta>0$
$$\lim_n\mu(A\cap\{|f_n-f|>\delta\})=0$$ -
Conversely, if $\mu$ is $\sigma$–finite and $f_n$ converges to $f$ weakly in measure, then there exists a subsequence $f_{n_k}$ such that $f_{n_k}$ converges point-wise $\mu$-a.s. to $f$.
An immediate consequence of (1) and (2) is the following
Proposition: If $\mu$ is $\sigma$-finite and $f_n$ converges to $f$ $\mu$-a.s. and $f_n$ converges to $g$ weakly in measure, then $f=g$ $\mu$-a.s.
My questions are: Does the Proposition above hold when $\mu$ is not $\sigma$-finite? Are there any semi-finite measures for which the Proposition does not hold?
This is motivated by a seemingly (at first sight) trivial question that appeared here. As it is pointed out in the comments to that question, the problem is indeed trivial when $\mu$ is $\sigma$-finite.
Note: $\mu$ is semi-finite if for any $A\in\mathscr{F}$, $\mu(A)>0$ implies that there is $B\in\mathscr{B}$, $B\subset A$, such that $0<\mu(B)<\infty$.
Edit: If $\mu$ is semi-finite, the conclusion of the Proposition above holds: Let $f_n$,$f$ and $g$ be as in the Proposition above, and suppose $\mu(|f-g|>0)>0$. Then there exists $A\in\mathscr{F}$ with $0<\mu(A)<\infty$ on which $f\neq g$. Since $f_n$ converges point wise $\mu$-a.s. to $f$, by (1) we have that for any $\delta>0$
$\mu(A\cap\{|f_n-f|>\delta\})\xrightarrow{n\rightarrow\infty}0$
Hence
$$\mu(A\cap\{|f-g|>2\delta\})\leq\mu(A\cap\{|f_n-g|>\delta\})+\mu(A\cap\{|f_n-f|>\delta\})\xrightarrow{n\rightarrow\infty}0$$
This means that $\mu(A\cap\{|f-g|>0\})=0$ which is a contradiction. $\Box$
The first question, in the case where $\mu$ admits atoms of infinite mass still escapes me. Recall that every measure $\mu$ can be expressed as the sum of two measures $\mu_0$ and $\nu$ where $\mu_0$ is semi-finite, and $\nu(A)=\infty$ if $A\in\mathscr{F}$ contains an atom of infinite mass, and $\nu(A)=0$ other wise. Then $\mu_0(|f-g|>0)=0$.
Best Answer
Here is a counter example that shows that the proposition given in the OP may fail when $\mu$ is not semi finite. On the Borel space $([0,1],\mathscr{B}([0,1])$ define the measure $$\mu=\infty\cdot\delta_0 +\lambda_1$$ where $\lambda_1$ is restriction to $(0,1]$ of the Lebesgue measure on the real line, and $\delta_0$ is the measure that assigns the value $1$ to any set that contains $0$ and the value $0$ otherwise.
It is obvious that $\mu$ is not semi finite since $\{0\}$ is an atom with $\mu(\{0\})=\infty$. For $n\in\mathbb{N}$ define $f_n=\frac{1}{n}\mathbb{1}_{(0,1]}$, and let $f\equiv0$ and $g=\mathbb{1}_{\{0\}}$. It is readily seen that $\lim_{n\rightarrow0}f_n(x)=f(x)=0$ for all $x\in [0,1]$, and that for any set $B\in\mathbb{B}((0,1])$ and $\delta>0$ $$\lim_{n\rightarrow\infty}\mu\big(x\in B: |f_n(x)-g(x)|>\delta\big)=0$$ However $$\mu\big(x\in[0,1]:f(x)\neq g(x)\big)=\mu(\{0\})=\infty$$.