# Strong convergence and convergence of integral

A sequence of probability measures $$\{\mu_n\}$$ converges strongly if for each measurable set $$A$$

$$\mu_{n}(A) \rightarrow \mu(A)$$

(notice that this convergence is different than the total variation)

when does this convergence imply

$$\int f \,d\mu_n\to \int f \,d\mu$$

for every bounded measurable function? I'm interested in the case in which the space is not Polish.

• Are all the $\mu_n$-s probabilities?
– user239203
Oct 11, 2019 at 18:05

$$\newcommand{nrm}[1]{\left\lVert{#1}\right\rVert}\newcommand{abs}[1]{\left\lvert{#1}\right\rvert}$$It always does. I'll use the notation $$\mu(f):=\int f\,d\mu$$ throughout. Essentially, you can use the fact that all the $$\mu$$-s are probabilities on $$X$$ (or the slightly weaker fact that $$M=\sup_{n\in\Bbb N}\mu_n(X)$$ and $$\mu(X)$$ are finite).

The result is obvious for simple functions. Now, recall that every bounded measurable function is uniform limit of simple functions. Namely, $$f_n:=\sum_{k=-2n}^{2n-1} \frac kn\cdot 1_{f^{-1}\left[\frac kn,\frac{k+1}n\right)}$$ satisfies $$\nrm{f-f_n}_\infty\le\frac1n$$ for all $$n>\sup_{x\in X}\abs{f(x)}$$.

Finally, for any fixed $$\delta>0$$ consider some simple function $$g$$ such that $$\nrm{f-g}_\infty<\delta$$ and observe that, for all $$n$$, $$\abs{\mu(f)-\mu_n(f)}\le\abs{\mu(f-g)}+\abs{\mu(g)-\mu_n(g)}+\abs{\mu_n(g-f)}\le\\\le (\mu(X)+\mu_n(X))\nrm{f-g}_\infty+\mu_n(g-f)\le (1+M)\delta+\abs{\mu(g)-\mu_n(g)}$$

The last quantity is a sequence converging to $$(1+M)\delta$$. Therefore, for all $$\delta>0$$ $$\limsup\limits_{n\to\infty}\abs{\mu(f)-\mu_n(f)}\le (1+M)\delta$$

Therefore $$\limsup\limits_{n\to\infty}\abs{\mu(f)-\mu_n(f)}=0$$.

• This seems good. Some clarifications. What is $M$ in the second to last displayed equation? Also the last two equations should probably have $f$’s instead of $X$’s. Oct 11, 2019 at 18:44
• @Condor5 $M$ is $\sup_n\mu_n(X)$.
– user239203
Oct 11, 2019 at 19:06
• Thanks, then the term that is left after the second inequality should be $\mu(g)-\mu_n(g)$ (and the last equation, after therefore should also have an $f$) Oct 11, 2019 at 19:10
• @Condor5 true, thanks.
– user239203
Oct 11, 2019 at 19:18