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Suppose $\mu_n$ and $\mu$ are probability measures such that $\mu_n \to \mu$ in total variation. I'm curious to what extent we can say $$\int f \ \mathrm{d}\mu_n \to \int f \ \mathrm{d}\mu,$$ when $\mu_n \to \mu$ in total variation - recall that this means $$\lim_{n \to \infty} \sup_{A \in \mathcal F}|\mu_n (A) - \mu(A)| \to 0,$$ where $\mathcal F$ is the underlying $\sigma$-algebra and $f$ of course is a Borel function. I've looked around a bit (admittedly not too hard - I thumbed through the TOC and index of Billingsley's "Convergence of Probability Measures" and looked at the Wikipedia page) and it seems to me I ought to be able to say that this should happen under very mild requirements. Under weak convergence of probability measures, of course, $\int f \ \mathrm{d}\mu_n \to \int f \ \mathrm{d}\mu$ holds for all bounded, continuous functions, which competely characterizes this mode of convergence. Surely I can get away with more with total variation convergence - what I hope "getting away with more" entails is that I can widen the class of $f$ such that this happens. For example, since $\mu_n (A) \to \mu(A)$ for all $A$ rather than just those $A$ with boundary probability $0$ I would hope we could loosen the continuity assumption on $f$.

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unbounded? discontinuous? both? – cactus314 Feb 19 '13 at 19:02
@johnmangual mainly I care about removing continuity (boundedness is fine), but I'm asking for the most general conditions available. – guy Feb 19 '13 at 19:18
up vote 2 down vote accepted

If $f$ is a bounded, measurable real function, and $\mu_n\to\mu$ in total variation, then $\int f \, d\mu_n\to \int f\, d\mu$. The reason is just that $f$ can be uniformly approximated by simple functions.

If $f$ is not bounded $\int f\, d\mu_n$ need not converge to $\int f\, d\mu$. For a counterexample, look at measures on the real line, let $\mu_n$ give $\{n\}$ probability $\frac1n$ and $\{0\}$ probability $1-\frac1n$, let $\mu$ give $\{0\}$ probability $1$, and let $f$ be the identity. Then $\mu_n\to\mu$ in total variation but, for all $n$, $\int f\, d\mu_n=1\ne \int f \,d\mu=0$.

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