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I asked this question on StackExchange a few days ago but didn't get any response, so I thought I would try here.

The Wikipedia article on convergence of measures defines three kinds of convergence: total variation, strong, and weak. For weak convergence, a number of equivalent formulations are given, but not for strong or total variation convergence. I have two questions.

  1. Are there some equivalent formulations for those notions of convergence too? In particular, is it true that $\mu_n$ strongly converges to $\mu$ iff $E_{\mu_n} f\to E_\mu f$ for every bounded measurable function $f$? According to an answer to this question, the left-to-right direction holds if strong convergence is strengthened to total variation convergence.

  2. Are there some nice characterizations of these topologies that are not stated in terms of convergent sequences?

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  • $\begingroup$ Do you haev any assumption on the base space? That is, it may be any measurable space, or is it a topological space, compact, metric...? $\endgroup$ – Pietro Majer Sep 19 '13 at 7:48
  • $\begingroup$ Not compact, but otherwise you can assume it is a nice space if it helps. $\endgroup$ – user39080 Sep 19 '13 at 10:56
  • $\begingroup$ The condition in 1 does not seem sufficient (wrto the total variation norm of measures) : E.g. if $\lambda$ is the Lebesgue measure on $[0,1]$ and $\mu$ and all $\mu_n$ are a.c. wrto $\lambda$ with densities $g_n$ and $g$ wrto $\lambda$, the condition means $g_n\to g$ weakly in $L^1$, whereas $\mu_n\to\mu$ in norm means $g_n\to g$ in the norm of $L^1$. $\endgroup$ – Pietro Majer Sep 19 '13 at 15:05
  • $\begingroup$ If by "strong convergence" you mean the same as in the wiki article, then I think the implication L to R holds if the sequence μn is also bounded in total variation norm (which is of course true if $\mu_n$ are positive measures). $\endgroup$ – Pietro Majer Sep 19 '13 at 16:48
  • $\begingroup$ I guess you meant to say "probability measures"? That is my interest, and I am glad if L to R holds. $\endgroup$ – user39080 Sep 19 '13 at 17:53
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To summarize the situation. Let $(X,\mathcal{F})$ a measurable space. The space $M(X,\mathcal{F})$ of all real-valued signed measures on $(X,\mathcal{F})$ is a Banach space wrto the total variation norm $\|\mu\|:=|\mu|(X)$ where the (non-negative) measure $|\mu|:= \mu_+ +\mu_-$ is the variation of $\mu$. The space $M(X,\mathcal{F})$may be isometrically embedded as a norm-closed subspace of the dual space $B(X,\mathcal{F})^*$ of the Banach space $B(X,\mathcal{F})$ of all bounded measurable functions on $X$, $B(X,\mathcal{F})$ with the uniform norm $\|\cdot\|_\infty$ ( that dual is in general much larger, since it also contains all additive measures). Note that the space of simple functions $S(X,\mathcal{F})$, linear span of characteristic functions of measurable sets, is norm dense in $B(X,\mathcal{F})$ via the usual approximation $f_n(x):= \lfloor nf(x)\rfloor/n$.

So the above "strong convergence of measures", that is with test functions in $B(X,\mathcal{F})$, is the weak* convergence of $B(X,\mathcal{F})^*$, in the particular cas eof sequences in $M(X,\mathcal{F})$. In particular any such convergent sequence of measures is norm bounded (as any w* convergent sequence of elements in a dual space), and it is "weakly convergent" in the above sense, that is with characteristic functions, hence also with simple functions as test. It is not norm convergent in general (as an example, take as said e.g. $\mu_n$ absolutely continuous w.r.to the Lebesgue measure on $X:=[0,1]$ and with densities $g_n\in L^1$ weakly convergent but not norm convergent).

Conversely, a norm bounded sequence of measures $\mu_n$ that weakly converges to $\mu$ wrto test functions $g$ in $S(X,\mathcal{F})$ also converges with test functions $f$ in $B(X,\mathcal{F})$, again an elementary and general fact. You can see it writing $$\langle \mu_n, f\rangle - \langle\mu, f\rangle=\langle\mu_n-\mu, g\rangle+ \langle\mu_n-\mu, g -f\rangle $$ with a simple function $g$, so that $$\limsup_{n\to\infty}|\langle\mu_n, f\rangle-\langle\mu, f\rangle|\le \limsup_{n\to\infty}|\langle\mu_n-\mu, g\rangle|+ \big(\sup_n \|\mu_n\|+\|\mu\|\big)\|g -f\|_\infty $$ whence $$\limsup_{n\to\infty}|\langle\mu_n, f\rangle-\langle\mu, f\rangle|=0,$$

since simple functions are $\|\cdot\|_\infty$ dense.

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