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Consider a separable Banach space $X$. The Borel $\sigma$-algebra $\mathcal{B}$ of $X$ is the same when taken with respect to the weak or strong topology. Hence, the space of probability measures over $X$ is the same with respect to the weak or the strong topology.

What differs is the topology of this space right? The weak convergence of a sequence of measures $\mu^n$ to $\mu$ is given by the convergence of integrals $$\int_X f(x) \mu^n(dx) \to \int_X f(x) \mu(dx)$$ for all weak-continuous functions or for all strong-continuous functions.


Now, if $(\mu^n)$ is a tight family of probability measures on $X$ with respect to the weak topology, then $\mu^n$ is relatively compact (see this notes by Kallianpur). Hence, there is a weakly convergent subsequence to a probability measure $\mu$, i.e. for all weak continuous f $${\int_X f(x) \mu^{n_k}(dx) \xrightarrow{k\to\infty} \int_X f(x) \mu(dx)}\,.$$


Let $(\mu^n$) be a sequence of regular measures on $X$. If for any $\varepsilon > 0$ there is a metrizable compact subset $K_\varepsilon \subset X$ such that $$\inf_{n} \mu^n(K_\varepsilon) > 1 - \varepsilon\,,$$ then the sequence $(\mu^n)_n$ admits a weakly convergent subsequence $(\mu^{n_k})_k$ (see Theorem 8.6.7 in V.I. Bogachev's "Measure Theory V.2."). Hence, there is a prob. measure $\mu$ such that for any weak continuous map $f$ $$\int_X f(x) \mu^{n_k}(dx) \xrightarrow{k\to\infty} \int_X f(x) \mu(dx)\,.$$

My question is: In what cases can I show that for a sequentially continuous function $f$ we have $$\int_X f(x) \mu^{n_k}(dx) \xrightarrow{k\to\infty} \int_X f(x) \mu(dx) \,?$$

I was trying the following: For fixed $\varepsilon > 0$, find $B_R = \{x \in X | \|x\| \leq R\}$ such that $\mu^n(B_R) > 1 - \varepsilon$ for all $n > 0$.

I was trying to restrict $\mu^{n_k}$ to $B_R$ and show convergence there, but I run into problems when dealing with $\mu^{n_k}(B_R) \to \mu(B_R)$. With respect to the strong topology I can pick R slightly larger avoiding any atom in the boundary of $B_R$, i.e. there are countably many $R$ dense in $[0, \infty)$ such that $\mu(\partial B_R) = 0$, hence $B_R$ is a continuity set. This is not possible with respect to the weak topology because the boundary of $B_R$ is $B_R$ itself.

My intuition is that a sequence of measure can not capture the difference between sequential continuous and continuous functions, but I can put that down into rigorous math.

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  • $\begingroup$ Is $X$ assumed separable? I am not sure that the weak and strong topologies induce the same Borel $\sigma$-algebra otherwise. $\endgroup$ – Nate Eldredge May 11 '17 at 18:17
  • $\begingroup$ Yes, $X$ is separable, I will add that. Thanks. $\endgroup$ – Jorge E. Cardona May 11 '17 at 18:23
  • $\begingroup$ Also, given that $\mu^n$ is tight, how do you conclude it has a weakly convergent subsequence? The usual version of Prohorov's theorem requires you are working on a separable metric space, but the weak topology isn't necessarily metrizable, not even on balls. Unless you also want to assume $X$ is reflexive? $\endgroup$ – Nate Eldredge May 11 '17 at 18:28
  • $\begingroup$ Hi, I'm basing this from the notes of Kallianpur in projecteuclid.org/euclid.lnms/1215451870 Theorem 2.2.1, the direction tight $\implies$ relatively weakly compact holds for Hausdorff spaces. The full Prokhorov requires Polish. Also, $X$ is reflexive as well. $\endgroup$ – Jorge E. Cardona May 11 '17 at 18:48
  • $\begingroup$ Related to the remark by @NateEldredge: That a set of probability measures is relatively compact does not imply that the set is sequentially relatively compact. So there might be a sequence of probability measures that has no convergent subsequence but onlya convergent subnet, and a subnet of a sequence need not be representable by a subsequence. Maybe you can rule that out, but there should be some argument then. $\endgroup$ – Michael Greinecker May 11 '17 at 21:45

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