# weak convergence and unbounded functions with bounded moment

I want to prove the following: Given a topological space (it is a Lusin space, but I think that does not matter) $$\Omega$$, a function $$f \in C(\Omega,\mathbb{R})$$ and a sequence of Radon measures $$P^{N}$$ defined on it that converges weakly to a measure $$P$$, then $$\mathbb{E}^{P^{N}}\left[ f \right] = \int_{\Omega} f dP^{N} \rightarrow \int_{\Omega} f dP = \mathbb{E}^{P}\left[f\right].$$ The problem is that the function $$f$$ is not bounded, i.e. $$f \notin C_{b}(\Omega)$$ but instead satisfies the condition $$\sup_{N} \mathbb{E}^{P^{N}}\left[\left|f\right|^{1+\varepsilon}\right] \leq C$$ for an $$\varepsilon > 0$$. I read this claim in a paper, but unfortunately with neither a proof nor a reference. A similar question was posed in this thread: weak convergence of probability measures and unbounded functions with bounded expectation but there we had $$\varepsilon = 0$$. As far as I can see, the counterexample posted there is not a counterexample here. The result appears to me rather elementary but I haven't found it anywhere yet...

• You tagged the question "probability". Are all the involved measures probability measures? If $\Omega$ is $\mathbb R$ with the usual topology, then we pick $R$ and a function $\chi_R$ with values in the unit interval, value $1$ on $[-R,R]$ and $0$ outside $[-R-1,R+1]$. The assumption on $f$ show that the contribution of $\int_{\mathbb R}f(1-\chi_R)$ is not important, and you can use the weak convergence for the part $\int_{\mathbb R}f\chi_R$. May 27, 2016 at 12:19

I give an answer for probability measures, as it seems from your questions that this is the setting you are interested into.

This follows from the following theorem contained in van der Vaart - Asymptotic Statistics. I will cite directly from that book:

A sequence of random variables $$Y_n$$ is called asymptotically uniformly integrable if $$\lim _{M \rightarrow \infty} \limsup _{n \rightarrow \infty} \mathrm{E}\left|Y_n\right| 1\left\{\left|Y_n\right|>M\right\}=0 .$$ Uniform integrability is the missing link between convergence in distribution and convergence of moments.

Theorem. Let $$f: \mathbb{R}^k \mapsto \mathbb{R}$$ be measurable and continuous at every point in a set C. Let $$X_n \rightsquigarrow X$$ where $$X$$ takes its values in $$C$$. Then $$\mathrm{E} f\left(X_n\right) \rightarrow \mathrm{E} f(X)$$ if and only if the sequence of random variables $$f\left(X_n\right)$$ is asymptotically uniformly integrable.

Example. Suppose $$X_n$$ is a sequence of random variables such that $$X_n \rightsquigarrow X$$ and $$\lim \sup \mathrm{E}\left|X_n\right|^p<\infty$$ for some $$p$$. Then all moments of order strictly less than $$p$$ converge also: $$\mathrm{E} X_n^k \rightarrow \mathrm{E} X^k$$ for every $$k.

By the preceding theorem, it suffices to prove that the sequence $$X_n^k$$ is asymptotically uniformly integrable. By Markov's inequality $$\mathrm{E}\left|X_n\right|^k 1\left\{\left|X_n\right|^k \geq M\right\} \leq M^{1-p / k} \mathrm{E}\left|X_n\right|^p .$$ The limit superior, as $$n \rightarrow \infty$$ followed by $$M \rightarrow \infty$$, of the right side is zero if $$k.

Therefore your condition implies asymptotic uniform integrability, that in turn gives weak convergence even if the $$f$$'s are unbounded. This is true as long as the moments are bounded at least as you specified.