Weak convergence of a sequence of probability measures implies integrability of the limiting probability measure

Question

Let $(X_{n})_{n \in \mathbb{N}}$ be a sequence of uniformly integrable random vectors with values in some normed vector space $V$ with $\mathbb{E}[\|X_n\|] < \infty$. This means that $$ \lim_{C \to \infty} \sup_{n \in \mathbb{N}} \mathbb{E}[ \| X_{n} \| \mathbb{1}_{\{\| X_{n} \| > C\}} ] = 0. $$ Furthermore, suppose that $\mu$ is a probability measure on $V$ and the sequence $(X_{n})_{n \in \mathbb{N}}$ converges weakly to $\mu$. This means that for every bounded Lipschitz continuous function $f \colon V \rightarrow \mathbb{R}$ it holds $$ \lim_{n \rightarrow \infty} \int_{V}f dP_{X_{n}} = \int_{V} f d\mu. $$ How can I show that then $\int_{V} \| x \| d\mu(x) < \infty$?

Answer 1

By the uniform integrability of $(X_n)_{n \in \mathbb{N}}$ we have $$M := \sup_{n \geq 1} \mathbb{E}(|X_n|) < \infty. \tag{1}$$ On the other hand, the weak convergence of $X_n \to X$ entails that

$$\mathbb{E}f(X_n) \to \mathbb{E}f(X) \tag{2}$$

for any bounded Lipschitz continuous function; in particular, we can choose $f(x) := \min\{|x|,k\}$ for fixed $k \geq 1$ to get

$$\mathbb{E}\min\{|X|,k\} \stackrel{(2)}{=} \lim_{n \to \infty} \mathbb{E}\min\{|X_n|,k\} \leq \sup_{n \in \mathbb{N}} \mathbb{E}(|X_n|)=M.$$

Applying the monotone convergence theorem we conclude that

$$\mathbb{E}(|X|) = \sup_{k \geq 1} \mathbb{E}\min\{|X|,k\} \leq M < \infty.$$

Answer 2

Fix $N$ and let $f(x)=\min \{||x||,N|\}$. Claim $f$ is a bounded Lipschitz function. (The only case when this is not obvious is when $||x||<N<||y||$. In this case $|f(x)-f(y)|=|||x||-N|=N-||x||\leq ||y||-||x|| \leq ||x-y||$ by triangle inequlaity). Hence $\int f(x)d\mu(x) =\lim E(\min \{||X_n||,N\})\leq \sup_n E||X_n||$. Note that $E||X_n|| \leq E||X_n|| I_{||X_n|| >C} + E||X_n|| I_{||X_n|| \leq C}$. We can make the first term $<1$ by choosing $C$ appropriately. The second term does not exceed $C$. We have proved that $\int \min \{|X||,N|\} d\mu(x) \leq C+1$ and the bound does not depend on $N$. Letting $n \to \infty$ we get $\int |x|| d\mu (x) <\infty$.