continuity of a map on $M(\mathbb{R}^n)$ Let $M:=M(\mathbb{R}^n)$ be the space of probability measures on $\mathbb{R}^n$ with respect to the Borel $\sigma$-algebra. Let $K\subset M$ be a compact convex subset. $K$ carries a natural topological structure, i.e. the weak topology induced by the bounded continuous functions. I have a continuous function $f:\mathbb{R}^n\to\mathbb{R}$ given such that 
$$|f(x_1,\dots,x_n)|\le K(1+\sum_{i=1}^n|x_i|)$$
for some constant $K$. I want to verify the continuity of $F:K\to\mathbb{R}$ on $K$, where
$$F(\mu):=\int_{\mathbb{R}^n}f d\mu$$
We know that every $\mu\in K$ has marginals $\rho_1,\dots,\rho_n$ with finite first moments. I have two questions:


*

*From the finite first moment it should follow: $\int_{\mathbb{R}^n\backslash[-a,a]^n}fd\mu\to 0$ uniformly in $\mu\in K$. Why is this the case? I've never heard the term finite first moment for a measure (just for r.v.)

*This should prove the continuity of $F$. How exactly? 

 A: The measure $\nu$ on the Borel subsets of the real line has a finite first moment if $$\int_{\mathbb R}|x|\mathrm d\nu<\infty.$$
It's equivalent to say that the real valued random variable $X$ has a finite first moment and $\mu_X$, the measure associated with $X$, has a finite moment. 


*

*Using the inequality about $f$, it's enough to show that 
$$\lim_{a\to \infty}\sup_{\mu\in K}\int_{\mathbb R^n\setminus[-a,a]^n}|x_j|\mathrm d\mu=0, \quad j\in\{1,\dots,n\}.$$
Using the fact that each element of $K$ has marginals $\rho_j$, we are reduced to prove that 
$$\lim_{a\to \infty}\int_{\mathbb R\setminus [-a,a]}|x|\mathrm d\rho_j=0.$$
It's the case by monotone convergence. 

*In this context, weak convergence is metrizable, so we only have to check sequential continuity. Take $\{\mu_k\}\subset K$ which converges in distribution to $\mu$ (an element of $K$). Fix $\varepsilon>0$ and $a$ such that $\sup_k\left|\int_{\mathbb R^n\setminus[-a,a]^n}f\mathrm d\mu_k\right|<\varepsilon$. Consider $\eta\colon\mathbb R^n\to \mathbb R$ a continuous function with compact support such that $\eta(x)=1$ if $x_j\in [-a,a]$. Then 
$$\left|\int_{\mathbb R^n}f(x)\mathrm d\nu_k-\int_{\mathbb R^n}f(x)\mathrm d\nu\right|\leqslant 2\varepsilon+\left|\int_{\mathbb R^n}f(x)\eta(x)\mathrm d\nu_k-\int_{\mathbb R^n}f(x)\eta(x)\mathrm d\nu\right|,$$
and we conclude taking $\limsup_{k\to \infty}$ (the last term vanishes because we now have a continuous bounded function). 
