Exercise: Limits and Probability Measure Let $\mu$ be a probability measure on $X$ (closed but unbounded), so that $\int_X \mu(dx) = 1$.
Let functions $f_i:X \rightarrow \mathbb{R}_{\geq 0}$,  $i = 1,2,...$, be Uniformly Integrable.
Prove that
$$ \limsup_{i\rightarrow \infty} \int_X  f_i(x) \mu(dx) \ - \  \lim_{n \rightarrow \infty} \limsup_{i \rightarrow \infty} \int_{X_n} f_i(x) \mu(dx) \ \ = \ \ 0   $$
for some sequence of compact sets $\{X_n\}_{n=1}^{\infty}$ converging to $X$ ($\lim_{n \rightarrow \infty} X_n = X$).
 A: We have $\int_{X_n}f_i(x)\mu(dx)=\int_Xf_i(x)\mu(dx)-\int_{X\setminus X_n}f_i(x)\mu(dx)$. Fix $\varepsilon>0$; thanks to the uniform integrability we can find $R>0$ such that for all $i$, $\int_{f_i\geq R}f_i(x)\mu(dx)\leq \varepsilon$. We get 
$$\int_{X\setminus X_n}f_i(x)\mu(dx)=\int_{X\setminus X_n\cap\{f_i\geq R\}}f_i(x)\mu(dx)+\int_{X\setminus X_n\cap\{f_i<R\}}f_i(x)\mu(dx)\leq \varepsilon+R\mu(X\setminus X_n)$$
so 
$$\int_{X_n}f_i(x)\mu(dx)\geq \int_{X}f_i(x)\mu(dx)-\varepsilon-R\mu(X\setminus X_n)$$
and taking the $\limsup$:
$$\limsup_i\int_{X_n}f_i(x)\mu(dx)\geq \limsup_i\int_{X}f_i(x)\mu(dx)-\varepsilon-R\mu(X\setminus X_n),$$
so for all $n$ and all $\varepsilon$
$$0\leq \limsup_i\int_{X}f_i(x)\mu(dx)-\limsup_i\int_{X_n}f_i(x)\mu(dx)\leq \varepsilon+R\mu(X\setminus X_n).$$
Letting $n\to +\infty$ it gives that for all $\varepsilon>0$:
$$0\leq \limsup_i\int_{X}f_i(x)\mu(dx)-\limsup_i\int_{X_n}f_i(x)\mu(dx)\leq \varepsilon,$$
so 
$$\lim_{n\to\infty}\left(\limsup_i\int_{X}f_i(x)\mu(dx)-\limsup_i\int_{X_n}f_i(x)\mu(dx)\right)=0.$$
