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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 $X$ ($\lim_{n \rightarrow \infty} X_n = X$).

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A $\lim_{n\to\infty}$ is missing, otherwise if $f_i=f$ integrable the result may be not true. – Davide Giraudo Mar 7 '12 at 11:13
up vote 1 down vote accepted

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 $$\lim_{n\to\infty}\left(\limsup_i\int_{X}f_i(x)\mu(dx)-\limsup_i\int_{X_n}f_i(x)\mu(dx)\right)=0.$$

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Looking at the result, say if we can also claim the following (again $\{f_i\}$ UI, $X_n \rightarrow X$. For any $\epsilon > 0$, the fact that $$ \int_{X_n} f_i(x) \mu(dx) \leq \epsilon \ \text{ for any compact set } X_n $$ implies $$ \int_{X} f_i(x) \mu(dx) \leq \epsilon $$ – Adam Mar 7 '12 at 18:02

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