Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
1  
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

1 Answer 1

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

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.