3

Assumptions:

1) $h(x,\omega)$ is defined on $\forall\omega\in\Omega, \forall x\in R\setminus\{x_0\}$ where $\omega$ is a random variable.

2) Let $h(x_0,\omega)=\lim\limits_{x\rightarrow x_0}h(x,\omega)$ (pointwise convergence).

3) $h(x,\omega)$ is integrable $\forall x\in R$.

4) $\lim\limits_{x\rightarrow x_0}\int_\Omega h(x,\omega)dP(\omega)$ exists.

5) The family $h(x,\omega)$ of random variables for all $x$ sufficiently close to $x_0$ is uniformly integrable. That is to say, for each $\epsilon>0$, there exists $C(\epsilon)$ such that $\limsup\limits_{x\rightarrow x_0}\int_{\{|h(x,\omega)|>C(\epsilon)\}}|h(x,\omega)|dP(\omega)<\epsilon$.

Now is this true?

$\int_\Omega h(x_0,\omega)dP(\omega)=\lim\limits_{x\rightarrow x_0}\int_\Omega h(x,\omega)dP(\omega)$

I can't see how to apply Lebesgue dominated convergence theorem here because it's hard to find a function to dominate all $h(x,\omega)$. Any ideas will be appreciated.

GNUSupporter 8964民主女神 地下教會
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Zhenduo Cao
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1 Answers1

1

It suffices to show that for all sequence $\left(x_n\right)_{n\geqslant 1}$ converging to $x_0$, the convergence $$ \int_\Omega h(x_0,\omega)dP(\omega)=\lim\limits_{n\rightarrow +\infty}\int_\Omega h(x_n,\omega)dP(\omega). $$ Let us define the random variables $Y_n\colon \omega\mapsto h(x_n,\omega)$, $n\geqslant 0$. Item 6 of this answer applies.

Davide Giraudo
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