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Let $P$ a probability and $X_n$ a random variable that is uniformly bounded, i.e. $\sup_n X_n<\infty $. We suppose $X_n\to X$. Do we have that $$\lim_{n\to \infty }\int_{\Omega } X_n dP=\int_{\Omega } \lim_{n\to \infty }X_n dP\ \ ?$$

To me it's almost bounded convergence theorem, but the bounded convergence theorem that I know is only valid on set of finite measure. So, $\Omega $ may be not bounded, but since $P(\Omega )=1$ maybe it also works.

I recall the bounded convergence theorem that I know :

If $f_n(x)\to f(x)$ a.e. $(f_n)$ is uniformly bounded and $m(E)$ is finite, then $$\lim_{n\to \infty }\int_E f_n=\int_E f.$$

Here it's a little bit different. But I have the intuition that it's almost the same. Do you have an explanation ?

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    $\begingroup$ It is the same. $P$ is taking the role of $m$ in your statement of the Bounded Convergence Theorem, and, as you said, $P(\Omega)=1<\infty$. $\endgroup$
    – Hayden
    Commented Jan 15, 2017 at 21:48

1 Answer 1

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$\{X_n\}$ is a sequence of measurable functions, i.e. $\{f_n\}$ in the theorem statement. Since we suppose $X_n \to X$ pointwise, we also have $X_n \to X$ a.e. Since $P(\Omega) < \infty$, by the theorem you've quoted, we have \begin{align*} \lim_{n \to \infty} \int_{\Omega} X_n dP \to \int_{\Omega} \lim_{n \to \infty}X_n dP = \int_{\Omega} X dP \end{align*} So your conclusion is correct.

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