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 ?

  • 2
    $\begingroup$ 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 Jan 15 '17 at 21:48

$\{X_n\}$ 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.


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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