# Proving that if $X_n$ is uniformly bounded, and if $X = \lim_{n}X_n$ with probability 1, then $E[X] = \lim_{n} E[X_n]$?

In Billingsley's Probability and Measure Book, he has a result otherwise known as the Bounded Convergence Theorem: Normally we would use the Lebesgue integral in lieu of the expectation and also use the almost everywhere condition. However, he uses convergence in probability instead. What confuses me is how he is able to put a bound on $\mid X \mid$. Is that a deterministic bound or is it under the assumption that:

$$P(\mid X \mid \leq K) = 1$$

Thanks.

• Because $X=lim_n X_n$ a.s., $X$ is bounded a.s., i.e. $P(|X|\leq K)=1$. – Danielsen Nov 29 '15 at 3:53
• I think I remember a result in analysis saying that if a sequence is convergence it is bounded. If true, not sure if it applies – BCLC Nov 29 '15 at 4:10
• @user136503 $X=\lim_n X_n$ a.s. means exist a set $\Omega^*\subset\Omega$ with $P(\Omega^*)=1$, such that for every $\omega\in\Omega^*$ $X(\omega)=\lim_n X_n(\omega)$. So $-K\leq X(\omega)\leq K$ for every $\omega\in\Omega^*$. Then we have $P(|X|\leq K)=1$. – Danielsen Nov 29 '15 at 4:21
• @user136503 By fixing $\omega$, it is just the limit of sequence of numbers. – Danielsen Nov 29 '15 at 4:55
• @user136503 $P(|X|\leq K)=1$ and $|X|\leq K$ a.s. are the same thing. You can not get $|X(\omega)|\leq K$ for all $\omega\in\Omega$. – Danielsen Nov 29 '15 at 5:00