Bounded Almost Sure convergence implies convergence in pth mean

Question

A book I'm reading gave the following result. If $X_n \to X $ a.s. and $|X_n|^p \le Z$ for some random variable $Z$ with finite expectation, then we have convergence in $p$th mean. I was wondering, if we have an $X_n$ that is bounded, then for any $p$ we can find a number $N$ such that $|X_n|^p \le N$, and of course a fixed number has finite expectation. Does this mean we have convergence in $p$th mean for all $p$?

Answer 1

Yes, that is correct. The result you quoted is an application of Dominated convergence theorem. On a finite measure space (such as a probability space), a constant function is integrable and therefore can serve as dominating function in the theorem.

Thus: if $X_n\to X$ a.s. and there exists a number $M$ such that $|X_n|\le M$ for all $n$, then we have $E(|X_n|^p) \to E(|X|^p)$ for every $p$.

More generally, for every continuous function $\phi:\mathbb R\to\mathbb R$ we have $E(\phi(X_n))\to E(\phi(X))$, for the same reason.