# Convergence in measure and bounded $L^p$ norm implies convergence in $L^p$

Let $1<p<\infty$, and let $\mu$ be a finite measure. Prove that if a sequence $\{f_n\}_{n=1}^{\infty} \subset L^p(\mu)$ satisfies
1. $f_n \rightarrow f$ in measure and
2. $\sup_{n \in \mathbb{N}} \|f_n\|_p < +\infty$ where $\|f_n\|_p = (\int |f_n|^p \,\mathrm{d}\mu)^{\frac{1}{p}}$ is the $L^p$ norm.
then $\int f_n \rightarrow \int f$ in $L^p(\mu)$.

I know that convergence in measure implies that there exists a subsequence converges a.e.,but how does it relate to convergence in $L^p(\mu)$.

Update:
Thanks for @carmichael561, the conclusion of the question was incorrect. It should be the convergence of integration.

The two conditions stated are not enough to imply convergence in $L^p$. Indeed, if $X=[0,1]$ and $\mu$ is Lebesgue measure, then the sequence of functions $$f_n=n1_{[0,\frac{1}{n^2}]}$$ converges to zero in measure and is bounded in $L^2$ because $$\int_X|f_n|^2\;d\mu=n^2\cdot\frac{1}{n^2}=1$$ for all $n$. But $f_n\not\to 0$ in $L^2$.
• It seems we need a stronger condition instead of $\sup_{n}\|f_n\|_p<\infty$. Perhaps $(f_n)$ is uniformly integrable? – ntt Apr 11 '17 at 19:27
• Yes, if $\{|f_n|^p\}$ is uniformly integrable then things should be ok. – carmichael561 Apr 11 '17 at 19:28