real analysis - Convergence in measure and bounded $L^p$ norm implies convergence in $L^p$

Let $11. $f_n \rightarrow f$ 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.

Answer

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 $L^2$.