First, please note the difference between this and the question asked here: $L^p$ Dominated Convergence Theorem
Let $f_n$ be a sequence of measurable functions on $E$ and $p\ge 1$. Suppose $f_n\to f$ in measure on $E$ and there exists an $L^p$-integrable function $g$ such that $|f_n|\leq g$. Prove that:
(1) $f_n$ and $f$ are $L^p$ integrable.
(2) $f_n\to f$ on $E$ in $L^p$.
Here is my proof:
I first prove that convergence in measure implies that there exists a subsequence $f_{n_k}$ converging to $f$ pointwise a.e.
$\mu\{|f_n-f|\ge\frac{1}{k}\}\leq\frac{1}{2^k}$. Define $E_k=\{|f_{n_k}-f|>\frac{1}{k}\}$ and $H_m=\bigcup_{k=m}^{\infty}E_k$.
It is clear that $\mu(E_k)<\frac{1}{2^k}$ and $\mu(H_m)\leq\sum_{k=m}^{\infty}\frac{1}{2^k}=\frac{1}{2^{m-1}}$.
Let $Z=\bigcap_{m=1}^{\infty}H_m$, then $\mu(Z)\leq\mu(H_m)\leq\frac{1}{2^{m-1}}\to 0$ as $m\to\infty$.
If $x\notin Z$, $x\notin H_m$, $x\notin E_k$ for all $k\ge m$, i.e. $|f_{n_k}-f|\leq\frac{1}{k}$ for all $k\ge m$. That means $f_{n_k}\to f$ for all $x\notin Z$ and $\mu(Z)=0$, thus pointwise convergence is proved.
To prove (1), since $|f_n|\leq g$ and $p\ge 1$,
$\int_{E}|f_n|^pd\mu\leq\int_{E}g^pd\mu<\infty$, hence $f_n$ is $L^p$-integrable.
By Fatou's lemma, $0\leq\int_{E}f^pd\mu\leq\liminf_{n\to\infty}\int_{E}f_n^{p}d\mu<\infty$, hence $f$ is $L^p$-integrable.
To prove (2), since pointwise convergence a.e. is proved, $|f_n|\leq g$ a.e. implies $|f|\leq g$. Since $(f_n-f)^p\to 0$ a.e., and $|f_n-f|^p\leq 2^p(|f|^p+|f_n|^p)\leq 2^{p+1}|g|^p$. So by Dominated Convergence Theorem for $L^1$, $\int_{E}|f_n-f|^p\to 0$.
Can anyone help me to check my proof? My question is whether the subsequence converges pointwise implies the sequence converges pointwise, and whether I can use DCT. I doubt this step.
Thanks!
