Let $\{f_n\}\subseteq L^p(\Omega)$, $\{|f_n|^p\}$ be uniformly integrable, and $f_n\to f$ in measure. Show $f\in L^p$ and $f_n\to f$ in $L^p$ Let $\{f_n\}\subseteq L^p(\Omega,\mathcal{F},\mu), p\in(0,\infty)$, where $\mu(\Omega)<\infty$, $\{|f_n|^p\}_{n\geq 1}$ is UI, and $f_n\to f$ in measure. Show that $f\in L^p(\mu)$ and $f_n\to f$ in $L^p(\mu).$
So, I had a few ideas on this one, but they don't seem to be panning out as neatly as I had hoped.
At first I wanted to say that since $\{|f_n|^p\}$ is uniformly integrable, then $\sup_{n}\int|f_n|^p<\infty, $ and from here apply Fatou's lemma. However, I realized that fatou's lemma requires point wise a.e. convergence, and all we have is convergence in measure. So my first question is; is it possible to salvage this approach? If not, what else might one do to establish this claim?
Regarding $f_n\to f$ in $L^p,$ I feel like I have a rough sketch of an idea, but I'm pretty sure it's misguided, so I could use some guidance here as well. My basic approach for this part was the following;
let $\epsilon^{1/p} >0$.
Let $A=\{x:|f_n(x)-f(x)|<\epsilon^{1/p}\} = \{x:|f_n(x)-f(x)|^p<\epsilon\}$ and $B=\{x:|f_n(x)-f(x)|\geq\epsilon^{1/p}\}=\{x:|f_n(x)-f(x)|^p\geq\epsilon\}$.
Then we have $$\int_\Omega|f_n-f|^p = \int_A|f_n-f|^p+\int_B|f_n-f|^p \leq \epsilon\mu(A)+\int_B|f_n-f|^p$$
Since $f_n\to f$ in measure, I can drive $\mu(B)$ to $0$ by letting $n\to\infty$. I'd like to be able to use this to get $\int_B|f_n-f|^p\to0$, but it's not clear to me that this should be true; perhaps $|f_n-f|$ don't converge fast enough.
Any thoughts would be greatly appreciated.
Thanks in advance.
 A: Step 1: We go to show that $f\in L^{p}.$ Since $\mu(\Omega)<\infty$,
by Vitali Theorem, $f_{n}\rightarrow f$ in measure $\Rightarrow$
there exists a subsequence $(f_{n_{k}})_{k}$ such that $f_{n_{k}}\rightarrow f$
a.e.
Since $\{|f_{n}|^{p}\}$ is uniformly integrable, $\sup_{n}\int|f_{n}|^{p}\,d\mu<\infty$.
By Fatou lemma, we have
\begin{eqnarray*}
\int|f|^{p}\,d\mu & \leq & \liminf_{k}\int|f_{n_{k}}|^{p}\,d\mu\\
 & \leq & \sup_{n}\int|f_{n}|^{p}\,d\mu\\
 & < & \infty.
\end{eqnarray*}
Observe that $|f_{n}-f|^{p}\leq2^{p}\left(|f_{n}|^{p}+|f|^{p}\right),$
so $\{|f_{n}-f|^{p}\mid n\in\mathbb{N}\}$ is uniformly integrable,
a fact that will be used at later time.
Step 2: $f_{n}\rightarrow f$ in $L^{p}$. Denote $\alpha_{n}=\int|f_{n}-f|^{p}$.
To show that $\alpha_{n}\rightarrow0$, we invoke that fact: $\alpha_{n}\rightarrow0$
iff every subsequence $(\alpha_{n_{k}})_{k}$ of $(\alpha_{n})$ has
further a subsequence that converges to $0$. Let $(\alpha_{n_{k}})_{k}$
be an arbitrarily given subsequence of $(\alpha_{n})$. Denote $f_{n_{k}}=g_{k}$.
Clearly $g_{k}\rightarrow f$ in measure. By Vitali Theorem again,
there exists a subsequence $(g_{k_{l}})_{l}$ such that $g_{k_{l}}\rightarrow f$
a.e. Let $\varepsilon>0$ be given. By uniformly integrability, there
exists $c>0$ such that
$$
\sup_{l}\int_{[|g_{k_{l}}-f|^{p}\geq c]}|g_{k_{l}}-f|^{p}<\varepsilon.
$$
Denote $A_{l}=[|g_{k_{l}}-f|^{p}\geq c]$. Observe that $1_{A_{l}^{c}}|g_{k_{l}}-f|^{p}\rightarrow0$
a.e. as $l\rightarrow\infty$ and $1_{A_{l}^{c}}|g_{k_{l}}-f|^{p}\leq c^{p}.$
By Dominated Convergence Theorem, $\int1_{A_{l}^{c}}|g_{k_{l}}-f|^{p}\rightarrow0$
as $l\rightarrow\infty.$ Choose $L$ such that $\int1_{A_{l}^{c}}|g_{k_{l}}-f|^{p}<\varepsilon$
whenever $l\geq L$. Finally, let $l\geq L$, then we have that
\begin{eqnarray*}
 &  & \int|g_{k_{l}}-f|^{p}\\
 & = & \int_{A_{l}}|g_{k_{l}}-f|^{p}+\int_{A_{l}^{c}}|g_{k_{l}}-f|^{p}\\
 & < & 2\varepsilon.
\end{eqnarray*}
This show that $\alpha_{n_{k_{l}}}\rightarrow0$ as $l\rightarrow0$.
That is, $f_{n}\rightarrow f$ in $L^{p}$.
A: Hints:
Step I:
Convergence in measure implies a.e. convergence for  a subsequence. If every subsequence of $f_n$  has a further subsequence which tends to $f$ in $L^{p} $ then $f_n \to f$ in $L^{p}$. From these conclude that there is no loss of generality in assuming that $f_n \to f$ a.e.
STEP II
Next use Fatou's Lemma  to conclude that if $(f_n)$ is UI and $f_n \to f$ a.e. then $(f_n-f)$ is UI. Conclude now that the proof can be reduced to the case when $f=0$.
FINAL STEP
Now $\int|f_n|^{p} =\int_A |f_n|^{p}+\int_B |f_n|^{p}$ where $A=\{x: |f_n(x)| <M\}$ and $B=\{x: |f_n(x)| \geq M\}$. Choose $M$ such that the second term is less than $\epsilon$. Use DCT for the  first term.
