If $\mu(X) < \infty$, $f_n \to f$ a.e., and $\int f_n^2 \leq C$, then $f_n \to f$ in $L^1$ I should be able to get this problem...I'm studying for a qualifying exam and the question is to show that if $\mu(X) < \infty$, $f_n \to f$ a.e., and $\int f_n^2 \leq C$, then $f_n \to f$ in $L^1$.  
I know $f \in L^1$ since this follows if $f^2 \in L^1$ since $\mu(X) < \infty$; then $\int |f|^2 = \int \liminf |f|^2 \leq \liminf \int |f_n|^2 \leq C$.  I then use Egoroff to reduce to showing $\int |f - f_n|$ can be made small on a set as small as I please.  But I cannot finish the argument.  The square is messing me up.  Can anyone provide the one-two punch that I need here?  
 A: From what you have you can do the following. Let $B$ be the set where $f_n\to f$ uniformly on $B$, and $\mu(B^c) <\epsilon$. You have
$$
 \int_X |f-f_n| = \int_B|f-f_n| + \int_{B^c}|f-f_n|.
$$
The first integral limits to $0$ by the uniform convergence, hence can be made smaller than $\epsilon$ for large enough $n$. For the second integral, apply the triangle inequality and Cauchy-Schwarz:
$$
 \int_{B^c} |f-f_n| \leq \int_{B^c}|f| +\int_{B^c} |f_n| \leq \mu(B^c)^{1/2}( ||f||_2 + ||f_n||_2)\leq 2\sqrt{C} \sqrt{\epsilon},
$$
which should finish it.
A: Alternative Solution: 
We need to show that $$\int_X |f_n-f| \ d\mu \to 0$$ as $n\to\infty$. We know that $\mu(X)<\infty$. Fix $\epsilon>0$. Let us write the above integral as sum of two components as follows: $$\int_X |f_n-f|\mathbb{I}\{|f_n-f|\le M\} \ d\mu+|f_n-f|\mathbb{I}\{|f_n-f|>M\} \ d\mu \ \ \ \ (1)$$. Now, consider the 2nd part: By Cauchy-Schwarz's inequality,followed by Chebychev's Inequality, we have : $$\int_X|f_n-f|\mathbb{I}\{|f_n-f|>M\} \ d\mu\le\left(\int_X (f_n-f)^2\mu\{x:|f_n-f|>M\}\right)^{1/2}\le\frac{\int_X |f_n-f|^2 \ d\mu}{M}$$. Since, $\int_{X}|f_n-f|^2\le\int_X 2(f_n^2+f^2)\le 2C$, choose $M$ large enogh so that 2nd part of (1) becomes less that $\epsilon$. Now, for the 1st part of (1), since $|f_n-f|\le M$ and $|f_n-f|\to 0$ and $\mu(X)<\infty$, by DCT, it converges to $0$ as $n\to\infty$. Therefore, we have, $\limsup\int_X |f_n-f|\le\epsilon$ for all $\epsilon>0$, which forces that $f_n\to f$ in $L^1$. $\square$
A: It turns out that $\int_{X} |f_{n}|^{2} \, d\mu \leq C$ implies $(f_{n})$ is uniformly integrable.  Convergence then follows from Vitali's Theorem.
To see that $(f_{n})$ is uniformly integrable, suppose $A$ is a measurable set.  For each $n$, we have by Holder's inequality,
\begin{align*}
\int_{A} |f_{n}| \, d\mu &= \int_{X} |f_{n}| \chi_{A} \, d\mu \\
&\leq \|f_{n}\|_{2} \|\chi_{A}\|_{2} \leq \sqrt{C \mu(A)}
\end{align*}
Therefore, $(f_{n})$ is uniformly integrable.
