Convergence pointwise of positive functions and convergence in $L^{1}(X)$ implies convergence in $L^{1}(\text{measble subset})$.

Question

Let $f_{n}, f \geq 0$ be measurable on a measure space $(X, \mathcal{M}, \mu)$ with $f_{n} \rightarrow f$ pointwise, and $\int f_{n} \rightarrow \int f < \infty$. Show that $\int_{E} f_{n} \rightarrow \int_{E} f$ for all measurable $E$.

What I've Tried: I tried using Egorov's theorem and found a (false) solution, only to realize later that $\mu$ isn't necessarily finite. Now I'm stuck. I want $\int_{E} |f_{n} - f| d\mu$ to be small. It's at most $\int |f_{n} - f| d\mu$, but that's maybe not so helpful. I feel a little silly I can't solve this seemingly easy problem... a hint would be appreciated.

Answer 1

As User8128 has pointed out, write $f = \lim \left( \frac{1}{2}(f_{n} + f) - \frac{1}{2}|f_{n} - f|\right)$. Then Fatou gives us

\begin{align} \int_{E} f &\leq \liminf \int_{E} \left( \frac{1}{2}(f_{n} + f) - \frac{1}{2}|f_{n} - f|\right) \\ &= \frac{1}{2}\liminf \int_{E} f_{n} + \frac{1}{2}\int_{E} f - \frac{1}{2}\limsup\int_{E} |f_{n} - f| \\ &= \frac{1}{2}\int_{E} f + \frac{1}{2}\int_{E} f - \frac{1}{2}\limsup\int_{E} |f_{n} - f| \\ &\implies 0 = \limsup\int_{E} |f_{n} - f|, \end{align}

whence $\lim \int_{E} |f_{n} - f| = 0$.

Answer 2

Hint: Fatou's lemma and $f = \lim_{n\to \infty } \left( \frac 1 2( f_n + f) - \frac 1 2\lvert f_n - f\rvert\right).$