Let $f_n:X\to [0,\infty]$ be measurable functions such that $f_n\to f$ pointwise. Then prove that if $\int f_nd\mu \to \int f d \mu <\infty$ then for all measurable sets $E$ we have $$\int _E f_nd\mu \to \int _E f d\mu$$
My attempt: I tried to prove something like $\lim \sup \int_E f_nd\mu\leq \int_Ef_nd\mu\leq \lim\inf \int_Ef_nd\mu$
Second inequality is immediate by Fatou's Lema. For finding the first I tried to find some integrable $g$ such that $|f_n|\leq g$ and use Fatou-Lebesgue ($\int\lim\sup f_nd\mu\geq\lim\sup\int f_nd\mu$)… but if I found some $g$ like that I can just use LDCT so I wouldn't need the first argument at all… and I couldn't think another way.
I don't think I'll find a $g$ like that :/
I'd appreciate some hint that just give another nice way to proceed!
Thanks in advance!
Best Answer
Since $\int fd\mu < \infty$, we can wlog assume $f$ is of $L^1$
Since $\int f_n d\mu \to \int f d\mu$ and $\int fd\mu < \infty$, for sufficiently large $n$'s we have $\int f_n d\mu <\infty$.
Hence, wlog, assume that all $f_n$ and $f$ are of $L^1$.
Now, by Scheffé's lemma, we have $\int |f_n - f| d\mu \to 0$.
Since $\int_E |f_n -f| d\mu \leq \int |f_n - f| d\mu$ for each measuralbe $E$, we are done.