# If $\int f_nd\mu \to \int f d \mu <\infty$ then for all measurable $E$, $\int _E f_nd\mu \to \int _E f d\mu$

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!

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.