Show that $\int\limits_{E}f=\lim\limits_{n\to\infty}\int\limits_E f_n$ For any measurable set $E$.

Question

Let $f_n:R\rightarrow R$ be a sequence of nonnegative measurable functions that converges pointwise to an integrable function $f$. Show that if $$\int\limits_{R}f=\lim\limits_{n\to\infty}\int\limits_R f_n$$ then $\int\limits_{E}f=\lim\limits_{n\to\infty}\int\limits_E f_n$ For any measurable set $E$.
Hint: Use the Fatou's Lemma twice.
Also Show that the implication is no longer true if we do not assume that the functions $\{f_n\}$ are non negative.

From the hint I was trying to apply the fatous Lemma for the set $E$.
That is $\int\limits_{E}f\leq\liminf\limits_{n\to\infty}\int\limits_E f_n$
Also since $f_n$ is nonnegative we will have $\int\limits_{E}f_n\le\int\limits_{R}f_n$
But that didn't help me to obtain the solution.

Also I appreciate a counter example for the negative case too

Answer 1

Apply Fatou's Lemma to $f-(f-f_n)^{+}$. We get $\int f \leq \lim \inf [\int f -\int (f-f_n)^{+}]$. This can be written as $\lim \sup \int (f-f_n)^{+} \leq 0$. Thus, $\int (f-f_n)^{+} \to 0$. It is also given that $\int [f-f_n] \to 0$. Combine these two to get $\int (f-f_n)^{-} \to 0$. Then conclude that $\int |f-f_n| \to 0$. This implies $|\int_E f -\int_E f_n| \leq \int |f-f_n| \to 0$.

For the second part take $f_n=nI_{(0,\frac 1 n)}-nI_{(-\frac 1 n,0)}$ and $f=0$. Take $E=(0,\infty)$.