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

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

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 $$E=(0,\infty)$$.