Properties of pointwise convergence on $\mathbb{R}$ translating to any measurable set $E$. Let $\{f_n\}$ be a sequence of nonnegative measurable function on $\mathbb{R}$ that converges pointwise on $\mathbb{R}$ to $f$ and $f$ is integrable over $\mathbb{R}$.  Show that if $\int_\mathbb{R} f = \lim\limits_{n \to \infty} \int_\mathbb{R} f_n$, then $\int_E f = \lim\limits_{n \to \infty} \int_E f_n$ for any measurable set $E$.
While I was reading this problem my first reaction was Fatou's lemma, however we have "$=$" and not "$\leq$".
I'm not sure what heavy machinery I could throw at this.
 A: We will show that actually, we have convergence in $L^1$. 


*

*We can assume that $\int_{\Bbb R}f=\int_{\Bbb R}f_n=1$ for all $n$. 

*Fix $\varepsilon>0$. Let $A$ of finite measure such that $\int_{\Bbb R\setminus A}f\leqslant \varepsilon$ and $\sup_{x\in A}|f_n(x)-f(x)|\to 0$. To get such a set, first fix $B$ of finite measure such that $\int_B fd\lambda\leqslant \varepsilon$. Let $\delta$ such that if $\mu(S)<\delta$ then $\int_S fd\lambda<\varepsilon$ (using again integrability of $f$). 
By Egoroff's theorem, let $A\subset B$ such that $\mu(B\setminus A)<\delta$ and there is convergence of $\{f_n\}$ uniformly on $A$. Then $\int_{\Bbb R\setminus A}fd\lambda\leqslant 2\varepsilon$.  

*We can find $n_1$ such that for $n\geqslant n_1$, $\int_{X\setminus A}f_nd\lambda\leqslant 2\varepsilon$. Indeed, we have 
$$\int_{X\setminus A}f_nd\lambda=1-\int_Af_nd\lambda\leqslant 1+\mu(A)\sup_{x\in A}|f_n(x)-f(x)|-\int_Afd\lambda\\ \leqslant\varepsilon+\mu(A)\sup_{x\in A}|f_n(x)-f(x)|.$$

*We conclude. For $n\geqslant n_1$, 
\begin{align}
\int_{\Bbb R}|f-f_n|d\lambda&=\int_A|f-f_n|d\lambda+\int_{\Bbb R\setminus A}|f-f_n|d\lambda\\
&\leqslant \mu(A)\sup_{x\in A}|f_n(x)-f(x)|+3\varepsilon.
\end{align} 

Some remarks:


*

*We didn't use the fact we were working on $\Bbb R$, we can replace it by any measure space. 

*We can replace pointwise convergence by almost sure convergence. 

