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Let $\left\{f_{n}\right\}$ be a sequence of non-negtive measurable functions on $(-\infty,\infty)$,such that $f_{n}\rightarrow f$ a.e,and suppose that $$\int f_{n}\rightarrow \int f$$ Prove that for each measurable set $E$,$$\int_{E}f_{n}\rightarrow \int _{E} f$$.

Is the statement correct without the non-negativity condition ? Either prove or give a counterexample.

Since the sequence of functions are nonnegative,so from fatou lemma,we have $\int_{E}f\leq\liminf\int_{E}f_{n}$,but I don't know how to derive the inverse inequality from $\int f_{n} \rightarrow \int f$ and construct a counter example without non-negativity condition.

Jack
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The non-negativity condition is indeed necessary. Let $f_n(x)=\begin{cases}-\frac{1}{n}&x\in[-2n,-n]\\\frac{1}{n}&x\in[n,2n]\end{cases}$. Clearly $f_n\rightarrow0$ and $\int f_n=0$ but letting $E=[0,\infty)$, $\int_Ef_n=1$ for all $n$ so $\int_Ef_n\not\rightarrow0$.

Edit: I realized I never actually provided a full solution to this so here goes: A famous result says that for a sequence of $L^p$ functions, convergence of norms (that is, $||f_n||_p\rightarrow||f||_p$)and pointwise convergence almost everywhere implies convergence in norm. See here for instance. Because all of our functions are positive, $||f_n||_1=\int f_n$ so that the condition $\int f_n\rightarrow \int f$ is equivalent to convergence of norms. In particular $f_n\rightarrow f$ weakly. The function $\chi_E$ is in $L_\infty=L_1^*$. It follows from weak convergence that $\int_E f_n=\int \chi_Ef_n\rightarrow\int\chi_Ef=\int_Ef$.

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