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Let $\{f_n\}$ be a sequence of nonnegative lebesgue measurable functions on $\mathbb{R}$ and $f \in L_1(\mathbb{R}) $.
if $f_n \rightarrow f$ a.e on $\mathbb{R}$ and $$\int_\mathbb{R} f_n dm \rightarrow \int_\mathbb{R} f dm \quad \text{as} \quad n \rightarrow \infty $$ prove that $$\int_E f_n dm \rightarrow \int_E f dm \quad \text{as} \quad n \rightarrow \infty .$$ for all measurable subsets E

Here are a couple of ideas that I have.

$f_n \rightarrow f$ a.e on $\mathbb{R}$ implies

if we let $A:=\{x\in \mathbb{R}:|f_n (x)-f(x)|>\epsilon\}$. then $m(A)=0$

Let $\mathbb{R}=A\cup B$ where $B=\mathbb{R}\backslash A$

then $E \subset \mathbb{R} = (E\cap A) \cup ( E\cap B)$

$$lim_{n \rightarrow \infty}\int_\mathbb{R} f_n dm=lim_{n \rightarrow \infty}\big[\int_A f_n dm+\int_B f_n dm\big]=lim_{n \rightarrow \infty}\int_B f_n dm = \int_\mathbb{R} f_n dm$$ i.e since $m(A)=0$ , $\int_A f_n dm \overset{?}= 0$ $\forall n$

On the other hand $$lim_{n \rightarrow \infty}\int_E f_n dm=lim_{n \rightarrow \infty}\int_\mathbb{R} f_n \chi_E= lim_{n \rightarrow \infty}\int_\mathbb{R} f_n \chi_{(E\cap A) \cup ( E\cap B)}= lim_{n \rightarrow \infty}\big[\int_\mathbb{R} f_n \chi_{(E\cap A) }dm+\int_\mathbb{R} f_n\chi_{(E\cap B)}dm\big]$$

But $(E\cap A) \subset A $ so $m((E\cap A))=0$ $\Rightarrow \int_\mathbb{R} f_n \chi_{(E\cap A)}=0$ $\quad$ Hence $$=lim_{n \rightarrow \infty}\int_\mathbb{R} f_n\chi_{(E\cap B)}$$

If a sequence $a_n$ converges to $a$, then every subsequence $a_{n_k}$ of $a_n$ converges to $a$ [is this correct correct?] So $$lim_{n \rightarrow \infty}\int_\mathbb{R} f_n\chi_{(E\cap B)}=\int_\mathbb{R} f dm$$.

can someone Kindly verify or correct it, or produce a proof? Thank you.

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  • $\begingroup$ See this. $\endgroup$ – David Mitra Dec 30 '16 at 7:20
  • $\begingroup$ You should add, in the end of the statement of the theorem, "for all measurable subsets $E\subset R$". $\endgroup$ – kobe Dec 30 '16 at 7:20
  • $\begingroup$ okay, I just did. I understand your proof below. but does mine make sense? $\endgroup$ – J. Kyei Dec 30 '16 at 7:33
  • $\begingroup$ Duplicate of math.stackexchange.com/q/678282 $\endgroup$ – saz Dec 30 '16 at 7:33
  • $\begingroup$ No, your proof is not correct since $m(A)$ does not equal $0$. $\endgroup$ – saz Dec 30 '16 at 7:36
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By Fatou's lemma,

$$\int_{\Bbb R\setminus E} f\, dm \le \liminf\int_{\Bbb R\setminus E} f_n\,dm$$

or $$\int_{\Bbb R} f\, dm - \int_E f\, dm \le \liminf \left(\int_{\Bbb R} f_n\, dm - \int_E f_n\, dm\right)$$

Using the condition $\int_{\Bbb R} f_n \, dm \to \int_{\Bbb R} f\, dm$, the inequality becomes

$$\int_{\Bbb R} f\, dm - \int_E f\, dm \le \int_{\Bbb R} f\, dm - \limsup \int_E f_n\, dm$$

Since $\int_{\Bbb R} f\, dm$ is finite, we deduce

$$\int_E f\, dm \ge \limsup \int_E f_n\, dm$$

On the other hand, Fatou's lemma also gives

$$\liminf \int_E f_n\, dm \ge \int_E f\, dm$$

Therefore $\int_E f_n\, dm \to \int_E f\, dm$.

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