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.

  • $\begingroup$ See this. $\endgroup$ 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

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

  • $\begingroup$ How do we know that $\int_{\Bbb R\setminus E}f_n \ dm \to \int_{\Bbb R\setminus E}f \ dm$ so that we may apply Fatou's lemma? $\endgroup$
    – Addem
    Oct 3 at 0:28
  • $\begingroup$ @Addem it is assumed that $\{f_n\}$ is a sequence of nonnegative measurable functions converging pointwise a.e. to $f$. So Fatou's lemma may be applied to obtain $\int_{\Bbb R\setminus E} f\, dm \le \liminf\int_{\Bbb R\setminus E} f_n\ , dm$. $\endgroup$
    – kobe
    Oct 3 at 3:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.