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This is problem 4T in Bartle's The elements of integration and Lebesgue measure.

Suppose $f_n$ are non-negative measurable function such that $(f_n)$ converges to $f$, and that $$\int fd\mu = \lim \int f_n d\mu<\infty.$$ Prove that for all measurable set $E$, we have $$\int_E fd\mu = \lim \int_E f_n d\mu.$$

I found a solution here.

The solution used the reverse Fatou's lemma, which needs the hypothesis of dominated boundedness, that is there exists a measurable function $g$ such that $f_n\le g$ for all $n$ and $\displaystyle \int g <\infty$. I cannot see how can we get this function $g$.

Thank you very much.

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  • $\begingroup$ such function exists by convergence. $\endgroup$ – Henricus V. Feb 14 '16 at 17:04
  • $\begingroup$ Hint: It's enough to show that $\lim_n\int|f-f_n|\,d\mu=0$. To that end write $|f-f_n|$ as $2(f-f_n)^+-f+f_n$, and observe that $(f-f_n)^+\le f$ which is integrable. $\endgroup$ – John Dawkins Feb 14 '16 at 17:10
  • $\begingroup$ @JohnDawkins If we find $g$ integrable such as $|f-f_n|\le g$ for any $n$ large enaugh, we can indeed conclude using the dominated convergence theorem. I understand that $(f-f_n)^+\le f$. But we need to dominate $|f-f_n|=(f-f_n)^++(f-f_n)^-$, and I couldn't do it. Written in the form you gave, we get $|f-f_n|\le 2f-f+f_n =f+f_n$. I also tried to dominate $(f-f_n)^-$ and finding that is it smaller than $f_n$, but we get to the same point. Would you please help me? $\endgroup$ – Scientifica Nov 14 '16 at 12:46
  • $\begingroup$ @HenryW. Would you please give me a hint on how to find such function? I explained what I tried in the comment above. $\endgroup$ – Scientifica Nov 15 '16 at 14:38
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The proof you mention seems wrong to me, since it uses the inequality $$ \limsup\int_Ef_n\le\int_E\limsup f_n $$ You may complete the proof applying Fatou's lemma to $E^c=\mathbb{R}\setminus E$. $$ \begin{multline} \int_Ef=\int_{\mathbb R}f-\int_{E^c}f\ge\int_{\mathbb R}f-\liminf\int_{E^c}f_n\\=\int_{\mathbb R}f-\liminf\Bigl(\int_{\mathbb R}f_n-\int_{E}f_n\Bigr)=\limsup\int_Ef. \end{multline}$$

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  • $\begingroup$ I mean the inequality $\lim \sup \int f_n \le \int \lim \sup f_n$. $\endgroup$ – Tien Kha Pham Feb 14 '16 at 17:24
  • $\begingroup$ Yes, I realized that. I have edited the answer. $\endgroup$ – Julián Aguirre Feb 14 '16 at 17:32
  • $\begingroup$ This an old(ish) answer, but would you mind clearing up: a) why the use of the inequality you mention is wrong; and b) whether you meant to write $f_n$ in place of $f$ at the very end of the answer? Thank you! $\endgroup$ – mlaci Oct 10 '17 at 3:52
  • $\begingroup$ The inequality is incorrect; take $f_n$ the characteristic function of $[n,n+1]$. And yes, I meant $f_n$. $\endgroup$ – Julián Aguirre Oct 12 '17 at 13:38
  • $\begingroup$ You're saying that, in that case, $lim\ sup\ 1 \leq \int_{E} lim\ sup\ f_n = \int_{E} 0$, correct? But wouldn't this example function break the requirement in the hypothesis that $\int fd\mu = \lim \int f_n d\mu<\infty$, since $\int 0 d\mu \neq lim\ 1$? Thanks again! $\endgroup$ – mlaci Oct 15 '17 at 20:31

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