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

Let $f_n$, $f$ be nonnegative measurable functions on $\mathbb{R}$ such that $f_n\to\ f$ for every real number (pointwise convergence). Suppose that $lim_{n\to \infty}\int_\mathbb{R}f_n=\int_\mathbb{R}f$.

Show that if $\int_\mathbb{R}f<\infty$, then $lim_{n\to \infty}\int_Ef_n=\int_Ef$ for every measurable set $E\subset\mathbb{R}$

Important theorems that I'm allowed to use: Theorem of Monotone Convergence, Fatou's Lemma, and Theorem of Dominated Convergence. I've been attempting to use dominated convergence by trying to define a sequence of functions bounded by f on $E$, but I haven't been able to come up with such a sequence.

Any hints would be greatly appreciated.

salvador
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1 Answers1

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For any measurable set $E \subset \mathbb{R},$ we have $\displaystyle \int_Ef \leqslant \int_{\mathbb{R}}f < \infty$ and $\displaystyle \int_{E}f_n \leqslant \int_{\mathbb{R}}f_n \to \int_{\mathbb{R}}f$.

Hence, for sufficiently large $n$, we have

$$\displaystyle \int_Ef_n < \infty.$$

Using Fatou's Lemma,

$$\int_E f = \int_E \lim f_n=\int_E \liminf f_n \leqslant \liminf \int_E f_n$$

and reverse Fatou's Lemma,

$$\liminf \int_E f_n \leqslant \limsup\int_E f_n \leqslant \int_E \limsup f_n = \int_E \lim f_n = \int_Ef.$$

Hence,

$$\tag{*}\int_Ef \leqslant \liminf \int_E f_n \leqslant \limsup\int_E f_n \leqslant\int_Ef,$$

and

$$\liminf \int_E f_n = \limsup\int_E f_n=\lim \int_E f_n = \int_Ef.$$

Update:

We can avoid the argument based on reverse Fatou as follows.

Note that since $f_n \to f$ and $\int_{\mathbb{R}} f_n \to \int_{\mathbb{R}} f$ we have

$$\begin{align}\limsup \int_E f_n &= -\liminf \left(-\int_E f_n \right) \\&= -\liminf \left(\int_{\mathbb{R}\setminus E} f_n-\int_{\mathbb{R}} f_n \right) \\ &= -\liminf \int _{\mathbb{R}\setminus E} f_n+\liminf \int_{\mathbb{R}} f_n \\ &\leqslant -\int_{\mathbb{R} \setminus E} \liminf f_n + \int_{\mathbb{R}} f \\ &= -\int_{\mathbb{R} \setminus E} f + \int_{\mathbb{R}} f \\ &= \int_E f\end{align} $$

The chain of inequalities (*) now follows.

Aloizio Macedo
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RRL
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  • @salvador: You're welcome. – RRL Mar 14 '15 at 05:48
  • I think this step $$\limsup\int_E f_n \leqslant \int_E \limsup f_n$$ is incorrect ..... or in case I am wrong just tell me why is this inequality correct? – Idonotknow Oct 15 '19 at 02:15
  • @Idonotknow: The reverse Fatou lemma involving $\limsup$ holds if there is a dominating integrable function $g$ such that $f_n \leqslant g$ for all $n$. I think this can be found given the other hypotheses. As this is over 4 years old I have to spend some time trying to reconstruct my thinking that led me to believe it to be true. – RRL Oct 15 '19 at 06:20
  • So, I am sorry for this question I know the question is long ago but I can not see anything about reverse Fatou 's Lemma in Royden 4th edition "Real Analysis" – Idonotknow Oct 15 '19 at 07:11
  • We can also show $\limsup \int_E f_n \leqslant \int_E f$ directly with the forward Fatou lemma and get the required inequality chain appearing above after “Hence,...”. I can add it as a comment or edit the answer if you like. – RRL Oct 15 '19 at 07:12
  • yes, please add it as a comment. – Idonotknow Oct 15 '19 at 07:16
  • Also, I am wondering if this question was asked differently on this site ? – Idonotknow Oct 15 '19 at 07:19
  • Ohhh I am sorry could you edit your answer instead as I am unable to read the comment – Idonotknow Oct 15 '19 at 07:26
  • thank you soo much! – Idonotknow Oct 15 '19 at 07:27
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    Update added. I still think I had found the necessary dominating function to apply the reverse Fatou lemma, but I have to work on that further. – RRL Oct 15 '19 at 07:47
  • why is $\lim f_{n} = \lim inf f$? – Idonotknow Oct 15 '19 at 12:28
  • Why is $\limsup \int_E f_n = -\liminf \left(-\int_E f_n \right) $? here we are dealing with functions and not sets @CalvinKhor , so why there is an equality? is not it should be inequality? –  Oct 15 '19 at 14:12
  • @Smart: We are dealing with sets. Given the sequence $a_n = \int_E f_n$ we have $\limsup a_n = \inf_n \sup_{k \geqslant n} a_k = \inf {\sup { a_k: k \geqslant n}: , n \in \mathbb{N} } $. – RRL Oct 15 '19 at 18:15
  • Why is the third equality in your update correct? are not the integrals sets and addition between sets means what? does it mean componentwise addition ? –  Oct 22 '19 at 01:00
  • @Smart: Either use $\liminf(a_n + b_n) \geqslant \liminf a_n + \liminf b_n$ so the third equality (noting the -1 factor ) is replaced with $\leqslant$ and we can proceed as before. Otherwise note that if $b_n$ converges we have $\liminf(a_n + b_n) = \liminf a_n + \liminf b_n$. Here we have $b_n = - \int_{\mathbb{R}} f_n$ and $\liminf b_n = \lim b_n = -\lim \int_{\mathbb{R}} f_n = - \liminf \int_{\mathbb{R}} f_n $. – RRL Oct 22 '19 at 02:13