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Let $(f_n)$ be a sequence in $L_p[0,1]$ ($1<p<\infty$) such that $f_n\to f\in L_p[0,1]$ pointwise almost everywhere. If there exists $M>0$ such that $\|f_n\|_p\le M$ for all $n$, and $g\in L^q[0,1]$, where $\dfrac{1}{p}+\dfrac{1}{q}=1$, then:

$$\int fg=\lim_{n\to\infty}\int f_ng.$$

I tried this with the Hölder's inequality:

$$\left|\int fg-\int f_ng\right|\le \|f_n-f\|_p\|g\|_q.$$

I'm not sure we can say directly that $\|f_n-f\|_p\to 0$. Perhaps with uniformly convergence/Egorov's theorem?

By Egorov's theorem, if $\epsilon>0$ there exists $A\subseteq[0,1]$ such that $\lambda(A^c)<\epsilon$ and $f_n\to f$ uniformly on $A$. It is clear then $$\int _Afg=\lim_{n\to\infty}\int_A f_ng,$$ since uniformly convergence implies $L_p$-convergence.

But I don't know how to do it on $A^c$. Also, why is $\|f_n\|_p\le M$ required?

Thank you.

Edited wrong title. Sorry.

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marked as duplicate by Guy Fsone, Rolf Hoyer, Krish, Claude Leibovici, Arnaud D. Nov 23 '17 at 9:43

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    $\begingroup$ Instead of taking $\lambda(A)<\epsilon$, choose $\delta>0$ so that $\lambda(A)<\delta$ implies $\int_A|g|^q<\epsilon^q$, and then use Egoroff to get $\lambda(A)<\delta$... $\endgroup$ – David C. Ullrich Jan 14 '16 at 3:33
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    $\begingroup$ Can you use the dominated convergence theorem? $\endgroup$ – mathematician Jan 14 '16 at 3:39
  • $\begingroup$ @user1952009 Holder's inequality. $\endgroup$ – David C. Ullrich Jan 14 '16 at 3:48
  • $\begingroup$ Oops. My $A$ is the same as your $A^c$. $\endgroup$ – David C. Ullrich Jan 14 '16 at 3:52
  • $\begingroup$ ok, the relation $1/p+1/q = 1$ appearing frequently in inequalities seems to come from the Young's inequality on numbers $\endgroup$ – reuns Jan 14 '16 at 3:52
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Use absolute integrability, Egorov theorem and Fatou lemma.

We aim to show $\int|f_ng-fg|<\epsilon$. Note since $g^q\in L^1(0,1)$, there exists $\delta>0$, such that for all $E$ with $|E|<\delta$, we have $\int_E |g|^q<\epsilon$.

Since $\mu([0,1])<\infty$, by Egorov theorem, there exist $E$ such that $|E|<\delta$ and $f_ng\to fg$ uniformly except on $E$. Hence for large enough $n$, we only need to consider $\int_E|f_ng-fg|=\int_E|f_n-f||g|$

By Holder's inequality $\int_E|f_n-f||g|\le (\int_E|f_n-f|^p)^{\frac{1}{p}}(\int_E|g|^q)^{\frac{1}{q}}\le (\int|f_n-f|^p)^{\frac{1}{p}}\epsilon^{\frac{1}{q}}$.

Also $\int |f_n-f|^p\le 2^p\int |f_n|^p+|f|^p\le 2^p(M^p+\limsup \int |f_n|^p)\le 2^{p+1}M^p$,by Fatou lemma. Thus for large $n$, we have $\int_E|f_ng-fg|\le 2^{\frac{p+1}{p}}M \epsilon^{\frac{1}{q}}$

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