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Convergence of integrals in $L^p$

Let $\{f_n\}$ be sequence of functions in $L^p$, $1\lt p \lt \infty$, which converge a.e. to a function $f\in L^p$. Suppose that there is a $M$ such that $\|f_n\| \leq M$ for all $n$. Then I would like to show that for each function $g\in L^q$ we have $$ \int fg = \lim \int f_ng.$$

Would the result hold true if $p =1$?

My Attempt:

I want to show $\left|\int (f_ng-fg)\right|\lt \varepsilon.$ I observe that

$$\left|\int (f_ng-fg)\right| \lt \int \left| f_n-f\right||g| \leq \|f_n-f\|_p \|g\|_q,$$

and

$$\|f_n-f\|_{L^p(E)}\leq M + \|f\|_{L^p(E)}.$$

Since $g\in L^q$,for every $\varepsilon \gt 0,\exists \delta \gt 0$ such if $\mu(E)\lt \delta$,

$$ \|g\| = \left(\int _E |g|^q\right)^{1/q} \lt \frac{\varepsilon}{2\left(M + \|f\|_{L^p(E)}\right)}.$$

I can also call on Egoroff, to find a set $E$ of measure less that $\varepsilon$ such $f_n$ converges uniformly on $E^c$. Thus for some $n\gt N$,

$$|f_n-f| \lt \frac{\varepsilon}{2(\mu(E^c))^{1/p}\|g\|_{L^q(E)}},$$ so that

$$\|f_n-f\|_{L^p(E^c)} \lt \frac{\varepsilon}{2\|g\|_{L^q(E^c)}}.$$

Thus, $$ \begin{align*} \|f_n-f\|_p \|g\|_q & = \|f_n-f\|_{L^p(E)}\|g\|_{L^q(E)} + \|f_n-f\|_{L^p(E^c)}\|g\|_{L^q(E^c)} \\ & \leq \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. \end{align*} $$

Please, is what I've done okay. I welcome criticisms and corrections. Thanks.

I also need help with the case $p=1$.

PS: This is not homework. It's a question in Royden's third edition, Chapter six.

Community
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Kuku
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3 Answers3

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Your penultimate line is not correct. You can't split $\Vert f_n-f\Vert_p \Vert g\Vert_q$ up as you did. What you wrote implies (since what you wrote would be true for any sufficiently large $n$) that $f_n$ converges to $f$ in norm. In general, this isn't the case.

But it's easily fixed. Start off by writing: $$ \Bigl|\,\int fg-\int f_n g\, \Bigr|\le \int |f_n-f||g| =\int_E |f_n-f||g|+\int_{E^C}|f_n-f||g| $$ and then apply Holder's inequality as you did to the two integrals on the right separately.


Some other errors:

  • You should have fixed $\epsilon$ at the start of your argument. Let $\bf \epsilon>0$.
  • Get rid of the clause "for every $\epsilon>0$" in the fifth line and instead say "there is a $\delta>0$ such that if $\mu(E)<\delta$..."
  • When applying Egoroff, you want $\mu(E)<\delta$, not $\mu(E)<\epsilon$.
  • After using Egoroff, you want to say "there is an $N$ so that for all $n>N$", not "for some $n>N$".


Your result is true for an arbitrary measure space. Is your measure space finite? If so, then you're ok.. If not, then your argument falls apart when you try to apply Egoroff's Theorem. But, you can fix things by first taking a set $A$ of finite measure for which $\int_{A^C} |g|^q$ is small (I'll leave it for you to decide how small) and then proceeding with your argument "inside $A$". Note that your estimate will have three terms now...


Finally, the result is false for $p=1$, even with a finite measure space. Take $f_n=n\chi_{[0,1/n]}$ and $g=\chi_{[0,1]}$. Then $f_n$ converges pointwise to $f=0$, the norm of $f_n$ in $L_1$ is 1 for each $n$, but $\int f_n g=1$ for each $n$ while $\int fg=0$.

David Mitra
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  • Thanks once again. From the question it is not immediately clear whether or the measure space is finite. Am I supposed to assume its not finite? – Kuku Mar 26 '12 at 18:34
  • I would think you would be told explicitly if you could assume finiteness or that the problem would be in a section that dealt with only finite measure spaces if finiteness was to be assumed. – David Mitra Mar 26 '12 at 18:37
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The result is a consequence of the dominated convergence theorem.

Since $f_n\in L^p$ and $g\in L^q$, $f_ng$ is integrable. Moreover $\lim_{n\to\infty}f_n(x)g(x)=f(x)g(x)$ almost everywhere. Finally, $$ \int_E|f_ng|\le\|f_n\|_p\|g\|_q\le M\|g\|_q. $$ Edit

This is not a correct proof. There is no dominating function.

Julián Aguirre
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Another approach is by way of the Banach-Saks theorem, which says that if $(h_n)$ is a norm-bounded sequence in $L^p$ (with $1 < p < \infty$) then $(h_n)$ has a subsequence whose Cesaro means converge strongly. When this result is applied to the sequence $(f_n)$ of the problem (or to a subsequence thereof) the a.e. convergence of $f_n$ to $f$ means that the strong limit (of the Cesaro means) is equal to $f$, regardless of the subsequence. Thus, the sequence of real numbers $\int f_ng$, $n\ge 1$, has the property that each of its subsequences has a further subsequence whose Cesaro means converge to $\int fg$. This easily implies that $\lim_n\int f_ng =\int f_g$.

John Dawkins
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