The Unapologetic Mathematician

Mathematics for the interested outsider

Mean Convergence is Complete

I evidently have to avoid saying L^1 in post titles because WordPress’ pingbacks can’t handle the unicode character ¹…

Anyhow, today we can show that the L^1 norm gives us a complete metric structure on the space of all integrable functions on X. But first:

If \{f_n\} is a mean Cauchy sequence of integrable simple functions converging in measure to f, then it converges in the mean to f as well. Indeed, for any fixed m the sequence \{\lvert f_n-f_m\rvert\} is a mean Cauchy sequence of integrable simple functions converging in measure to \lvert f-f_m\rvert. Thus we find

\displaystyle\int\lvert f-f_m\rvert\,d\mu=\lim\limits_{n\to\infty}\int\vert f_n-f_m\rvert\,d\mu

But the fact that \{f_n\} is mean Cauchy means that the integral on the right gets arbitrarily small as we take large enough m and n. And so on the left, the integral gets arbitrarily small as we take a large enough m. That is,

\displaystyle\lim\limits_{m\to\infty}\int\vert f-f_m\rvert\,d\mu=0

and \{f_n\} converges in the mean to 0.

As a quick corollary, given any integrable function f and positive number \epsilon, there is some integrable simple function g with \lVert f-g\rVert_1<\epsilon. Indeed, since f is integrable there must be some sequence \{f_n\} of integrable simple functions converging in measure to it, and we can pick g=f_n for some sufficiently large n.

Now, if \{f_n\} is a mean Cauchy sequence of integrable functions — any integrable functions — then there is some integrable function f to which the sequence converges in the mean. The previous corollary tells us that for every f_n there’s some integrable simple g_n so that \lVert f_n-g_n\rVert_1<\frac{1}{n}. This gives us a new sequence \{g_n\}, which is itself mean Cauchy. Indeed, for any \epsilon>0 choose an N>\frac{3}{\epsilon} large enough so that \lVert f_n-f_m\rVert_1<\frac{\epsilon}{3}. Then for n,m\geq N we have

\displaystyle\lVert g_n-g_m\rVert_1\leq\lVert g_n-f_n\rVert_1+\lVert f_n-f_m\rVert_1+\lVert f_m-g_m\rVert_1\leq\frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}=\epsilon

Since \{g_n\} is Cauchy in measure it converges in measure to some function f. Then our first result shows that \{g_n\} also converges in the mean to f — that \lVert g_n-f\rVert_1 goes to zero as n goes to \infty. But we can also write

\displaystyle\lVert f_n-f\rVert_1\leq\lVert f_n-g_n\rVert_1+\lVert g_n-f\rVert_1\leq\frac{1}{n}+\lVert g_n-f\rVert_1

so if we choose a large enough n, this will get arbitrarily small as well. That is, \{f_n\} also converges in the mean to f.

This shows that any mean Cauchy sequence of integrable functions is also mean convergent to some function, and thus the space of integrable functions equipped with the L^1 norm is a Banach space.

June 8, 2010 - Posted by | Analysis, Measure Theory


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  5. Hi, thank you for the post. I’m studying real analysis and I find you blog through a google search!

    I have a quick question about the first equation. How can I get from the convergence in measure to the convergence of the integral?

    We know that |f_n-f_m| converge in measure to |f-f_m|, we have |f_n-f_m| is a mean Cauchy sequence and |f_n – f_m| are simple functions. But I have trouble putting the conditions together to get the convergence of the integral…

    It seems that it repeated the conditions for the corollary to be proofed.

    Comment by vulpeculazhou | December 22, 2010 | Reply

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