# The Unapologetic Mathematician

## 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