## Mean Convergence is Complete

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

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

If is a mean Cauchy sequence of integrable simple functions converging in measure to , then it converges in the mean to as well. Indeed, for any fixed the sequence is a mean Cauchy sequence of integrable simple functions converging in measure to . Thus we find

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

and converges in the mean to .

As a quick corollary, given any integrable function and positive number , there is some integrable *simple* function with . Indeed, since is integrable there must be *some* sequence of integrable simple functions converging in measure to it, and we can pick for some sufficiently large .

Now, if is a mean Cauchy sequence of integrable functions — *any* integrable functions — then there is some integrable function to which the sequence converges in the mean. The previous corollary tells us that for every there’s some integrable simple so that . This gives us a new sequence , which is itself mean Cauchy. Indeed, for any choose an large enough so that . Then for we have

Since is Cauchy in measure it converges in measure to some function . Then our first result shows that also converges in the mean to — that goes to zero as goes to . But we can also write

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

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 norm is a Banach space.

[…] is, the sequence is Cauchy in the mean. But we know that the norm is complete, and so converges in the mean to some function . But this convergence […]

Pingback by Equicontinuity, Convergence in Measure, and Convergence in Mean « The Unapologetic Mathematician | June 9, 2010 |

[…] sets is Cauchy in if and only if its sequence of characteristic functions is mean Cauchy. Since mean convergence is complete, the sequence of characteristic functions must converge in mean to some function . But mean […]

Pingback by Completeness of the Metric Space of a Measure Space « The Unapologetic Mathematician | August 9, 2010 |

[…] functions on a measure space , which we called or . We’ve seen that this gives us a complete normed vector space — a Banach space. This is what we’d like to […]

Pingback by Hölder’s Inequality « The Unapologetic Mathematician | August 26, 2010 |

[…] spaces, we need to show that they’re complete. As it turns out, we can adapt the proof that mean convergence is complete, but we will take a somewhat different approach. It suffices to show that for any sequence of […]

Pingback by Some Banach Spaces « The Unapologetic Mathematician | August 31, 2010 |

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 |