The Unapologetic Mathematician

Mathematics for the interested outsider

Mean Convergence Properties of Integrals

Okay, we’ve got our general definition of integrable functions, and we’ve reestablished a bunch of our basic properties in this setting. Let’s consider some properties that involve the L^1 norm.

First off, the basic definitions of the L^1 norm carries across: we set

\displaystyle\lVert f\rVert_1=\int f\,d\mu

and we say that a sequence \{f_n\} of integrable functions is Cauchy in the mean or is mean Cauchy if \lVert f_n-f_m\rVert_1 goes to zero for sufficiently large m and n. We immediately find that any sequence that is Cauchy in the mean is also Cauchy in measure, since our earlier proof didn’t really depend on the functions being simple.

We also have a couple more results whose proofs don’t really depend on the simplicity of f_n, and so these carry across without change. If \{f_n\} is mean Cauchy, with indefinite integrals \{\nu_n\}, then the collection of set functions \{\nu_n\} is uniformly absolutely continuous. Further, we can define the limiting set function

\displaystyle\nu(E)=\lim\limits_{n\to\infty}\nu_n(E)

for every measurable E\subseteq X, and we find that this set function \nu is finite-valued and countably additive.

So now we can formally define the obvious notion: the sequence \{f_n\} of integrable functions “converges in the mean” or is “mean convergent” to an integrable function f if \lVert f_n-f\rVert_1 goes to zero as n goes to \infty. And, as we might expect, if \{f_n\} converges in the mean to f then it converges in measure as well.

Indeed, for every \epsilon>0 we define

\displaystyle E_n=\left\{x\in X\big\vert\lvert f_n-f\rvert\geq\epsilon\right\}

and then we see that

\displaystyle\int\lvert f_n-f\rvert\,d\mu\geq\int\limits_{E_n}\lvert f_n-f\rvert\,d\mu\geq\epsilon\mu(E_n)

Thus \mu(E_n) must go to 0 as n\to\infty, and so \{f_n\} converges in measure to f.

About these ads

June 4, 2010 - Posted by | Analysis, Measure Theory

3 Comments »

  1. [...] is, converges in mean to , which we know means it converges in measure to as well. But we picked this sequence to converge in measure to , [...]

    Pingback by When is an Integral Zero? « The Unapologetic Mathematician | June 7, 2010 | Reply

  2. [...] already shown that convergence in mean implies convergence in measure, and we’ve shown that convergence in [...]

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

  3. [...] provides a nice tool to make sure certain sequences of integrable functions converge (in the mean) to integrable limits. Yes, we have the definition and the characterization in terms of convergence [...]

    Pingback by Lebesgue’s Dominated Convergence Theorem « The Unapologetic Mathematician | June 10, 2010 | Reply


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Follow

Get every new post delivered to your Inbox.

Join 391 other followers

%d bloggers like this: