# The Unapologetic Mathematician

## The L¹ Norm

We can now introduce a norm to our space of integrable simple functions, making it into a normed vector space. We define

$\displaystyle\lVert f\rVert_1=\int\lvert f\rvert\,d\mu$

Don’t worry about that little $1$ dangling off of the norm, or why we call this the “$L^1$ norm”. That will become clear later when we generalize.

We can easily verify that $\lVert cf\rVert_1=\lvert c\rvert\lVert f\rVert_1$ and that $\lVert f+g\rVert_1\leq\lVert f\rVert_1+\lVert g\rVert_1$, using our properties of integrals. The catch is that $\lVert f\rVert_1=0$ doesn’t imply that $f$ is identically zero, but only that $f=0$ almost everywhere. But really throughout our treatment of integration we’re considering two functions that are equal a.e. to be equivalent, and so this isn’t really a problem — $\lVert f\rVert_1=0$ implies that $f$ is equivalent to the constant zero function for our purposes.

Of course, a norm gives rise to a metric:

$\displaystyle\rho(f,g)=\lVert g-f\rVert_1=\int\lvert g-f\rvert\,d\mu$

and this gives us a topology on the space of integrable simple functions. And with a topology comes a notion of convergence!

We say that a sequence $\{f_n\}$ of integrable functions is “Cauchy in the mean” or is “mean Cauchy” if $\lVert f_n-f_m\rVert_1\to0$ as $m$ and $n$ get arbitrarily large. We won’t talk quite yet about convergence because our situation is sort of like the one with rational numbers; we have a sense of when functions are getting close to each other, but most of these mean Cauchy sequences actually don’t converge within our space. That is, the normed vector space is not a Banach space.

However we can say some things about this notion of convergence. For one, a sequence $\{f_n\}$ that is Cauchy in the mean is Cauchy in measure as well. Indeed, for any $\epsilon>0$ we can define the sets

$\displaystyle E_{mn}=\left\{x\in X\big\vert\lvert f_n(x)-f_m(x)\rvert\geq\epsilon\right\}$

And then we find that

$\displaystyle\lVert f_n-f_m\rVert=\int\lvert f_n-f_m\rvert\,d\mu\geq\int\limits_{E_mn}\lvert f_n-f_m\rvert\,d\mu\geq\epsilon\mu(E_mn)$

As $m$ and $n$ get arbitrarily large, the fact that the sequence is mean Cauchy tells us that the left hand side of this inequality gets pushed down to zero, and so the right hand side must as well.

This notion of convergence will play a major role in our study of integration.

May 28, 2010