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
Don’t worry about that little 
dangling off of the norm, or why we call this the “
norm”. That will become clear later when we generalize.
We can easily verify that 
and that 
, using our properties of integrals. The catch is that 
doesn’t imply that 
is identically zero, but only that 
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 — implies that is equivalent to the constant zero function for our purposes.
Of course, a norm gives rise to a metric:
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 
of integrable functions is “Cauchy in the mean” or is “mean Cauchy” if 
as 
and 
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 that is Cauchy in the mean is Cauchy in measure as well. Indeed, for any 
we can define the sets
And then we find that
As and 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.
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May 28, 2010 -
Posted by John Armstrong |
Analysis, Functional Analysis, Measure Theory
### That is, the normed vector space is not a Banach space. ###
But L^1 is certainly a Banach space…
Be careful, Cristi: like I said, the situation is sort of like with the rational numbers, which are not complete. The space we have ahold of now is not the full space
, but only the space of integrable simple functions.
It’s easy to construct a sequence of integrable simple functions which is mean Cauchy, but which do not converge in the mean to an integrable simple function.
You’re right, of course. But why not define the norm for all good functions (like i first thought you did)?
That’s exactly the point of Lebesgue’s approach to integration, Cristi: first you integrate simple functions, and then you approach other functions as limits of simple functions. You can’t define the
norm on “integrable functions” until you know what an “integrable function” is.
Yes, but you can define the integral of simple functions, then the integral of positive/integrable functions, and derive the properties of all integrals directly, instead of spending too much time on simple functions only. This is just my taste (based on Rudin), maybe you want to take a more pedagogical approach.
Okay, Cristi, so sketch it out: how do you go from simple integrable to positive integrable functions? Make sure that you know the integral of a positive integrable function doesn’t depend on which sequence of simple integrable functions you pick to converge to it.
You define the integral of a positive measurable function f to be the supremum of the integrals of those simple functions that are at most f. Then you prove the Monotone Convergence Theorem. This is how Rudin does it in the first chapter of R&C and it’s very quick. But not everybody likes his style.
One thing I’ve wondered is why we do not define the measure of any non-negative f as the measure of { (x,y) : y <= f(x) } in the R^2 product measure.
Well, Tom, for one thing we haven’t talked about multiple Lebesgue integrals and the analogue of Fubini’s theorem. In practice, once we do it should work out to be just the measure you refer to.
Hi John, I understand that you want to take a particular order, but is there a big drawback to my suggestion which is why all authors (that I know of) avoid it?
It’s possible to define the
norm like that, yes, but then you basically have to do integrals anyway. It doesn’t really simplify anything.