Re: Convergence of integrals of uniformly convergent functions

On 28 Mar 2007 14:10:51 -0700, "Amanda" <sca18@xxxxxxxxxxx> wrote:

I'm studying measure theory and in an exercise we were asked to use
Lebesgue's techniques to prove the following theorem about Riemann
integrals:

If f_n is a sequence of real valued Riemann integrable function
defined on a compact interval I and converges uniformly to a function
f, then f is Riemann integrable and Int f(x) dx = lim Int f_n(x) dx
(integrals over I = [a, b]).

[...]

I think this is a beautiful exercise, but I got disappointed. Lebesgue
techniques usually make it much simpler to prove theorems about
Riemann integrals, but in this case the standard proof, based on
partitions and upper and lower sums, is much simpler, at least than
the proof I could come up with.

Before seeing this last paragraph I was planning on pointing
out that simply using the definition of the Riemann integral
is actually much simpler. But you realize that...

I think that it's simply a slightly silly exercise, if it
includes a stipulation that you must use the Lebesgue
integral. You shouldn't be disappointed at the fact that
a direct proof is simpler: The Lebesgue integral is much
better in many ways, so that if you have a problem about
integrals in general, where they could be taken to be
Lebesgue integrals _or_ Riemann integrals, then taking
them to be Lebesgue integrals will often be a good idea.
But there's no reason that the Lebesgue integral should
be better for a problem that's specifically _about_
Riemann integrals.

Here's a problem that should make you happier:

Suppose that f_n is Riemann integrable on [a,b],
the f_n are uniformly bounded and tend to 0 at
every point. Show that int_a^b f_n -> 0.

Of course that's trivial using the Lebesgue
integral. If you can give a proof using only
Riemann integrals, without sneaking a lot of
measure theory into the proof in disguise,
that would be something.

Thank you

Amanda.


************************

David C. Ullrich
.

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