Re: Calculus 102

On Wed, 18 Jul 2007 01:03:10 -0700, William Elliot
<marsh@xxxxxxxxxxxxxxxxxx> wrote:

From: David C. Ullrich <ullrich@xxxxxxxxxxxxxxxx>
<marsh@xxxxxxxxxxxxxxxxxx> wrote:

If integral (a,b) |f(x)| dx is finite,
then integral (a,b) f(x) dx is finite.
-oo <= a < b <= oo

This last is not true without some hypotheses on f.
Exactly what hypotheses would depend on whether you're
talking about the Riemann integral or the Lebesgue integral.

Whoops. Is this calculus 201?

What about, when 0 <= g([0,oo)), using
integral(a,b) g(x) dx = lim(x->b) integral(a,x) g(x) dx?

Was this supposed to be an explanation of what sort of
integral you're talking about? It doesn't answer the question
(also the notation makes no sense.)

What's the status with that theorem?

Since you won't simply tell me whether you're talking
about the Riemann or Lebesgue integral I'll explain
assuming it's the Lebesgue integral. Say E is a
non-measurable set contained in [0,1]. Define
f(x) = 1 for x in E, -1 for x not in E. Then
|f| = 1 so there's no problem with int_0^1 |f|,
but int_0^1 f does not exist.

If you're talking about the Riemann integral then
let E be the rationals in [0,1].

For the Lebesgue integral you need to assume that
f is measurable. For the Riemann integral you
need to assume _something_ (for example assuming
that f is continuous would be enough.)


----


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

David C. Ullrich
.

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