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October 19th, 2016, 09:08 AM   #1
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product of two functions as a subset of the union

Hi all,

Setting: f,g A --> R integrable

My question is: why is it true that the set {x in A | (f*g)(x) is not continuous} is a subset of {x in A | f(x) is not continuous} united with {x in A | g(x) is not continuous} ?

I'm trying to solve problem 3.14 of Spivak's Calculus on Manifolds. I have to show that when f and g are integrable so is the product of them.

Greetings and thanks,
T
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October 19th, 2016, 11:44 AM   #2
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If f and g are integrable they have a finite number of discontinuities and isolated points at infinity.

Then fg has a finite number of discontinuities and isolated points at infinity and so is also integrable.
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October 19th, 2016, 02:33 PM   #3
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f is integrable if $\displaystyle \sum f_{i} \Delta x_{i}$ exists and is unique for arbitrary subdivisions approaching zero.

$\displaystyle g_{i} \Delta x_{i}$ is an arbitrary subdivision approaching zero. Therefore,

$\displaystyle \sum f_{i}(g_{i} \Delta x_{i})$ exists and is unique.
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November 2nd, 2016, 10:27 PM   #4
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Quote:
Originally Posted by zylo View Post
If f and g are integrable they have a finite number of discontinuities and isolated points at infinity.

Then fg has a finite number of discontinuities and isolated points at infinity and so is also integrable.
This is definitely not true.

As for the original question, does f*g mean the pointwise product or the convolution? Assuming the former, I suggest you try proving this first in the case that both $f$ and $g$ are monotone functions. This seems much easier and once done you can use the result that integrable functions can be uniformly approximated by compactly supported functions with bounded variation which can be decomposed as the difference of two monotone functions.

It is worth noting (and may be helpful) that monotone functions have at most countably many discontinuities and they are all jump type.
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