A question on the definability of convolution

[zzzhhh]

In Apostol's "Mathematical Analysis", Page 328 (see the image below and the underlined sentence),

why does the Lebesgue integral (41) exist for

?[/URL]
The definition of convolution is as follows:

Thanks!

 

[JSuarez]

I'm not sure if I understand your question. For that special case, the Lebesgue integral doesn't have to exist for all $x \in \left[a,b\right]$, and that is precisely what the author says.

 

[zzzhhh]

My question comes from the underlined sentence. That the convolution in (41) can be defined means the Lebesgue integral in (41) exists and the author asserts this integral exists for all x in [a,b] under the given condition. But I do not understand why this integral

exists.

 

[JSuarez]

What the author (Apostol) asserts is that, if f and g are Riemann integrable, then that convolution is well defined, for all $x \in \left[a,b\right]$. On the other hand, if the functions are Lebesgue integrable (and this include much less regular functions), then then their (Lebesgue) integral may not be defined. In fact, the author gives an example immediately after.

 

[zzzhhh]

What the author (Apostol) asserts is that, if f and g are Riemann integrable, then that convolution is well defined, for all $x \in \left[a,b\right]$

My question is why this statement holds, not the statement on the other hand.

 

[JSuarez]

Because the product of two Riemann integrable functions is also (Riemann) integrable. This is a consequence of, for example, theorem 7.48 of Apostol's book.

 

[zzzhhh]

But f and g are Riemann-integrable only on [a,b] under the given condition, which does not insure f(t) and g(x-t), and in turn their product, are Riemann-integrable on [0,x].

 

[JSuarez]

Note that $x \in \left[a,b\right]$, then for a<0, there is no problem, because $\left[0,x\right]\subseteq \left[a,b\right]$; if a>0, you may always assume that the functions are 0 outside the interval.

 

[zzzhhh]

if a>0, you may always assume that the functions are 0 outside the interval.

Why? Why can we assume the functions are 0 outside the interval ([a,b] or [0,x])?