It's been a long time since I did any analysis, but I'll try to post some useful comments:

>How does one prove that if f is in Lp(Rn) (the lebesgue
>space of order p) and g is in L1(Rn) (that is, g is
>integrable) then the p-th power of the absolute value of
>their convolution ¦f*g¦^p is measurable?(p is greater than
>or equal to 1 and less than or equal to infinity)

First, just to clarify, you want to prove that the convolution is a measurable function, not an integrable function, right? Therefore, it has to satisfy the property that the inverse image of any open set is a measurable set, but not the condition that the lebesgue integrals over its regions of positive support and negative support are finite.

Because of this, there is no difference (for finite p>=1) between asserting the |f*g|^p is measurable and that |f*g| is measurable, because the function y-->y^p is bicontinuous for 0 <= y, so open sets in the range of |f*g| and |f*g|^p map onto each other.

>I have seen a proof where this is shown to hold if f and g
>are greater than or equal to 0, but I can't see how the
>result follows from this special case. I thought of
>decomoposing f and g into their positive and negative parts
>and applying the special result somehow but didn't get far.

I can't think off-hand how to prove this fact, but if you already have a proof for nonnegative valued f and g, then your idea of decomposing f and g into positive and negative parts should work.

Here is an idea: assuming that you are using the standard topology in R (the range space of the functions f and g), then since this topology has a countable subbasis of intervals (i.e. every open set can be written as a countable union of intervals), and since the measurable sets are a sigma algebra (closed under countable unions and intersections and complementation) you only need to prove the that the inverse image of an arbitrary open interval is measurable.

However, all open intervals are intersections of semi-infinite open intervals, so all you really need to prove is that the inverse image of {y:y>a} and of {y:y<b} are measurable. So let f be partitioned into f₊ and f_-, its positive and negative parts, and consider |(f₊ + f_-)*g| > a. This is the union of (f₊ + f_-)*g > a and (f₊ + f_-)*g < -a. You only need prove that one of these has a measurable inverse image, since the other proof will be very similar.

There is still probably some trickery involved, but this should get you started. If you are still working on this problem, please let us know how it turns out, and in turn, I'll let you know if I have any further ideas.

>Thank you

You're welcome!

--Stuart Anderson