Are you sure you have copied this correctly?Originally Posted by Fubini
when it exists is a constant.
RonL
My question is from Folland "Real Analysis" 2nd ed. p.77 Chapter 2, #56.
If f is lebesgue integrable on (0,a) and
g(x) = integral (from x to a) [t^-1 f(t)dt]
Prove then that:
1. g is integrable on (0,a) and
2. integral (from 0 to a) [g(x)dx] = integral (from 0 to a) [f(x)dx]
I was able to get this by following the advice of the book. First apply Tonelli's theorem then, Fubini's theorem, from the previous section on p.68.
int (0 to a) [ |g(x)| dx] is non-negative so we can switch the order of integration by Tonelli's theorem.
int (o to a) [ |g(x)|] = int (0 to a) int (x to a) [ |f(t)| t^-1 dt dx]
=int (0 to a int (0 to t) [ |f(t)| t ^-1 dx dt]
noting the regions (x<t<a, 0<x<a)
and (0<x<t, 0<t<a)
are identical.
evaluating the inside integral gives:
int (0 to a) [ {x |f(t)| t^-1 (x=t to x=0)} dt]
=int (0 to a) [ |f(t)| dt ]
< infinity (by assumtion, it is given f if Lebesgue
integrable on (0,a) ).
Hence g(x) is Lebesgue integrable on (0,a).
Now the equality is clear as the methodology on how to achieve it is given above (switching integration order) and apply Fubini's theorem for integrable functions.