1. ## Lebesgue integration

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]

2. Originally Posted by Fubini
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 0 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]
Are you sure you have copied this correctly?

$g(x)\ =\ \int_0^a\ \frac{f(t)}{t} dt$

when it exists is a constant.

RonL

3. ## Re: Lebesgue integration

You're absolutely right it is constant, it should be the integral from x to a. Sorry.

4. Originally Posted by Fubini
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'm still a bit confused by this

My Lebesgue theory is a bit rusty, but if this were true it should
also hold for a continuous bounded Riemann integrable function.

In fact should it not hold for $f(x)\ =\ x^n, n > 1\$
when $a < \infty ?$

RonL

5. ## solution

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.

6. It certainly works for f(t) = t^2 (a bounded Riemann integrable function)

0<a<infinity

one gets 1/3 a^3 either way.

Any Riemann integrable function is Lebesgue integrable on a bounded interval.