Problems in the integration limits to apply Fubini's theorem

Question

If $f:(0,a)\rightarrow\mathbb{R}$ integrable function and $$g(x)=\int_{x}^a \dfrac{f(t)}{t}dt.$$ Then $g$ is integrable and $\int_{0}^a g(t)dt=\int_{0}^a f(t)dt$.

I have to use Fubini's theorem but there is a problem with integration limits since

$$\int_{0}^{a} g(t)=\int_{0}^{a}\big( \int_{t}^a \dfrac{f(s)}{s}ds\big)dt,$$ and interchange integration limits isnt possible because the variable $x$ is in the limit of integration.

If take $\chi_{[t,a]}$ the caracteristic function for the interval $[t,a]$ then $$\int_{t}^a \dfrac{f(s)}{s}ds=\int_{0}^a \dfrac{f(s)}{s}\cdot\chi_{[t,a]}(s) ds$$ Is right to take the function in this way to apply Fubini's theorem?

Hint please.

Answer 1

This is a nice problem.

$$\int_0^a g(s)ds=\int_0^a\int_s^a\displaystyle\frac{f(t)}{t}dtds $$

Following your idea

$$\int_0^a\int_s^a\displaystyle\frac{f(t)}{t}dtds=\int_0^a\int_0^a\displaystyle\frac{f(t)}{t}\chi_{[s,a]}(t)dtds$$

Using Fubini we can change the order of integration:

$$\int_0^a\int_0^a\displaystyle\frac{f(t)}{t}\chi_{[s,a]}(t)dtds=\int_0^a\int_0^a\displaystyle\frac{f(t)}{t}\chi_{[s,a]}(t)dsdt$$

And then

$$\int_0^a\int_0^a\displaystyle\frac{f(t)}{t}\chi_{[s,a]}(t)dsdt=\int_0^a\displaystyle\frac{f(t)}{t}\int_0^a\chi_{[s,a]}(t)dsdt$$

Prove that $\chi_{[s,a]}(t)=\chi_{[0,t]}(s)$ this implies

$$\int_0^a\displaystyle\frac{f(t)}{t}\int_0^a\chi_{[s,a]}(t)dsdt=\int_0^a\displaystyle\frac{f(t)}{t}\int_0^a\chi_{[0,t]}(s)dsdt=\int_0^a\displaystyle\frac{f(t)}{t}\int_0^t dsdt=\int_0^a f(t)dt$$ Therefore $$\int_0^a g(s)ds=\int_0^a f(t)dt$$

I hope it help you.

Edited: The function $g$ is integrable by the following result

If $f$ is integrable on $[a,b]$, then the function $F$ defined by $$F(x)=\int_a^xf(t)dt$$ is a continous function of bounded variation on $[a,b]$ (Lemma 6 chapter 5 differentiation of an integral, Royden Real Analysis)

In particular the function $F$ is integrable. The function $g$ is defined in this manner.