Problems in the integration limits to apply Fubini's theorem 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.
 A: 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. 
