Let $f:[0,b] \to \mathbb{R}$ be Lebesgue integrable. We define $$g(x)=\int_x^b\frac{f(t)}{t}dt,\quad 0<x\le b.$$ Show that $g(x)$ is Lebesgue integrable in $[0,b]$. Also show that $$\int_0^bg(x)dx=\int_0^bf(t)dt.$$
If we show the asked equality then it is obvious that $g$ is Lesbegue integrable, since $f$ is. Now, I have try to show the equality using the fact that, $$\int_0^b\Bigl[g'(x)-\frac{f(x)}{x}\Bigr]dx=0$$, since $g$ is the antiderivative of $f$ but I had no success. I am pleased to know other ideas to approach this problem.
Hint.
Let $\Delta = \{ (x,t) \mid 0 \leq x \leq t \leq b \}$. Define $h(x,t)$ on $[0,b] \times [0,b]$ by $h(x,t) = \chi_{\Delta}(x,t) \frac{f(t)}{t}$. Apply Fubini's theorem to $h$ to integrate it in two different ways. (There will be an initial step involving $|h|$ to prove integrability of $h$.)
Edit: $\Delta$ would more accurately be changed in order to be a subset of $(0,b] \times (0,b]$.