Exercise 4 from Chapter 2 in Real Analysis by Elias M. Stein 
Suppose $f$ is integrable on $[0,b]$, and
  $$g(x) = \int_x^b \frac{f(t)}{t}dt$$ for $0 \lt x \le b$
  prove that $g$ is integrable on $[0,b]$, and $$\int_0^b g(x)dx= \int_0^bf(t)dt.$$

I know that the key point is how to prove slow increase of $\frac{f(t)}{t}$. I prepare to find a function sequence that has compact support and converge to $g$, and use the monotone convergence theorem. but I can't do that. Can anybody help me? 
 A: Hint. Note that by Fubini's Theorem, for $0<a<b$,
$$\int_a^b g(x)dx=\int_a^b \left(\int_x^b \frac{f(t)}{t}dt\right)dx=\int_a^b \frac{f(t)}{t}\left(\int_a^t dx\right) dt.$$
A: This is an answer to a comment made by the OP regarding a previous answer.
Strictly speaking you're right, we need to verify integrability before we can apply Fubini. But you're forgetting Tonelli's Theorem: If $F(x,y)\ge0$ is a measurable function on a product space then both iterated inyegrals and the integral with respect to the product measure are all equal.
The point is that there's no  integrability hypothesis in Tonelli; it applies even when all three integrals are infinite.
The following happens very very often in applications of Fubini: 


First we use Tonelli to verify that things are integrable, and then we apply Fubini.


In your problem it's clear that $|g(x)|\le\int_x^b|f(t)|/t\,dt$. So an application of Tonelli shows that $\int|g(x)|\le\int|f(x)|$, just as in the other answer.
(In Robert Z's defense: Speaking informally people often mean "Fubini + Tonelli" when they say "Fubini".)
