Suppose $f$ is integrable on $[0,b]$. Show that $$ g(x)= \int_{x}^{b} \frac{f(t)}{t} dt$$ is integrable.
$\textbf{My Attempt:}$ We want to show that $\int_{0}^b \mid g(x) \mid dx = \int_{0}^{b} \left ( \int_{x}^b \frac{\mid f(t) \mid}{\mid t \mid} dt \right) dx< \infty$.
Since $f$ is integrable, then we know that $ f(x) < \infty$ a.e $x$. Does this imply that $f$ is bounded for a.e $x$ ?
Also, if this is true could I use this fact to easily finish this proof?
Actually, $$\int_{0}^b \mid g(x) \mid dx \leqslant \int_{0}^{b} \left ( \int_{x}^b \frac{\mid f(t) \mid}{t} dt \right) dx=\int_{0}^{b}\frac{\mid f(t) \mid}{t} \left ( \int_0^t dx \right) dt=\int_{0}^{b}\mid f(t) \mid dt.$$