I need to prove:
If $f$ is Lebesgue integrable on $(0, a)$ and $g(x) = \int_x^a (f(t)/t)dt$, then g is integrable on $(0, a)$.
I know that since f is integrable on the interval $(0, a)$ I have $\int_0^a|f(t)|dt<\infty$ and indeed $\int_b^a|f(t)|dt<\infty$ for any $b\in(0,a)$. I need$\int_0^a\int_x^a|f(t)/t|dtdx<\infty$ (since the inner integral is the definition of g(x)).
My first instinct is to try and use the integrability of f to show that g is integrable, simply using some inequalities, but this approach has a problem. I know that since $t\geq0$ in the interval, $|f(t)/t|=|f(t)|/t$, and whenever $t\geq1$, $|f(t)|\geq|f(t)|/t$, but this doesn't seem particularly helpful since the integral begins at $0$, not $1$. The problem is I'm not clear on why the function $1/t$ doesn't "blow up" near $0$ and make this false for some $f(x)$.
Best Answer
Hint: Use the Fubini Tonelli theorem.
details:
it gives
$$ \int_0^a\left\{\int_x^a \frac{|f(t)|}{t} dt \right\} dx = \int_0^a \frac{|f(t)|}{t} \left\{\int_0^t dx \right\} dt = \int_0^a \frac{|f(t)|}{t}tdt = \int_0^a |f(t)| dt <\infty $$