I got stuck on this problem from Real Analysis by Folland. Can anybody give me any hints on how to solve this?
If $f$ is Lebesgue integrable on $(0, a)$ and $$ g(x) = \int_x^a{t^{-1}f(t)dt} $$ then $g$ is integrable on $(0, a)$ and
$$ \int_0^a{g(x)dx} = \int_0^a{f(x)dx} $$
Let $h(t)=\frac{f(t)}{t}1_{[y,a]}(t),\:0<y<a$. Since $0<1/t<M$ on $[y,a]$, $h(t)$ is integrable, i.e $$ \int_y^{a}{\int_x^a|h(t)|dt}\:dx<aM\int_x^a|f(t)|dt<\infty $$ By Fubini's theorem \begin{align} \int_y^{a}{\int_x^ah(t)dt}\:dx &=\int_y^a \frac{f(t)}{t}\int_y^{t}dx\:dt \\ &=\int_y^a (t-y)\frac{f(t)}{t}\:dt \\ &=\int_y^a f(t)\:dt-\int_y^a \frac{y}{t}f(t)\:dt\tag{1} \end{align} Note that bound for $x$ changes to [$y,t$] because $t$ is from $x$ to $a$. Since for any $y\in(0,t],\:y<t$ $$ \frac{y}{t}f(t)<f(t) $$ Thus $\frac{y}{t}f(t)$ is integrable for $f$ is integrable. Also for any $t\in(0,a]$, there is $$ \lim_{y\to0+}\frac{y}{t}f(t)=0\tag{2} $$ By Dominated convergence theorem, (1) and (2) $$ \lim_{y\to0+}\int_y^{a}{\int_x^ah(t)dt}\:dx =\int_0^a f(t)\:dt-\lim_{y\to0+}\int_y^a \frac{y}{t}f(t)\:dt=\int_0^a f(t)\:dt $$