suppose $f$ is integrable on $[0,b]$,and $g(x)=\int_{x}^{b}\frac{f(t)}{t}dt$ ,for $0 < x\leqslant b$, how to prove that $g$ is integrable over $[0,b]$.
And this is under the definition of Lebesgue integration.
suppose $f$ is integrable on $[0,b]$,and $g(x)=\int_{x}^{b}\frac{f(t)}{t}dt$ ,for $0 < x\leqslant b$, how to prove that $g$ is integrable over $[0,b]$.
And this is under the definition of Lebesgue integration.
Hint: When $f\geqslant0$, $\int_0^bg(x)\mathrm dx=\int_0^b\left(\int_x^b\frac{f(t)}t\mathrm dt\right)\,\mathrm dx\stackrel{\text{(Tonelli)}}{=}\int_0^b\left(\int_0^t\mathrm dx\right)\frac{f(t)}t\mathrm dt=\int_0^bf(t)\mathrm dt. $ Using Fubini theorem, adapt the argument to the general case.