Let $f$ is integrable function on $[0,1]$.
Define $g(x)=\int_x^b\frac{f(t)}{t}dt$ for $0
How can I show that $g(x)$ is integrable?
Let $f$ is integrable function on $[0,1]$.
Define $g(x)=\int_x^b\frac{f(t)}{t}dt$ for $0
How can I show that $g(x)$ is integrable?
Measurability follows from the continuity of the function $x\mapsto g(x)$ at every $x\ne0$. Integrability follows from Fubini theorem, namely $ \int_0^1|g(x)|\mathrm{d}x\le\int_0^1\int_x^1|f(y)|y^{-1}\mathrm{d}y\mathrm{d}x=(*). $ Now, the double integral is over the set $0\le x\le y\le 1$ and, for each $0\le y\le 1$, $ \int_0^y\mathrm{d}x=y, $ hence $ (*)=\int_0^1|f(y)|\mathrm{d}y, $ which is finite.