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Let $f$ is integrable function on $[0,1]$.

Define $g(x)=\int_x^b\frac{f(t)}{t}dt$ for $0<x\leq 1$ and $g(0)=0$.

How can I show that $g(x)$ is integrable?

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Could you use integration by parts, letting $u=1/t, dv = f(t)dt$? Then your integral becomes $\int \frac{F(t)}{t}dt + \frac{F(t)}{t^2}$, where $F(t)$ is the antiderivative of $f(t)$. – Michael Chen Apr 4 '11 at 7:50
Well, I know the calculus. But, I want to prove this only using the definition of measurable function defined in measure theory course. – user8484 Apr 4 '11 at 10:17
up vote 4 down vote accepted

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.

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