Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.