# proof related to convergence of a integral

i have the following condition $$0\le f(x)\le g(x)$$ and $$\int_{a}^{b}g(x)dx$$ is convergent for any $a$ and $b$ (which means $a$ or $b$ can tend to infinity) then prove that $$\int_{a}^{b}f(x)dx$$ also is convergent. now i realize that as $\int_{a}^{b}f(x)dx$ will be bounded as each term of the Riemann integral will be less than that of $g(x)$ and as for $a$ and $b$ finite there is no problem as integral is finite. the problem arises when either one or both of them tend to infinity. now i get a basic feel for the problem and i also realize that $\lim_{x\to\infty}f(x)=\lim_{x\to\infty}g(x)=0$ same is the case for $x=-\infty$ . but i am not able to prove that $f(x)$ will be convergent. help appreciated .

• Actually, $f$ need not be integrable at all. See my counterexample in the ensuing answer. – Mark Viola Apr 12 '15 at 18:08
• i wrote a comment below , the example does not work – avz2611 Apr 12 '15 at 18:08

Case 1 Assume $a$ and $b$ are real numbers.

Suppose that $f(x) =1$ when $x$ is irrational and $f(x) =0$ when $x$ is rational.

Take $g(x) =2$ for all $a\le x\le b$.

Then, clearly $0\le f \le g$ and $\int_a^b g(x)dx=2(b-a)$ is convergent.

But as a Riemann integral, $f$ is not integrable.

Case 2: $b=\infty$.

Suppose that $f(x) =1$ when $a\le x \le 2a$ is irrational and $f(x) =0$ when $a\le x\le 2a$ is rational and $f=0$ elsewhere.

Take $g(x) =2$ for all $a\le x \le 2a$ and $g=\frac{1}{x^2}$ elsewhere.

Then, clearly $0\le f \le g$ and $\int_a^b g(x)dx=2a+\frac{1}{2a}$ is convergent.

But as a Riemann integral, $f$ is not integrable.

• well that is not true when a or b are tending to infinity , and condition specified by me are for any $a$ and $b$ – avz2611 Apr 12 '15 at 18:07
• I interpreted $a$ and $b$ to be real numbers, not infinity. – Mark Viola Apr 12 '15 at 18:09
• i edited the question, it is clearer now – avz2611 Apr 12 '15 at 18:13
• I added a counter example for the case in which $b=\infty$. I hope that this helps. – Mark Viola Apr 12 '15 at 18:15
• this time i can take $a$ as $-\infty$ you need to consider all cases right – avz2611 Apr 12 '15 at 18:17