Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I know that if $f,g$ are continuous functions on $[a, b]$ and $f(x) \leq g(x)$ for all $x \in [a,b]$, $\int_a^b f(x) dx \leq \int_a^b g(x) dx$. Would it also be true that $\int_{-\infty} ^\infty f(x) dx \leq \int_{-\infty} ^\infty g(x) dx$? My intuition tells me this should be the case.

share|cite|improve this question
I believe this is the case as the area beneath $g(x)$ must be greater then or equal to that of $f(x)$. – Argon May 3 '12 at 0:18
I'm assuming both functions are positive-valued? If both improper integrals exist, then they're equal to the limit of the integrals over $[-a, a]$ as $a$ gets big, and then the result follows from the result you already are happy with.(It's enough for just the integral of $g$ to exist; that implies the integral of $f$ exists.) – user29743 May 3 '12 at 0:20
up vote 2 down vote accepted

Yes it is true. To see this, consider that

$$\int_0^R f(x) dx \leq \int_0^R g(x) dx \forall R >0 $$


$$\int_R^0 f(x) dx \leq \int_R^0 g(x) dx \forall R <0$$

Thus, taking the limits if they exists, yields your desired inequality.

P.S. Even if the limit don't exist, you can still prove exactly the same way that $$\int_{-\infty}^\infty g(x)-f(x)dx \geq 0 \,.$$

Note that this integral is either convergent or $+ \infty$ because $g-f \geq 0$.

share|cite|improve this answer
Very nice explanation! – Student May 3 '12 at 0:21

Yes, the Riemann definition of an improper integral is just what you get by sending the integration limits to infinity ($a_n \le b_n$ implies $\lim_{n\to\infty} a_n \le \lim_{n\to\infty} b_n$ if the limits exist).

For the Lebesgue case there is no distinction between the two types of integral (the answer is still yes).

share|cite|improve this answer

Your Answer


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.