Property of Improper Integrals

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.

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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

Yes it is true. To see this, consider that

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

and

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

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 $g-f \geq 0$.

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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).

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