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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'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.)2012-05-03

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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!2012-05-03
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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).