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Let $f$ be a not decreasing, Lebesgue integrable function with finite integral over $[0,+\infty)$.

It seems obvious to me that $\lim_{b\rightarrow\infty}\int_b^{\infty}f(x)\,dx=0$ then.

But how can one very formally show it?

And assumption about monotonicity is crucial here, yes?

Michael Hardy
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luka5z
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    Monotonicity is not needed here. Idea: $f=f^+ - f^-$. So $f^+=f^+ 1_{x \leq b}+f^+1_{x > b}$. What can you say about the integral of $f^+ 1_{x \leq b}$ as $b \to \infty$? – D Poole Sep 07 '14 at 18:23
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    Obviously, a better title is needed. –  Sep 07 '14 at 21:29

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Be $$F = \int_0^\infty f(x)\,\mathrm dx.$$ We have from the general rules of integration: $$F = \int_0^\infty f(x)\,\mathrm dx = \int_0^b f(x) \,\mathrm dx + \int_b^\infty f(x) \,\mathrm dx.$$ Therefore $$\int_b^\infty f(x) \,\mathrm dx = F - \int_0^b f(x) \,\mathrm dx$$ and thus $$\lim_{b\to\infty} \int_b^\infty f(x) \,\mathrm dx = F - \lim_{b\to\infty}\int_0^b f(x) \,\mathrm dx = F - \int_0^\infty f(x)\,\mathrm dx = F - F = 0.$$

celtschk
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    In the context of Lebesgue's theory, the fact that $\displaystyle\lim_{b\to\infty} \int_0^b f = \int_{[0,\infty)} f$ is not axiomatic, but derived from the axioms. I might have included something on how that's done. – Michael Hardy Sep 07 '14 at 19:45
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Lebesgue dominated convergence theorem.

Robert Israel
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