It seems obvious, but how to prove it formally?

Question

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?

Answer 1

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 $$\lim_{b\to\infty} \int_b^\infty f(x) \,\mathrm dx = F - \lim_{b\to\infty}\int_0^b f(x) \,\mathrm dx = F - \int_0^\infty f(x)\,\mathrm dx = F - F = 0.$$

Answer 2

Lebesgue dominated convergence theorem.