Trying to prove the convergence theorem for integrals.
Suppose $0\leq g(x) \leq f(x) \,\forall x \ge a$ and ($f,g$ both integrable). Prove that $\int_{a}^{\infty}g$ converges provided that $\int_{a}^{\infty}f$ converges.
my attempt:
I know $0 \le \int_{a}^{t}g \le \int_{a}^{t}f \, \forall t\ge a$ and so $0 \le \int_{a}^{\infty}g \le l$, where $l=\int_{a}^{\infty}f$. But how does this show that $\lim_{t \to \infty} \int_{a}^{t}g$ converges? My assumption is that I have to prove this limit exists, but how are we assured that the limit does not oscillate like crazy for values between $0$ and $l$? Any idea how to finish the proof?
Best Answer
You have $0 \le g(x) \; \forall \; x \ge a$. This means
$$h(t) = \int_{a}^{t}g(x)dx \tag{1}\label{eq1A}$$
is an increasing function in $t$. In particular, the value will not oscillate at all, as you are concerned about. Also, since it's bounded above by
$$l = \int_{a}^{\infty}f(x)dx \tag{2}\label{eq2A}$$
as stated in the question, then
$$\lim_{t \to \infty}\left(\int_{a}^{t}g(x)dx\right) \tag{3}\label{eq3A}$$
must converge to a limit point itself, call it $L$, with $L$ being such that
$$0 \le L \le l \tag{4}\label{eq4A}$$
One way to see this to consider a sequence $y_n = h(n)$ for integers $n \ge a$. This is a monotone, bounded (below by $0$ & above by $l$ in \eqref{eq2A}) sequence. Regarding proving, and other details, about these sequences always converging, there are proofs in books such as Elementary Analysis: The Theory of Calculus by Kenneth A. Ross. Also, there are also various posts here, such as at Bounded monotone sequences and convergence., Clarification on the proof that all bounded monotone sequences converge, Every bounded monotone sequence converges, and Question on proof concerning : A bounded, monotone sequence converges.