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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?

Saran Wrap
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1 Answers1

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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.

John Omielan
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  • Hmm, how do I show the "it must converge to a limit point itself" rigourously? ie. how do I show such an $L$ indeed exists? Intuitively it makes sense, I just want to see how it is done. – Saran Wrap Mar 22 '20 at 02:30
  • For instance, $\lim_{x \to 0}\sin(\frac{1}{x})$ is bounded between $0$ and $1$ but the limit does not exist. – Saran Wrap Mar 22 '20 at 02:32
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    It is because the function $h$ is increasing: an increasing function that is bounded above has a limit at $+\infty$ (or more generally left- and right-limits at any point). – Raoul Mar 22 '20 at 02:42
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    @SaranWrap The key issue here is the sequence is monotonic, non-decreasing, with an upper bound. I added some details about using the Bolzano-Weierstrass theorem, which I hope helps to explain why. – John Omielan Mar 22 '20 at 02:47
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    @SaranWrap You're welcome. Note I just did a site search of "bounded monotone sequence converge" (but without the double quotes) to find quite a few other posts here asking about this issue, with my providing links in my answer to $3$ of these. If you like, you may wish to read those, or other related posts, for more details about this issue. – John Omielan Mar 22 '20 at 02:52
  • @JohnOmielan I do not uderstand your argument with BW. Also, the result "increasing + bounded above => convergent" is often used to prove BW. – Raoul Mar 22 '20 at 07:22
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    @Raoul Thanks for your feedback. Re: using the argument with BW, if you have an infinite monotone subsequence converging, note any points not included, after the first included one, must have included points both before & after it, so it's bounded by those converging points, so these other points must also be converging. I only added this to help give a conceptual idea. Although, as I recall, the BW theorem can be proven in other ways, since it's not directly related to the answer, plus there's so much printed & online info., I removed that paragraph & added to my last one with more details. – John Omielan Mar 22 '20 at 18:16