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Let $f$ and $g$ be continuous functions on $(a,b)$ such that $0 \le f\left( x \right) \le g(x)$ for all $x \in \left( {a,b} \right)$; $a$ can be $ - \infty $ and $b$ can be $ + \infty $.

Prove: $$\mathop \int \limits_a^b g\left( x \right)dx < \infty \Rightarrow \mathop \int \limits_a^b f\left( x \right)dx < \infty $$

This is an exercise in Elementary Analysis by Ross. Chapter 36, exercise 6a. I think I'm almost able to prove this, using the definitions give in 36 and some theorems in 33. The thing is that I don't use that $0 \le f\left( x \right)$. Could anybody help me where in the proof I need to use this?

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    Did you prove carefully that $\int_a^b f(x)\,dx$ exists? What if $f(x)=\sin x -1$ on $(-\infty,\infty)$ (and $g$ nonnegative)?2012-12-06
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    That's a very good point.. I can prove that $\mathop \smallint \limits_c^\alpha f + \mathop \smallint \limits_\alpha ^d f$ exist, but when $c \to {a^ - }$ and $d \to {b^ - }$, I must prove that those limits exist, is that what you mean? If $f\le0$, than both integrals are increasing, and they are bounded by $g$ which proves that both limits converges. Is that correct ?2012-12-06
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    Well, in a certain sense the result is true without the assumption, but the definite integral of $f$ might not exist. Example: $g(x)=0$ everywhere, $f(x)=-1$ everywhere, integrating from $-\infty$ to $\infty$.2012-12-06
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    Yes, if I understand you correctly. (The first integral's value increases as $c\rightarrow a^-$, e.g.)2012-12-07
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    @DavidMitra Again, same question. I was thinking, not for every sequence $s_n$ with $\lim s_n =a$ and $s_n\in(a,\alpha)$ those integral values are increasing. So I don't think this proves that this limit of this integral converges.. Any better idea?2012-12-10
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    @Kasper Peulen It suffices to consider decreasing sequences that converge to $a$. Or, let $F(\alpha)=\int_\alpha^{(b-a)/2} f(x)\,dx$. Define $G(\alpha)$ in a similar manner. You have $F(\alpha)\le G(\alpha)$ which implies $\lim_{\alpha\rightarrow a^-} F(\alpha)$ is not infinite. Since $f\ge 0$, $F(\alpha_1)\ge F(\alpha_2)$ whenever $\alpha_1\le \alpha_2$. This implies the limit exists; and thus $f$ is improperly integrable "near $a$". You have to deal with $b$ too, of course.2012-12-10
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    @DavidMitra I understand that $\lim_{\alpha\rightarrow a^-} F(\alpha)$ is not infinite. And I understand that $F(\alpha_1)\ge F(\alpha_2)$ whenever $\alpha_1\le \alpha_2$. But I don't understand how this proves that the limit exist. Why does it suffices to consider only decreasing sequences that converge to a. I'm using Elementary Analysis by Ross. And in this book it says that I need to you prove it for every sequences $s_n$ with $\sin s_n=a$. I get intuitive idea that it suffices to only consider decreasing sequences with the left hand limit, but don't I need to prove that as well ?2012-12-11
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    @KasperPeulen Here is the proof that the limit exists (for arbitrary $(s_n)$): Suppose $\lim_{\alpha\rightarrow a^+} F(\alpha)$ does not exist. Let $L$ be the limit obtained by a decreasing subsequence of $(s_n)$ ($L$ exists by the Monotone Convergence Theorem). By assumption, we know $\lim_{\alpha\rightarrow a^+} F(\alpha)\ne L$. So there is an $\epsilon>0$ so that for every integer $M$, there is an integer $n(M)$ with $F(s_{n(M)}). From this, you can find a subsequence of $(s_n)$, call it $(q_n)$ such that $F(q_n) for all $n$.2012-12-11
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    @KasperPeulen Can you see how the existence of these two subsequences leads to a contradiction? $$ $$ Maybe it's easier to consider the limit at $b$. Here, you have an increasing function that is bounded above. The limit of the function at the "left of" $b$ must exist. (And,in my previous comments, I had some errors; At $a$, we want the limit from the right.)2012-12-11
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    Wow.. brilliant. Thank you so much :) It's sad that I can't upvote you, I really appreciate your answers.2012-12-11

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