# comparison test for improper integrals

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)? – David Mitra Dec 6 '12 at 23:44
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 ? – Kasper Dec 6 '12 at 23:53
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 $\infty$. – André Nicolas Dec 6 '12 at 23:59
Yes, if I understand you correctly. (The first integral's value increases as $c\rightarrow a^-$, e.g.) – David Mitra Dec 7 '12 at 0:01
@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? – Kasper Dec 10 '12 at 17:37