$ f,g $ are nonnegative and locally integrable on $ [a,b) $ and
$ L := \lim_{x\to b-}\frac{f(x)}{g(x)}\ $ exists as extended real number.
If $ 0 < L \le \infty $ and $g$ is not improperly integrable on $[a,b)$, prove that neither is $f$.
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Here is my line of thought:
If $g$ is not improperly integrable, then the $ lim_{x\to b-} g(x)$ is undefined or $= +\infty$, since $g$ is nonnegative. So if $f$ were improperly integrable on $[a,b)$, then it's limit would exist and be finite, therefore the limit of $ \frac{f(x)}{g(x)}$ would go to $0$, since g diverges. That is a contradiction since $ L > 0 $.
However, if we assume that $f$ is not improperly integrable, then the limit of $ \frac{f(x)}{g(x)}$ would be $ \frac{\infty}{\infty} $. I don't know what to do from there since that doesn't exactly show that $ 0 < L \le \infty $ right? Or does it, and is that the end of my proof?
$$\lim_{x\rightarrow+\infty}\frac{x^{-2}}{x^{-1}} = 0$$ So this limit exists as a real number. But x^{-2} is integrable and x^{-1} isnt. What you're trying to prove doesn't hold if the limit is zero.
Let's answer your question with the additionnal assumption that the limit is a strictly positive extended real number.
$f(x) = g(x)\frac{f(x)}{g(x)}$. This function $\frac{f(x)}{g(x)}$ converges toward a strictly positive extended real number. Thus by the definition of a limit (separating the cases where the limit is finite or infinite) we have $c>0$ and $d in [a,b)$ such that for $x \in [d,b)$, $\frac{f(x)}{g(x)}>c$.
So for $x \in [d,b)$, $f(x) = g(x)\frac{f(x)}{g(x)} > cg(x)$. Can you prove that $f$ isn't integrable now ?