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If we have two positive, decreasing functions $f,g:[a,\infty)\to\mathbb{R}$ such that $$\int_{a}^{\infty}f(x)\text{d}x<\infty\;\;\;\text{and}\;\;\;\int_{a}^{\infty}g(x)\text{d}x=\infty$$ can we infer that there exists a function $\widehat{f}$ with $\widehat{f}(x)=f(x)$ for almost all $x$, such that we have $\lim_{x\to\infty}\frac{\widehat{f}(x)}{g(x)}=0$?

Generally $\lim_{x\to\infty}\frac{f(x)}{g(x)}=0$ does not hold, a counterexample was provided here. But in this counterexample one can also choose a null set $A$, so that for $\widehat{f}:= f\mathbb{1}_{A^C}$ we have $\lim_{x\to\infty}\frac{\widehat{f}(x)}{g(x)}=0$.

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  • $\begingroup$ Oops, I missed "decreasing", sorry. $\endgroup$ Commented Mar 23, 2021 at 19:36

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No. Take $a=1$. We'll have $$f(x) = \sum_{n=1}^\infty a_n1_{[n,n+1)}(x)$$ $$g(x) = \sum_{n=1}^\infty b_n1_{[n,n+1)}(x)$$ for decreasing sequences $(a_n)_n,(b_n)_n$. If $\lim_{n \to \infty} \frac{a_n}{b_n} \not = 0$, then there does not exist $\widehat{f}$ with what you want. Let $a_n = \frac{1}{n^2}$ and just choose $(b_n)_n$ to sometimes match $a_n$ and then stabilize for a while.

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