# Comparing decay of integrable and non integrable function

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 $$\lim_{x\to\infty}\frac{\widehat{f}(x)}{g(x)}=0$$.

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

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.