Let $ f $ be a monotonically-decreasing non-negative function satisfying $ \displaystyle \lim\_{x \to \infty} f(x) = 0 $. Is it true that the following claim holds?



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> **Claim:** There exists a function $ g $ such that $ \displaystyle \int\_{0}^{\infty} g(x) ~ \mathrm{d}{x} = \infty $ and $ \displaystyle \int\_{0}^{\infty} f(x) g(x) ~\mathrm{d}{x} < \infty $.
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Note that we can assume w.l.o.g. that $ \displaystyle \int\_{0}^{\infty} f(x) ~\mathrm{d}{x} = \infty $, as otherwise $ g \equiv 1 $ does the job. My idea for a proof was to split $ [0,\infty) $ into intervals $ I\_{n} \stackrel{\text{df}}{=} [x\_{n},x\_{n + 1}) $, so that $ \displaystyle \int\_{I\_{n}} f(x) ~ \mathrm{d}{x} = \frac{1}{n} $, and define $ g $ by $ \displaystyle g(x) \stackrel{\text{df}}{=} \sum\_{n = 0}^{\infty} \frac{1}{\operatorname{length}(I\_{n}) \cdot n} \mathbf{1}\_{[x\_{n},x\_{n + 1})} $, where $ \operatorname{length}(I\_{n}) \stackrel{\text{df}}{=} x\_{n + 1} - x\_{n} $.


Unfortunately, this doesn’t seem to work as one can construct a piecewise-constant function $ f $ so that $ \operatorname{length}(I\_{n}) \rightarrow \infty $, which seems to kill my original approach (which, roughly speaking, was to construct $ g $ so that $ \displaystyle \int\_{0}^{\infty} f(x) g(x) ~ \mathrm{d}{x} \leq \sum\_{n = 1}^{\infty} \frac{1}{n^2} $).



For every monotonically-decreasing non-negative function $ f $, does there exist a function $ g $ so that $ f g $ is integrable?
On the positive half of the $x$-axis, build mutually disjoint triangular "spikes" of base $1/n$ and height $2n$ $(n=1, 2, 3, ...)$ spreading them sufficiently far in the positive direction of the $x$-axis so that when multiplied by $f(x)$, the area of the $n$-th flattened spike becomes smaller than $2^{-n}$. This can be done since $f(x)$ converges to $0$ as $x\to\infty$. There is your $g(x)$.


Remark: I just edited this answer. The assumption of monotonicity of $f$ is redundant. The construction of $g$ does not require it. It suffices that $\displaystyle\lim\_{x\to\infty}f(x)=0$.


