If $\int_0^{x} g \leq \int_0^x f$ and $\phi$ is nonincreasing then $\int_0^{\infty} \phi g \leq \int_0^\infty \phi f$

Question

Let $f, g$ be measurable real-valued functions on $[0, \infty)$, with $$\int_0^{x} g \leq \int_0^x f$$ for each $x$. Show that if $\phi: [0, \infty) \rightarrow [0, \infty)$ is nonincreasing, then also $$\int_0^{\infty} \phi g \leq \int_0^{\infty} \phi f$$ I just got no idea where to start. I know that $\phi$ must decrease to a limit as $x \rightarrow \infty$. Can someone please give me a hint?

Answer 1

Assume first that $g(x)\to0$ when $x\to+\infty$. Then, a powerful tool is that there exists some nonnegative measure $\mu$ such that, for every nonnegative $x$, $$\phi(x)=\int_x^\infty\mu(\mathrm dy).$$ Then, by Tonelli, $$\int_0^\infty\phi(x)g(x)\mathrm dx=\int_0^\infty g(x)\int_x^\infty\mu(\mathrm dy)\mathrm dx=\int_0^\infty\left(\int_0^yg(x)\mathrm dx\right)\mu(\mathrm dy),$$ and the comparison of the inner parenthesis with the analogous quantity for $f$, yields the result.

Finally, if $g(x)\to c$ when $x\to+\infty$ with $c\gt0$, this adds $c$ times the integrals on $(0,+\infty)$ of each function hence the comparison of the integrals carries through. (Alternatively, to subsume this in the first case, consider that $\mu$ may have an atom at $+\infty$.)