There is a question in my mathematical analysis book.
Suppose $f(x)$ is a monotonic function on $[0,+\infty)$, and $g(x)\not\equiv 0$ is a continuous and periodic function with period $T>0$ on $\mathbb R$. Prove that $\int_0^{+\infty} f(x) \,\text{d}x$ converges if and only if $\int_0^{+\infty} f(x)|g(x)|\,\text{d}x$ converges.
I can prove that if $\int_0^{+\infty} f(x) \,\text{d}x$ converges, we know $\int_0^{+\infty} f(x)|g(x)|\,\text{d}x$ converges. For the inverse direction, I have tried the following way, but I didn’t make it.
Since $g(x)\not\equiv 0$ is a continuous and periodic function with period $T>0$, we know there exist a $A>0$ and $[a,b]\subset (0,T)$ such that for any $x\in [a,b]$, there is $g(x)>A$. Then we know $$A\int_a^b f(x) \,\text{d}x \leq \int_a^b f(x)|g(x)|\,\text{d}x$$
That’s all what I do. Please do me a favor, thank you!
We define an integral $I(f)=\int_0^{\infty}f(x)\,\mathrm{d}x$ to converge iff the positive and negative halves $$I_+(f)=\int_0^{+\infty}\max(f(x),0)\,\mathrm{d}x\text{,}\quad I_-(f)=\int_0^{+\infty}\max(-f(x),0)\,\mathrm{d}x$$ both converge.
WOLOG suppose $f$ is positive and monotonically decreasing, then $$\int_0^Tf(x)\left|g(x)\right|\,\mathrm{d}x\geq\int_0^Tf(T)\left|g(x)\right|\,\mathrm{d}x=f(T)\int_0^T\left|g(x)\right|\,\mathrm{d}x$$ and as well $$\int_0^{T}f(x)\,\mathrm{d}x\leq\int_0^{T}f(0)\,\mathrm{d}x=f(0)T$$ which means $$\int_0^{kT}f(T)\,\mathrm{d}x\leq T\sum_{n=0}^{k-1}f(nT)\text{.}$$
You can finish the proof from here, using the contrapositive.