$|f(x)|<|g(x)|$ and $\int g(x)<\infty\Rightarrow\ \int f(x)<\infty$

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Let f,g continious functions on $[0,\infty)$ s.t $\forall x\in[0,\infty), |f(x)|\le|g(x)|$. Prove or give counterexample thtat if $\int_ 0^\infty g(x)dx<\infty$ then $\int_0^\infty f(x)<\infty$.

I think that's incorrect but I can't find any good example. If it'd be $|f(x)|\ge |g(x)|$, I could have given example of $g(x)=\frac {\sin{x}}{x}$ and $f(x)=\frac 1 x$. Which counterexample can I give? what is the intuition for such questions?

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Your example, with some modification, works. We just get rid of the $\frac{1}{x}$.

Let $g(x)=\frac{\sin x}{x}$, and let $f(x)=|g(x)|$. It seems clear that you are aware that $$\int_0^\infty \frac{\sin x}{x}\,dx$$ converges.

The fact that $$\int_0^\infty \left|\frac{\sin x}{x}\right|\,dx$$ diverges is likely familiar to you. If proof is needed, note that between $n\pi+\frac{\pi}{6}$ and $n\pi+\frac{\pi}{2}$, we have $|\sin x|\ge \frac{1}{2}$.