If $0\leq f(x)\leq g(x)$, and $\int^{\infty}_{a}g(x)dx$ diverges, so does $\int_a^\infty f(x)\,dx$?

Question

If $0\leq f(x)\leq g(x)$, and $\int^{\infty}_{a}g(x)dx$ diverges, so does $\int_a^\infty f(x)\,dx$?

I need to find a counterexample for this statement since it is False.

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The divergent theorem is actually the other way around, if $\int_a^\infty f(x)\,dx$ diverges, then $\int_a^\infty g(x)\,dx$ diverges. Thus, the statement in the question is False. I cannot think of any counterexamples. How should I proceed?

Answer 1

L. Lhier's second answer ($g(x) = 1$ and $f(x) = 0$) is probably the minimal one. But you could look at $p$-integrals too. Let $f(x) = \frac{1}{x^2}$ and $g(x) = \frac{1}{x}$. Then $0 \leq f(x) \leq g(x)$ for all $x \geq 1$, and $\int_1^\infty g(x)\,dx$ diverges, but $\int_1^\infty f(x)\,dx$ converges.

Intuitively speaking, an improper integral (at least for a nonnegative function) is attempting to measure the area of an infinite region: the region under the graph, above the $x$-axis, and to the right of $x=a$. If $0 \leq f(x) \leq g(x)$, the region below the graph of $f$ is contained in the region below the graph of $g$. Therefore, if the region below the graph of $g$ has finite area (the integral converges), the region below the graph of $f$ must also have finite area. Contrapositively, if the region below the graph of $f$ has infinite area (the integral diverges), the region below the graph of $g$ must have infinite area. But if the region below the graph of $g$ has infinite area, the region below the graph of $f$ could have finite or infinite area.

Answer 2

For $a \geq 1$ take $g(x)=1$ and $f(x)=e^{-x}$. Or should $a$ be arbitrary?