The integral of a function multiplied with an unbounded function converges if the product is a bounded function

Question

Suppose $f:[1,\infty]\rightarrow \mathbb{R}$ is such that $g(x):=x^2f(x)$ is a bounded function. Prove that $\int_1^\infty f$ converges.

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Intuitively I think it seems reasonable. As $x^2$ is increasing and unbounded above, $f(x)$ must be a decreasing function with higher absolute slope than $x^2$ so as $g(x)$ can be bounded. Not sure how to proceed further from here.

Answer 1

Let $|g(x)|\le M$ (assumption it is bounded). Then $|f(x)|\le \frac{M}{x^2}$ so $\int_1^\infty|f(x)|dx\le M\int_1^\infty\frac{dx}{x^2}=M$. Therefore $f(x)$ is integrable.