If f is continuous and 0 < f(x) < g(x) on the interval [0, ∞) and $\int_0^\infty g(x) dx = M < \infty$ then $\int_0^\infty f(x) dx$ exists?

Question

If f is continuous and $0 < f(x) < g(x)$ on the interval $[0, ∞)$ and $\int_0^\infty g(x) dx = M < \infty$ then $\int_0^\infty f(x) dx$ exists.

True or False, and why? I'm not sure what the M means or how to use it.

Answer 1

This is true. $M$ is a real number for the value of integral. Let $h(t)=\int_0^tf(x)\: dx$. Since $$ 0<h(t)=\int_0^t f(x)\: dx<\int_0^t g(x)\: dx<\int_0^\infty g(x)\: dx < \infty $$ $h(t)$ is monotone increasing with upper bound. So $$ \lim_{t\to\infty}h(t)=\int_0^\infty f(x)\: dx $$ exists.