Lebesgue integral preserves the inequality

Question

Suppose $f$ and $g$ are measurable functions such that $f\le g$ and the Lebesgue integral $\int g$ is finite. By finite I mean that $\int g$ is a real number. I want to show that $\int f$ is also finite. This result is intuitive but I don't know how to prove it formally. The only way that comes to my mind is restoring to the definition of Lebesgue integral but it seems that it's unnecessary lengthy proof. Is there any shorter proof for this result?

Answer 1

Take $g=0$ and $f(x) =-1$. Then $g$ is integrable and $f \leq g$ but $f$ is not integrable. ($\int_{\mathbb R} f(x)dx=-\infty$)

If $0 \leq f \leq g$ then $0 \leq \int fd\mu \leq gd\mu<\infty$.