Lebesgue measurable function, positive everywhere, has integral zero

Question

Suppose $f:[0,1] \rightarrow \mathbb{R}$ is a Lebesgue-measurable function. Also suppose that $f(x) > 0$ for all $x$. Is it possible that $\int_0^1 f(x) \, dx = 0$?

Answer 1

No, it's not. Consider the sets $E_n = \{x \in [0,1]: f(x) > 1/n\}$. Since $f(x) > 0$ for all $x \in [0,1]$ we can write $$[0,1] = \bigcup_{n=1}^{\infty} E_n.$$ Since $\lambda([0,1]) = 1 > 0$ there must exist a $m \in \mathbb{N}$ such that $\lambda(E_{m}) > 0$. Now notice that $$\int_{[0,1]} f \;d\lambda \geq \int_{E_{m}} f \;d\lambda \geq\int_{E_{m}}\frac{1}{m} = \frac{\lambda(E_m)}{m} > 0$$