Extending continuous functions from almost everywhere to everywhere

Question

If I have a function $f$ from $f: \mathbb{R}^d \to \mathbb{R}$ such that the integral of $f$ over every measurable set $E$ is greater than 0 then $f \ge 0$ almost every where.

If I have the assertion that $f$ is continuous can I extend this to be $f\ge 0$ everywhere and how do I show this?

Answer 1

Outline of the proof:

• Since you know that $f(x) \ge 0$ a.e., then another way of saying this is that $\{ x \in \mathbb{R}^d \mid f(x) \ge 0 \}$ has complement of measure 0. This implies that $\{ x \in \mathbb{R}^d \mid f(x) \ge 0 \}$ is dense in $\mathbb{R}^d$ (since otherwise, the complement would contain some nonempty open subset of $\mathbb{R}^d$, and would thus have positive measure).
• Also assuming $f$ is continuous, then we get that $\{ x \in \mathbb{R}^d \mid f(x) \ge 0 \} = f^{-1}([0, \infty))$ is closed.
• The only subset of $\mathbb{R}^d$ which is both dense and closed is all of $\mathbb{R}^d$.