Prove that if $f$ is integrable on $\mathbb{R}^d$ and $\int_E f\geq 0$ for every $E$ measurable set prove $f(x)\geq 0$ almost everywhere.
Defining $F=\{x:f(x)<0\}$ I have managed to show that $\int_Ff=0$. However, there is a line a proof I am reading that follows on from $\int_Ff=0$ saying "Since $f<0 \text{ for all } x \in F$, we conclude that $m(F)=0$" and that doesn't really make sense to me.
Question: I know that $\int_Ff=0$ implies $f=0$ almost everywhere in $F$. But how does one get from that to $m(F)=0$?
Proof I am reading:
You know that $\int_Ffdx=0$ and that $f(x)<0$ for all $x\in F$. This also means that $f(x)\neq0$ for all $x\in F$. The fact that the integral over $F$ vanishes tells two things: $f$ is zero a.e. or...