If $f:\mathbb{R}\to \mathbb{R}$ is a measurable function such that $$\int_{[a,b]}f\, dm>0$$ for all interval $[a,b]$ implies $f>0$ almost everywhere.
My attempt so far:
$$A:=\{x\in[a,b]: f(x)<0\} = \bigcup_{n=1}^\infty \{x\in[a,b]:f(x)< \frac{-1}{n} \}$$ If we assume $m(A)>0$, then there exists $m\in \mathbb{N}$ such that $$m(A_n) = m(\{x\in[a,b]:f(x)< \frac{-1}{n} \})>0$$ This implies $$ \int_{A_n}f \,dm \leq \int_{A_n}\frac{-1}{n} \,dm <0$$ But this is not a contradiction because $A_n$ is not a closed interval. Any suggestions?
Let $\mu$ be a set function defined on the Lebesgue measurable sets that $\mu(A)=\displaystyle\int_{A}fdm$, then $\mu(A)>0$ for any finite interval. Let $\lambda$ be the restriction of $\mu$ to finite open intervals. Then the Caratheodory extension $\lambda^{\ast}=\mu$ on finite open intervals, so they agree on the sigma algebra generated by finite open intervals, which is essentially the Borel sets. So $\mu(B)=\lambda^{\ast}(B)>0$ for any Borel set. For Lebesgue measurable set $S$, write $S=B\cup N$ with null set $N$, and now we have $\mu(S)=\mu(B)>0$.