Borel measure generated by right-continuous increasing function Let $F:I→R_+$ be a right-continuous increasing function and $F'=0$ almost everywhere, and let $μ_F$ be the Borel measure generated by $F$ so that $μ_F(a,b] = F(b)−F(a)$ for all $a,b \in I$. We want to show that there exist a Lebesgue measurable set $E$ of measure $1$ with $μ_F(E)=0$.
I can guess that $E$ should be $E=\{F'=0\}$, and I have computed that $μ_F(a,b)=F(b^-)−F(a)$, $μ_F[a,b]=F(b)−F(a^−)$, and $μ_F\{a\}=F(a)−F(a^−)$. But I don't know how to use the condition $F'=0$ almost everywhere to help proving the thing. Can someone help?
 A: Given how you state what you want to prove I'm guessing that $I=(0,1)$. You shouldn't make us guess about these things...
Suppose $K\subset E$ is compact and $\epsilon>0$. Since $\mu_F$ is regular it's enough to show that $\mu_F(K)<\epsilon$. Since $K$ is compact and $F'=0$ at every point of $K$, you can write $$K\subset\bigcup_{j=1}^n(a_j,b_j)$$where $$\frac{F(b_j)-F(a_j)}{b_j-a_j}<\epsilon$$and $(a_j,b_j)\subset I$. If three open intervals have a point in common then one of them is contained in the union of the other two; thus we may assume that no point is in more than two of the intervals, so that $$\sum(b_j-a_j)=\int_0^1\sum\chi_{(a_j,b_j)}\le\int_0^12=2.$$ So $$\mu_F(K)\le\sum\mu_F((a_j,b_j])=\sum(F(b_j)-F(a_j)<\epsilon\sum(b_j-a_j)\le2\epsilon.$$

For the sake of completeness, a proof that if three open intervals have a point in common then one of them is contained in the union of the other two: Suppose that $x\in\bigcap_{j=1}^3(a_j,b_j)$. Let $a_n=\min(a_1,a_2,a_3)$ and $b_m=\max(b_1,b_2,b_3)$. Then $$\bigcup_{j=1}^3(a_j,b_j)=\bigcup_{j=1}^3((a_j,x]\cup[x,b_j))=(a_n,x]\cup[x,b_m)\subset(a_n,b_n)\cup(a_m,b_m).$$
