A lebesgue measurable set $E$ with $\mu(E)=1$ and $\mu_F(E) = 0$.

Question

I have to prove the following: suppose that $F : [0, 1] → \mathbb{R}_+$ is a right continuous increasing function and $F' = 0$ almost everywhere. Let $\mu_F$ be such that $\mu_F((a, b]) = F(b) − F(a)$ for all $0 ≤ a < b ≤ 1$. Show there is a Lebesgue measurable set $E$ with $\mu(E)=1$ such that $\mu_F(E) = 0$.

Following a hint, I have been trying to use Vitali’s covering lemma on the set $\{ F'=0\}$ to show that for any positive $\varepsilon>0$ and $\delta>0$ there is a set $G$ such that $\mu(G) ≥ 1 − \varepsilon$ and $\mu_F(G) ≤ \delta$, but I have not been able to go much further.

Answer 1

I found that the $5r$-covering lemma may help here.

Lemma: Let $\mathcal{B}$ be a family of balls in a separable metric space such that $$ \sup\{diam(B) : B \in \mathcal{B}\} < \infty. $$ Then there is a countable, pairwise disjoint sequence $\{B_i\}_{i=1}^\infty$ s.t. $$ \bigcup_{B\in \mathcal{B}}B\subseteq \bigcup_{i=1}^\infty 5B_i. $$

Now, we may show the following result (which can be applied to your case):

Claim: If $F$ is a distribution function and $E$ is a measurable subset of $(0,1)$ s.t. $F'\mid_E=0$, then $\mu_F(E)=0$.

Proof. Fix $\varepsilon>0$. Since $F'\mid_E=0$ for $x\in E$, there exists $t_x>0$ s.t. $|F(x+t)-F(x)|\le \varepsilon|t|$ for all $t\in[-t_x,t_x]$. Then take $0<u_x\le Ct_x$ for some $C<1/5$ and consider a family of balls

$$ \mathcal{B}=\left\{B_{u_x}(x):x\in E\right\}. $$

By the lemma there exists a countable pairwise disjoint sequence of balls s.t. $E\subseteq \bigcup_{i=1}^\infty B_{5u_i}(x_i)$. Thus,

$$ \mu_F(E)\le \sum_{i=1}^{\infty}\mu_F\left(\left(x_i-5u_i,x_i+5u_i\right]\right) \\ = \sum_{i=1}^{\infty}\left[F(x_i+5u_i)-F(x_i-5u_i)\right] \le 5\varepsilon\sum_{i=1}^{\infty}2u_i\le 5\varepsilon. $$

Sending $\varepsilon\downarrow 0$ implies the result.