Singular Lebesgue-Stieltjes measure

Question

I read that if $F:\mathbb{R}\to\mathbb{R}$ is a non-decreasing singular function, i.e. a non-decreasing continuous function such that $F′(x)=0$ almost everywhere, then the Lebesgue-Stieltjes measure $\mu_F$, defined as the Lebesgue extension of the interval measure $m(a,b)=m[a,b)=m(a,b]=m[a,b]=F(b)−F(a)$, is such that $\mu_F(A)=0$ for the set $A:=\{x\in\mathbb{R}:F′(x)=0\}$ (which is complement of a set $B$ of Lebesgue measure $\mu(B)=0$) where its derivative is 0.

By definition of $\mu_F$ I understand that $$\mu_F(A)=\inf\sum_k F(b_k)−F(a_k)$$ where $[a_k,b_k]$ are intervals containing $A$, but I do not know how to see that this infimum is 0. If we could find such intervals such that $F′$ is continuous on $[a_k,b_k]$ the desired property would follow from $F(b_k)−F(a_k)=\int_a^b F′(t)dt=0$ where the integral is Riemann's. By using the Lebesgue integral instead, I can only see that, since $F$ is monotonic, $F(b_k)−F(a_k)\geq\int_{[a_k,b_k]}F′d\mu=0$ which does not help me to show what I want... Does anybody know how to prove that such $μ_F$ is zero where $F′$ is zero? I thank you so much!

Answer 1

If you had $\mu_F(A)>0$, then also $\mu_F(K)>0$ for a compact subset $K\subset A$ by regularity. For each $x\in K$, find an interval $I_x$ centered at $x$ with $\mu_F(I_x)<\epsilon |I_x|$, and cover $K$ by finitely many of these. By deleting intervals if necessary, we can also assume that no point of $\bigcup I_{x_j}$ lies in more than two intervals, and we can keep the diameter of this union bounded.

It now follows that $$ \mu_F(K)\le\mu_F\left( \bigcup I_{x_j}\right) \le \epsilon \sum |I_x| \le 2\epsilon \left| \bigcup I_{x_j} \right| \le C\epsilon . $$ Since we can do this for any $\epsilon>0$, the claim follows.

Answer 2

Some more explanation on how to find the $I_x$ in the answer by @user138530: We know that $F'(x) = 0$ at $x \in K$, then the lower limit and upper limit are equal, i.e., $\lim\limits_{\vert h\vert \to 0}\sup_{0<\vert t \vert<\vert h\vert}\frac{F(x+t)-F(x)}{t} = \lim\limits_{\vert h\vert \to 0}\sup_{0<\vert t \vert<\vert h\vert}\frac{F(x+t)-F(x)}{t} = F'(x) = 0$. Taking $\vert h \vert$ small we can find a small interval around $x$ and t small enough, where $\vert\frac{F(x+t) - F(x)}{t} \vert < \epsilon$ (this is by the limit of the fraction is 0), and thus $\mu_{F}((x,x+t]) = \mu_{F}((x,x+t)) \text{ (since $F$ is even continuous everywhere in this case, so Stieltjes measure has no atom)} = \vert F(x+t)-F(x)\vert<\epsilon t = \epsilon \vert (x,x+t) \vert$ Apply the same argument on $\vert F(x)-F(x-t)\vert$ and we get the desired result.

To choose the constant $C$, we can take some large interval $[-K,K] $ to bound $[0,1]$, and we can also WLOG assume the interval cover has union inside $[-K,K]$ (because removing parts outside of $[-K,K]$ makes no change to the argument) for this fixed $K$.

In user138530's answer, we can even WLOG assume the finite subcover $\lbrace I_{x_j} \rbrace_{j=1}^{n}$ has empty intersection among any two, since $F$ is both left and right continuous, then $\mu_F$ has no atom: After excluding the common part of the two, we can delete finitely many single points, say, the boundary points, without changing the resulting measure.