A measure zero uncountable set for the Lebesgue-Stieltjes measure $\mu_F$.

Question

Let $\mu_F$ be the Lebesgue-Stieltjes measure on $\Bbb R$ associated with the increasing function $F:\Bbb R\to \Bbb R$. Construct an uncountable set of measure 0 for $\mu_F$.

When $F(x)=x$ and we get the Lebesgue measure on $\Bbb R$ and the Cantor set in the interval $[0,1]$ is an example.

My guess is that the same argument can be repeated for $\mu_F$, if there is an interval $[a,b]$ where $F$ is continuous (hence making sure that any singleton set has measure 0), and construct a "Cantor set" there.

But an increasing function can have a dense set of discontinuities as shown in here.

What can one do in this case?

Answer 1

Radon-Nikodym ensures that for the Lebesgue measure $m$, one can express $\mu_F=\lambda+\nu$ where $\lambda\perp m$ on some set $E$ with $m(E^c)=0=\lambda(E)$ and $\nu\ll m$. By regularity, there exists a compact $K\subset E$ with $m(K)>0$. Define $f(x)=m(K\cap (-\infty,x])$ and notice that for $x<y$, $f(y)-f(x)=m(K\cap (x,y])\le y-x$ so $f$ is continuous. Since $K$ is compact, $f(x)=0$ for large negative $x$ and $f(x)=m(K)$ for large positive $x$. By the IVT, there exist $a<b$ s.t. $f(a)=\frac{m(K)}3$ and $f(b)=\frac{2m(K)}3$, in particular, $m(K\cap(-\infty,a])=m(K\cap[b,\infty))=\frac{m(K)}3$. Thus $K\cap (-\infty,a]$ and $K\cap [b,\infty)$ are compact with positive measure. Repeating this process inductively yields a $C\subset E$ analogous to the Cantor set (uncountable Lebesgue null set) and therefore $\mu_F(C)\le \lambda(E)+\nu(C)=0$.