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

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?

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 $$x, $$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 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 $$\mu_F(C)\le \lambda(E)+\nu(C)=0$$.
• (+1) In particular, $\lambda$ is the measure supported on the set of (countable) discontinuities $D$ of $F$ and $\nu$ is the continuous part. For any Borel set $A \in \mathcal{B}(\mathbb{R})$ we have $\mu_F(A) = m^\ast(F(A)) + \underbrace{\sum\limits_{p \in A \cap D} \mu_F(\{p\})}_{ = \lambda(A)}$.
• @r9m I don't think this description of $\lambda$ is accurate. If $F$ is the Cantor function (which is continuous) then $\lambda=\mu_F$ and $\nu=0$. (And $E^c$ is the Cantor set.) Commented Oct 24, 2020 at 7:15