BV[a,b]$\cap$C[a,b]$\neq$AC[a,b]

Question

Take an $f\in$BV[0,1]$\cap$C[0,1] e.g. the Cantor function. I take the Lebesgue Stiltigies measure of $f$:

$$ \mu_f((a,b])=f(b)-f(a). $$

Now I have a finite positive measure, so I can do the Radon-Nikodim decomposition:

$$ \mu_f=\mu_1+\mu_2 $$

(I denote | $\ $| the Lebesgue measure)

With $\mu_1\perp |\ \ |$ and $\mu_2<<|\ \ |$. So $\mu_2= h $ d$|\ \ |$

I have that $\mu_f(\{x\})=0$ for all $x\in[0,1]$ because $f\in$C[0,1].

• Can I say that $\mu_1=0$?

• In this case, why have I that $f$ is not AC[0,1]? $f(x)=\mu_2([0,x])=\int_0^xh $ d$|\ \ |$ for all $x\in[0,1]$.

Thank you in advance! I think that the point is that $\mu_1\neq 0$ but I don't understand how a measure without 'atoms' could have the singular part not $0$.

Answer 1

A basic property of the Cantor function $f$ is $f'(x)=0$ a.e. (w.r.t Lebesgue measure). If $\mu_1=0$ then $\mu_f << ||$ which is equivalent to the fact that $f$ is an absoluetly continuous function. If $f$ is absolutely continuous then $f(b)-f(a)=\int_a^{b}f'(t)dt=0$, a contradiction.