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

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 $$0$$.

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.
• Only now I understand that could exist a measure $\mu_1$ that has not atoms but is $\perp\ |\ \ |$ – marco 2 days ago
• Like the function $\mu_1$ in the example that has support in the Cantor set (and $|C|=0$) – marco 2 days ago