# the Lebesgue-Stieltjes measure induced by the Cantor-Lebesgue-Vitali function

Considering the extension on the whole $\mathbb{R}$ of the Cantor-Lebesgue-Vitali function (obtained defining $f(x)=0 \quad \forall x \leq 0$ and $f(x)=1\quad \forall x\geq 1$), I have to prove that the Lebesgue Stieltjes measure $\lambda_f$ induced by $f$ on $\mathbb{R}$, is singular mutually with $\mathcal{L^1}$.

Since I prooved that $\exists f'(x)=0 \quad q.o. x \in [0,1]$ (and so it is q.o. in $\mathbb{R}$ since $f$ is constant outside $[0,1]$) , I deduced that $$0=\frac{d \lambda_f}{d \mathcal{L}^1}(x) \quad q.o. x \in \mathbb{R}$$ But then I conclude that $$\lambda_f(\mathbb{R})=\int_\mathbb{R} \frac{d \lambda_f}{d \mathcal{L}^1}(x) dx=0$$ So the Lebesgue-Stiltejes measure induced by the Cantor-Lebesgue-Vitali function is the null measure almost everywhere (and so the thesis is prooved)...but it seems to me a strong sentence, so maybe I'm wrong somewhere...

Could somebody give me a little help please?

The problem is that $\lambda_{f}$ is not absolutely continuous w.r.t. $\mathcal{L}^{1}$. Indeed, if $f\colon\mathbb{R}\to\mathbb{R}$ is a bounded nondecreasing function, $f$ satisfies the Integral Calculus Identity, that is $f(y)-f(x)=\int_{x}^{y}f'(t)dt$ for every $x,y\in\mathbb{R}$, with $x<y$, if and only if $f$ is continuous and $\lambda_{f}<<\mathcal{L}^{1}$. Now, the extension of the Cantor-Lebesgue-Vitali function is continuous but it does not satisfy ICI, thus $\lambda_{f}$ cannot be absolute continuous w.r.t. $\mathcal{L}^{1}$. In particular, you cannot apply the Radon-Nikodym's theorem on the whole measure $\lambda_{f}$, as you have done in the last step. Instead, you can use the RN theorem on the absolutely continuous part $\lambda_{f,ac}$. Then, you have $0=\frac{d\lambda_{f,ac}}{d\mathcal{L}^{1}}(x)$ $\mathcal{L}^{1}$-a.e. $x\in\mathbb{R}$, whence $\lambda_{f,ac}(\mathbb{R})=\displaystyle\int_{\mathbb{R}}\frac{d\lambda_{f,ac}}{d\mathcal{L}^{1}}(x)dx=0$, thus $\lambda_{f,ac}\equiv0$. Now, by the Lebesgue's decomposition theorem, there exists a unique decomposition of $\lambda_{f}$ into the absolutely continuous part $\lambda_{f,ac}$ w.r.t. $\mathcal{L}^{1}$ and the mutually singular part $\lambda_{f,s}$ w.r.t. $\mathcal{L}^{1}$, that is $\lambda_{f}=\lambda_{f,ac}+\lambda_{f,s}=0+\lambda_{f,s}=\lambda_{f,s}$, and the proof is accomplished.
Actually, the Cantor set $C$ is such that $\mathcal{L}^{1}(C)=\lambda_{f}(\mathbb{R}\setminus C)=0$.