2
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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 $\mathcal{L}^{1}(C)=\lambda_{f}(\mathbb{R}\setminus C)=0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.