Can some one please give me an example of continuous Singular measure on Reals which is not absolutely continuous to cantor-type sets. Thank you.
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Consider the probability measure $\mu_p$ on the Borel sigma-algebra on $[0,1]$, which is the image of the product $\mathop{\otimes}\limits_{n\geqslant1}\beta_p$ of the Bernoulli measures $\beta_p=p\delta_1+(1-p)\delta_0$ on $\{0,1\}$ by the function $U$ defined on $\{0,1\}^\mathbb N$ by $U(x)=\sum\limits_{n=1}^\infty x_n/2^n$ where $x=(x_n)_{n\geqslant1}$.
Then for every $p\ne\frac12$, the measure $\mu_p$ is purely singular, $\mu_p$ has no atom, and every interval with positive (Lebesgue) length has positive measure for $\mu_p$.
In words, under $\mu_p$, the bits of a real number in $[0,1]$ are i.i.d. Bernoulli random variables with probability $p$ of being $1$. If $p=\frac12$, the measure $\mu_\frac12$ is the restriction of the Lebesgue measure to $[0,1]$ hence it is densitable, and this is the exception. Actually, the measures $\mu_p$ and $\mu_q$ are mutually singular for every $p\ne q$.
Did
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Can you elaborate further? – Yiorgos S. Smyrlis Dec 19 '13 at 12:10
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1Here is an excellent article related to classification of Continuous Singular measures : http://www.imath.kiev.ua/~mfat/html/papers/2011/2/al_ko_pra_to/art.pdf – chandu1729 Dec 19 '13 at 12:31
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@YiorgosS.Smyrlis Sure. Which points? – Did Dec 19 '13 at 13:46
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@Did What kind of product are you refering to? – Yiorgos S. Smyrlis Dec 19 '13 at 13:54
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@YiorgosS.Smyrlis The product measure, naturally. – Did Dec 19 '13 at 13:56
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@Did The product of all measures $p\delta_1+(1-p)\delta_0$, $p\ne1/2$? Two many such measures! – Yiorgos S. Smyrlis Dec 19 '13 at 13:59
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@YiorgosS.Smyrlis No, the product $\otimes_{n\in\mathbb N}\mu_n$ where, for every $n\in\mathbb N$, $\mu_n=p\delta_1+(1-p)\delta_0$. – Did Dec 19 '13 at 14:02
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@chandu1729: by any chance do you know other references for classification of continuous singular measures, maybe a textbook on singular measures? – Sergio Parreiras May 15 '14 at 16:00
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So you're saying that you have $U:{0,1}^\mathbb{N}\rightarrow[0,1]$, where on ${0,1}^\mathbb{N}$ you have the measure $\beta=\otimes_p\beta_p$. This induces a measure $\mu_p$ on $[0,1]$ via preimage of $U$ as: $\mu_p(A)=\beta(U^{-1}(A))$? – rod Apr 22 '21 at 13:05
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Also, to prove that $\mu_p$ is singular w.r.t. the Lebesgue measure, can you observe that $\mathcal{L}(\mathbb{Q}\cap[0,1])=0$ but $\mu_p((\mathbb{Q}\cap[0,1])>0$? – rod Apr 22 '21 at 13:17
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Let $C$ be the usual Cantor set, and $\nu_C$ its probability measure. Clearly $\nu_C(A)=0$, where $A$ is the subset of $[0,1]$, consisted of the numbers in which the digit 1 appears in their ternary expansion, while the Lebesgue measure of $A$ is 1. The support of $\nu_C$ is the set of the numbers in which the digit 1 does not appears in their ternary expansion. It is not hard to modify $\nu_C$ so that its support is dense in $[0,1]$ and is still a singular measure. (Singular w.r.t. the Lebesgue and Dirac measures.). So for every Borel set $E\subset\mathbb R$ define: \begin{align} \nu_0(E)&=\nu_c(E) \\ \nu_1(E)&=\nu_0(E)+\tfrac{1}{2}\nu_C\big(\tfrac{1}{3}E+\tfrac{1}{3}\big) \\ \nu_2(E)&=\nu_1(E)+\tfrac{1}{2^3}\nu_C\big(\tfrac{1}{9}E+\tfrac{1}{9}\big)+\tfrac{1}{2^3}\nu_C\big(\tfrac{1}{9}E+\tfrac{7}{9}\big) \\ \nu_3(E)&=\nu_2(E)+\tfrac{1}{2^5}\left(\sum_{i=1}^4\nu_C\big(\tfrac{1}{3^3}E+\tfrac{k_{3,i}}{3^3}\big)\right) \end{align} where $k_{3,i}=1,7,19,25$, and in general $$ \nu_{n+1}(E)=\nu_n(E)+\tfrac{1}{2^{2n+1}}\left(\sum_{i=1}^{2^n}\nu_C\big(\tfrac{1}{3^{n+1}}E+\tfrac{k_{n+1,i}}{3^{n+1}}\big)\right), $$ where $k_{n+1,i}-1$ runs in the sets of the numbers $\{1,2,\ldots,3^n\}$, which miss 1 in their ternary expansion.
Clearly this sequence $\nu_n$ converges, the the measure variation norm, to a positive Borel measure $\nu$, supported in the set of the elements of $[0,1]$ with finite number of aces in their ternary expansion, with $\nu([0,1])=2$. This measure is singular, as its support has zero Lebesgue measure, and it does not contain any Dirac measures, and none of the $\nu_n$ does.
However, I am not sure that you are looking for something like that.
Yiorgos S. Smyrlis
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