2

Suppose that $(X,E,μ)$ is a non-atomic probability measure space. Let $\xi :X \to \mathbb{R}$ be a random variable. A Borel measure $\mu_{\xi}$ in $\mathbb{R}$ defined by $\mu_{\xi}(X)=\mu(\xi^{-1}(X))$ for $X \in B(\mathbb{R})$ is called a Borel probability measure on $\mathbb{R}$ defined by $\xi$.

Question: Does there exists a measurable function $\xi :X \to \mathbb{R}$ such that a Borel probability measure $\mu_{\xi}$ is non-atomic?

Conifold
  • 11,756
George
  • 1,541
  • 2
    There exists $\xi$ so that $\mu_\xi$ is any probability measure you want: http://math.stackexchange.com/questions/31962/how-to-split-an-integral-exactly-in-two-parts/31967#31967 –  Jul 17 '14 at 19:25
  • @Byron Schmuland Thank you very much for your information. – George Jul 18 '14 at 18:06
  • This answer has an interesting consequence: (AC). If $card(a)<c$ (here $c$ denotes the candinality of the continuum), then there does not exist a non-atomic probability measure $\mu$ defined on the power set of $a$. – George Jul 19 '14 at 09:56
  • [email protected] I specify mine comments:_$(AC)&(MA)&\neg (CH)$.If $card(a)<c$ (here $c$ denotes the candinality of the continuum), then there does not exist a non-atomic probability measure $μ$ defined on the power set of $a$. – George Jul 19 '14 at 16:51
  • Proof.Assume the contrary and $\mu$ and $a_0$ be such objects. Let $\xi: a \to R$ be a gaussian random variable on $a$. Then the cardinality of $\xi(a)$ is less then $c$ and by well known theorem it's Lebesgue measure is equal to zero. Let $B_0$ be a Borel subset of $R$ of Lebesgue measure zero which contains $\xi(a)$. Then it's Gaussian measure also will be zero and we get that $1=\mu({ x:\xi(x)\in R})=\mu({x:\xi(x)\in B_0})=\mu_{\xi}(B_0)=0$. This is a contradiction._ – George Jul 19 '14 at 16:52

0 Answers0