0
$\begingroup$

I encountered the following question in Kai Lai Chung's A Course in Probability Theory:

Let $X:\Omega\to\mathbb R$ be a random variable. Is it true that $X(\Omega)$ must be Borel measurable in $\mathbb R$? Or even weaker, Lebesgue measurable?

My intuition is that this is false, and I hope to use the classical Cantor set construction: Let $g(x)=f(x)+x$, $f(x)$ being the Cantor function. Then $g:[0,1]\to [0,2]$ is a homeomorphism, and $g(C)$ has Lebesgue measure 1. Take $N\subseteq g(C)$ be nonmeasurable, and then define $$ X(\omega):g(\omega)1_{g^{-1}(N)}(\omega) $$ Then $X(\Omega)=N$, but the problem is that $X$ is not Borel measurable.

Anyone knows how to modify the above example? Thanks!

  • 1
    What if you take $\Omega$ to be a non measurable set (Lebesgue or Borel), endowed with the power set so every function is measurable. Then define $X(\omega)=\omega$. – 2017-01-31
  • 0
    Thank you for your answer! It seems I made the question complicated. – 2017-01-31

0 Answers 0