Let F be a random variable defined on a probability space $(\Omega,\mathcal F, \mathbb P)$ with values on $(E,\mathcal E, \lambda)$, where $E=(-1,1)$ and $\lambda$ is the Lebesgue measure.

To prove that the law of $F$ is absolutely continuous (with respect to the Lebesgue measure on $\mathbb R$) I know that I should prove that for every measurable set $A\in\mathcal E$ such that $\lambda(A)=0$, we have $\mathbb P\circ F^{-1}(A)=0.$

It that equivalent to show that for any measurable function $g:(-1,1)\rightarrow [0,1]$ such that $\int_{-1}^1 g(x)dx=0$, we have $\mathbb E[g(F)]=0$ ? Why?

  • $\begingroup$ Radon–Nikodym theorem might help $\endgroup$ – Larsson Nov 23 '18 at 18:13
  • $\begingroup$ Definitely! :-) Thank you very much. Indeed one implication is obvious (if the second statement holds for any measurable function, it holds in particular for the indicator functions on measurable sets, so it implies the first statement). Viceversa, thank to Radon Nikodym, we know that a density exists, so I can write the measure of a set w.r.t $\mathbb P\circ F^{-1}$ in terms of the Lebesgue measure. Then it suffices to verify that the second statement is true for indicator functions and extend it to any measurable function through the standard method. $\endgroup$ – claudia Dec 3 '18 at 9:48

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