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?