Let $(\Omega,\mathcal{F},P)$ denote a probability space. Suppose $X: (\Omega, \mathcal{F}) \to (\mathbb{R}^n, \mathcal{B}^n)$ and $\epsilon: (\Omega, \mathcal{F}) \to (\mathbb{R}, \mathcal{B})$ are random variables. For any $e \in \mathbb{R}$, define $u:\mathbb{R} \times \Omega \to [0,1]$ as the probability that $\epsilon \leq e$ conditional on $X$. Thus $u(e,\cdot)$ is $\sigma(X)$-measurable and satisfies $$ \int_A u(e,\omega) \,dP = P(\{\epsilon \leq e\} \cap A) $$ for all $e \in \mathbb{R}$ and $A \in \sigma(X)$. Now define $v:\mathbb{R} \times \Omega \to [0,1]$ as the probability that $\epsilon \leq e - m(X)$ conditional on $X$. Then $v(e,\cdot)$ is $\sigma(X)$-measurable and satisfies $$ \int_A v(e,\omega) \,dP = P(\{\epsilon \leq e - m(X)\} \cap A) $$ for all $e \in \mathbb{R}$ and $A \in \sigma(X)$.

How can I show that $v(e,\omega) = u\left(e - m(X(\omega)),\omega\right)$ almost everywhere?


Start by proving that if $f(s,x)$ is jointly measurable and bounded, then $$ E[f(\epsilon,X)\,|\,X](\omega) = \int_{\Bbb R} f(s,X(\omega))\,d_s u(s,\omega) $$ for $P$-a.e. $\omega\in\Omega$. (This formula should be clear if $f$ has the special form $f(s,x)=g(s)h(x)$ with $g,h$ bounded and measurable; after this use the monotone class theorem for functions.) Now apply the formula to $f(s,x) = 1_{\{s\le e-m(x))\}}$.

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  • $\begingroup$ Thanks for your response. I'm still having trouble showing $\int_A \int_\mathbb{R} g(s) h(X(\omega)) d_s u(s,\omega) dP = \int_A g(\epsilon(\omega))h(X(\omega)) dP$ for $A \in \sigma(X)$ even when $g$ is simple. Am I doing something incorrectly? $\endgroup$ – David Sep 30 '15 at 14:52

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