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?