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For real random variables $X,Z$, if $Z$ is measurable with respect to the sigma-algebra generated by a random variable $X$, there is a Borel measurable function $g:\mathbb{R}\to \mathbb{R}$ with $Z=g(X)$. To prove the bounded case, approximate $Z$ pointwise by finite-valued simple functions $Z_m$. Each level set is $X^{-1}(B)$ for a Borel set $B$, so each approximation factors as $Z_m=g_m(X)$ with $g_m$ a Borel measurable function. Apply scalar clipping to $g_m$ with a common bounded interval and use its limit superior to obtain $g$. For unbounded finite $Z$, apply the bounded case to the one-to-one bounded continuous function $t\mapsto \arctan t$, then invert on $(-\pi /2,\pi /2)$ and assign arbitrary values elsewhere. Exact measurability gives exact factorization; assumptions after adjoining null sets give an almost surely equal version.

The sigma-algebra generated by a random variable $X$ is the smallest sigma-algebra on $\Omega$ making $X$ a measurable function into $(\mathbb{R},\mathcal{B}(\mathbb{R}))$. Equivalently, it is the intersection of all sigma-algebras containing the inverse images $X^{-1}(C)$ of Borel sets $C$.

Let $\mathcal{D}=\{X^{-1}(C):C\in \mathcal{B}(\mathbb{R})\}$. This collection is already a sigma-algebra: $X^{-1}(\mathbb{R})=\Omega$,The right-hand sets are in $\mathcal{D}$, since the Borel sigma-algebra is closed under the indicated operations. Any sigma-algebra making $X$ measurable must contain every member of $\mathcal{D}$. Conversely $\mathcal{D}$ makes $X$ measurable by definition. HenceThis equality holds exactly as a collection of sets, without completing the sigma-algebra.