> [!theorem] > > Let $(\Omega_1, \cf_1)$ and $(\Omega_2, \cf_2)$ be [[Measure Space|measurable spaces]], $T: \Omega_1 \to \Omega_2$, and $X: \Omega_1 \to \real$ be arbitrary mappings, then the following are equivalent: > 1. $X$ is $(\sigma(T), \cb(\real))$-measurable. > 2. There exists a $(\cb(\real), \cb(\real))$-measurable map $f$ such that $X = f \circ T$. > > *Proof (decomposition)*. Suppose that $X$ is $(\sigma(T), \cb(\real))$-measurable. Let $\alg$ be the collection of all maps for which the result applies. > > First assume that $X = \chi_{E}$ for some $E \in \sigma(T)$, then there exists $F \in \cb(\real)$ such that $E = T^{-1}(F)$. Taking the function $f = \chi_F$ yields the result. > > Now let $X, Y \in \alg$ and $c \in \real$. This allows decomposing $cX + Y$ as > $ > cX + Y = c f_X \circ T + f_Y \circ T = (cf_X + f_Y) \circ T > $ > meaning that $cX + Y \in \alg$. > > Lastly, let $\seq{X_n} \subset \alg$ be a bounded, increasing sequence of non-negative functions converging to $X = \sup_{n \in \nat}X_n$. Decompose > $ > \sup_{n \in \nat}X_n(x) = \underbrace{\sup_{n \in \nat}f_n}_{f_{X}} \circ T(x) > $ > > By a [[Monotone Class Argument]], this includes all measurable maps.