Given measurable functions \(f\), \(g\) on a common space \(\Omega\), we say that \(f\) is \(g\)-measurable if the induced \(\sigma\)-algebras satisfy \(\sigma(f) \subset \sigma(g)\). If \(f \colon \Omega \to S\) is a function and \((S, \mathcal{S})\) is a measurable space, then $$\sigma(f) = \left\{ f^{-1}A \mid A \in \mathcal{S} \right\}$$ is the smallest \(\sigma\)-algebra on \(\Omega\) such that \(f\) is \(\sigma(f)/\mathcal{S}\)-measurable.

Suppose we have another measurable function \(g \colon \Omega \to T\) where \((T, \mathcal{T})\) is a measurable space. Under what conditions is \(f\) is \(g\)-measurable?

This requires that \(\sigma(f) \subset \sigma(g)\), which means in some sense that the \(\sigma\)-algebra \(\mathcal{S}\) on \(S\) the domain of \(f\) cannot be too "coarse". The other condition turns out to be that \(f\) must be a function of \(g\).

Universal property of induced \(\sigma\)-algebras

Something like: Let \(X\) and \(Y\) be random variables. Then \(Y\) is \(\sigma(X)\)-measurable if and only if \(Y\) is a function of \(X\).