> [!theorem] > > Let $\Omega$ be a ground set and $\mathcal P$ be a $\pi$-system over it. Let $\alg$ be a collection of maps $\Omega \to \real$ such that > 1. $\chi_E \in \alg$ for all $E \in \mathcal{P}$. > 2. $\alg$ is a vector space. > 3. For any monotone sequence $\seq{f_k} \subset \alg$ such that $0 \le f_n \to f$ [[Pointwise Convergence|pointwise]] with $f$ being bounded[^1], the limit $f \in \alg$. > > Then $\alg$ contains all bounded $(\sigma(\mathcal P), \cb(\real))$-measurable maps. > > ### $\lambda$-system > > The set > $ > \Lambda = \bracs{E \subset \Omega: \chi_E \in \alg} > $ > is a $\lambda$-system. Since it's also a $\pi$-system, $\Lambda = \sigma(\mathcal P)$. > > *Proof*. If $E \subset F \in \Lambda$, then $\chi_{F \setminus E} = \chi_F - \chi_E$ by (2), and $F \setminus E \in \Lambda$. If $\seq{E_n} \subset \Lambda$ is an ascending sequence of set, then $\chi_{E_n} \upto \chi_{\bigcup_{n \in \nat}E_n}$ and $\bigcup_{n \in \nat}E_n \in \Lambda$ by (3). > > > ### Bounded Measurable Functions > > Let $f$ be a non-negative bounded $(\sigma(\mathcal P), \cb(\real))$-measurable function, then $f$ is in $\alg$. Extend this to all measurable maps by decomposing to positive and negative parts, then using the vector space property. > > *Proof*. $f$ is a uniform limit of simple $(\sigma(\mathcal P), \cb(\real))$-measurable functions, which are in $\alg$ by linearity. [^1]: Dropping this assumption leads to the result holding for unbounded functions as well.