Simple Function is Measurable

Theorem

Let $\struct {X, \Sigma}$ be a measurable space.

Let $f: X \to \R$ be a simple function.

Then $f$ is $\Sigma$-measurable.

Proof

Let $f$ be written in the following form:

$f = \ds \sum_{i \mathop = 1}^n a_i \chi_{S_i}$

where $a_i \in \R$ and the $S_i$ are $\Sigma$-measurable.

Next, for each ordered $n$-tuple $b$ of zeroes and ones define:

$\map {T_b} i := \begin{cases} S_i & : \text {if $\map b i = 0$}\\ X \setminus S_i & : \text {if $\map b i = 1$} \end{cases}$

and subsequently:

$T_b := \ds \bigcap_{i \mathop = 1}^n \map {T_b} i$

From Sigma-Algebra Closed under Intersection, $T_b \in \Sigma$ for all $b$.

Also, the $T_b$ are pairwise disjoint, and furthermore:

$f = \ds \sum_b a_b \chi_{T_b}$

where:

$a_b := \ds \sum_{i \mathop = 1}^n \map b i a_i$

This needs considerable tedious hard slog to complete it.
In particular: prove it, it's a messy business
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