# 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 businessTo discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Finish}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

Now we have, for all $\lambda \in \R$:

- $\set {x \in X: \map f x > \lambda} = \ds \bigcup \set {T_b: a_b > \lambda}$

which by Sigma-Algebra Closed under Union is a $\Sigma$-measurable set.

From Characterization of Measurable Functions: $(5)$ it follows that $f$ is measurable.

$\blacksquare$

## Sources

- 2005: René L. Schilling:
*Measures, Integrals and Martingales*... (previous) ... (next): $8.5 \ \text{(ii)}$