# Simple Function is Measurable

## Theorem

Let $\left({X, \Sigma}\right)$ 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:

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

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

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

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

and subsequently:

$T_b := \displaystyle \bigcap_{i \mathop = 1}^n T_b \left({i}\right)$

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

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

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

where:

$a_b := \displaystyle \sum_{i \mathop = 1}^n b \left({i}\right) a_i$

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

$\left\{{x \in X: f \left({x}\right) > \lambda}\right\} = \displaystyle \bigcup \left\{{T_b: a_b > \lambda}\right\}$

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

$\blacksquare$