Absolute Value of Simple Function is Simple Function/Proof 2

Theorem

Let $\left({X, \Sigma}\right)$ be a measurable space.

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

Then $\left\vert{f}\right\vert: X \to \R$, the absolute value of $f$, is also a simple function.

$(1): \quad f = \displaystyle \sum_{k \mathop = 0}^n a_k \chi_{E_k}$

Then, for all $x \in X$:

$\left\vert{f}\right\vert \left({x}\right) = \displaystyle \left\vert{ \sum_{k \mathop = 0}^n a_k \chi_{E_k} \left({x}\right) }\right\vert$

The fact that $(1)$ forms a standard representation ensures that for every $x \in X$, precisely one $k$ has $x \in E_k$.

Now suppose that $x \in E_l$.

Then $\chi_{E_l} \left({x}\right) = 0$ if and only if $k \ne l$ by definition of characteristic function.

It follows that $\left\vert{f}\right\vert \left({x}\right) = \left\vert{a_l \cdot 1}\right\vert = \left\vert{a_l}\right\vert$.

Now define $g: X \to \R$ by:

$g \left({x}\right) := \displaystyle \sum_{k \mathop = 0}^n \left\vert{a_k}\right\vert \chi_{E_k} \left({x}\right)$

By construction, $g$ is a simple function, and for $x \in E_l$, $g \left({x}\right) = \left\vert{a_l}\right\vert$.

Thus, since every $x$ is in $E_l$ for precisely one $l$, it is showed that $g = \left\vert{f}\right\vert$.

As $g$ was a simple function, it follows that $\left\vert{f}\right\vert$ is, too.

$\blacksquare$