Absolute Value of Simple Function is Simple Function/Proof 2

Proof
By Simple Function has Standard Representation, $f$ has a standard representation, say:


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

Then, for all $x \in X$:


 * $\map {\size f} x = \ds \size {\sum_{k \mathop = 0}^n a_k \map {\chi_{E_k} } x}$

by definition of pointwise absolute value.

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 $\map {\chi_{E_l} } x = 0$ $k \ne l$ by definition of characteristic function.

It follows that $\map {\size f} x = \size {a_l \cdot 1} = \size {a_l}$.

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


 * $\map g x := \ds \sum_{k \mathop = 0}^n \size {a_k} \map {\chi_{E_k} } x$

By construction, $g$ is a simple function, and for $x \in E_l$:
 * $\map g x = \size {a_l}$

Thus, since every $x$ is in $E_l$ for precisely one $l$, it has been shown that $g = \size f$.

As $g$ was a simple function, it follows that $\size f$ is also a simple function.