Pointwise Sum of Simple Functions is Simple Function

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Let $f,g : X \to \R$ be simple functions.


Then $f + g: X \to \R, \left({f + g}\right) \left({x}\right) := f \left({x}\right) + g \left({x}\right)$ is also a simple function.


Proof

We have $f + g = + \circ \left\langle{f, g}\right\rangle \circ \Delta_X$, where:

$\Delta_X: X \to X \times X$ is the diagonal mapping on $X$
$\left\langle{f, g}\right\rangle: X \times X \to \R \times \R, \left\langle{f, g}\right\rangle \left({x, y}\right) := \left({f \left({x}\right), g \left({y}\right)}\right)$
$+: \R \times \R \to \R$ is real addition

From this, we see that $+$ may be restricted to $\operatorname{Im} \left({\left\langle{f, g}\right\rangle}\right)$, the image of $\left\langle{f, g}\right\rangle$.

It is immediate that this image equals $\operatorname{Im} \left({f}\right) \times \operatorname{Im} \left({g}\right)$.


By Measurable Function is Simple Function iff Finite Image Set, $\operatorname{Im} \left({f}\right)$ and $\operatorname{Im} \left({g}\right)$ are finite.

Therefore, by Cardinality of Cartesian Product, $\operatorname{Im} \left({\left\langle{f, g}\right\rangle}\right)$ is also finite.

Next, from Corollary to Image of Composite Mapping:

$\operatorname{Im} \left({\left\langle{f, g}\right\rangle \circ \Delta_X}\right) \subseteq\operatorname{Im} \left({\left\langle{f, g}\right\rangle}\right)$

whence the former is finite.


Now $+$ is a surjection:

$+: \operatorname{Im} \left({\left\langle{f, g}\right\rangle \circ \Delta_X}\right) \to \operatorname{Im} \left({f + g}\right)$

by Restriction of Mapping to Image is Surjection.


By Cardinality of Surjection, this implies that $\operatorname{Im} \left({f + g}\right)$ is finite.

Whence Measurable Function is Simple Function iff Finite Image Set grants that $f + g$ is a simple function.

$\blacksquare$


Also see


Sources