Pointwise Sum of Simple Functions is Simple Function

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\struct {X, \Sigma}$ be a measurable space.

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


Then $f + g: X \to \R, \map {\paren {f + g} } x := \map f x + \map g x$ is also a simple function.


Proof

We have $f + g = + \circ \innerprod f g \circ \Delta_X$, where:

$\Delta_X: X \to X \times X$ is the diagonal mapping on $X$
$\innerprod f g: X \times X \to \R \times \R, \map {\innerprod f g} {x, y} := \tuple {\map f x, \map g y}$
$+: \R \times \R \to \R$ is real addition.



From this, we see that $+$ may be restricted to $\Img {\innerprod f g}$, the image of $\sequence {f, g}$.

It is immediate that this image equals $\Img f \times \Img g$.


By Measurable Function is Simple Function iff Finite Image Set, $\Img f$ and $\Img g$ are finite.

Therefore, by Cardinality of Cartesian Product, $\Img {\sequence {f, g} }$ is also finite.

Next, from Corollary to Image of Composite Mapping:

$\Img {\innerprod f g \circ \Delta_X} \subseteq \Img {\innerprod f g}$

whence the former is finite.


Now $+$ is a surjection:

$+: \Img {\innerprod f g \circ \Delta_X} \to \Img {f + g}$

by Restriction of Mapping to Image is Surjection.


By Cardinality of Surjection, this implies that $\Img {f + g}$ is finite.

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

$\blacksquare$


Also see


Sources