Pointwise Minimum of Simple Functions is Simple/Proof 1
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Theorem
Let $\struct {X, \Sigma}$ be a measurable space.
Let $f, g : X \to \R$ be simple functions.
Then the pointwise minimum $\min \set {f, g}: X \to \R$ is also simple function.
Proof
From Pointwise Sum of Simple Functions is Simple Function:
- $f + g$ is simple.
From Scalar Multiple of Simple Function is Simple Function:
- $-g$ is simple.
Then, from Pointwise Sum of Simple Functions is Simple Function we have:
- $f - g$ is simple.
From Absolute Value of Simple Function is Simple Function:
- $\size {f - g}$ is simple.
So, from Scalar Multiple of Simple Function is Simple Function, we have:
- $-\size {f - g}$ is simple.
From Pointwise Sum of Simple Functions is Simple Function, we then have:
- $\paren {f + g} - \size {f - g}$ is simple.
Finally, from Scalar Multiple of Simple Function is Simple Function, we have:
- $\dfrac 1 2 \paren {\paren {f + g} - \size {f - g} }$ is simple.
By Minimum Function in terms of Absolute Value, we have:
- $\ds \min \set {f, g} = \frac 1 2 \paren {\paren {f + g} - \size {f - g} }$
so:
- $\min \set {f, g}$ is simple.
$\blacksquare$