Piecewise Combination of Measurable Mappings is Measurable/Binary Case
Theorem
Let $\struct {X, \Sigma}$ and $\struct {X', \Sigma'}$ be measurable spaces.
Let $f, g: X \to X'$ be $\Sigma \, / \, \Sigma'$-measurable mappings.
Let $E \in \Sigma$ be a measurable set.
Define $h: X \to X'$ by:
- $\forall x \in X: \map h x := \begin{cases} \map f x & : \text {if $x \in E$} \\ \map g x & : \text {if $x \notin E$} \end{cases}$
Then $h$ is also a $\Sigma \, / \, \Sigma'$-measurable mapping.
Proof
Let $E' \in \Sigma'$ be a $\Sigma'$-measurable set.
Then by definition of preimage:
- $\map {h^{-1} } {E'} = \set {x \in X: \map h x \in E'}$
Expanding the definition of $h$, this translates into:
- $\map {h^{-1} } {E'} = \set {x \in E: \map f x \in E'} \cup \set {x \in \relcomp X E: \map g x \in E'}$
where $\complement$ denotes set complement.
That is, we have:
- $\map {h^{-1} } {E'} = \paren {E \cap \map {f^{-1} } {E'} } \cup \paren {\relcomp X E \cap \map {g^{-1} } {E'} }$
All sets on the right hand side are $\Sigma$-measurable.
By Sigma-Algebra Closed under Intersection and Sigma-Algebra Closed under Union, so is $\map {h^{-1} } {E'}$.
Since $E' \in \Sigma'$ was arbitrary, $h$ is a $\Sigma \, / \, \Sigma'$-measurable mapping.
$\blacksquare$
Sources
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $\S 8$: Problem $3 \ \text{(i)}$