Piecewise Combination of Measurable Mappings is Measurable

Theorem

Let $\struct {X, \Sigma}$ and $\struct {X', \Sigma'}$ be measurable spaces.

Binary Case

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.

General Case

Let $\sequence {E_n}_{n \mathop \in \N} \in \Sigma, \ds \bigcup_{n \mathop \in \N} E_n = X$ be a countable cover of $X$ by $\Sigma$-measurable sets.

For each $n \in \N$, let $f_n: E_n \to X'$ be a $\Sigma_{E_n} \, / \, \Sigma'$-measurable mapping.

Here, $\Sigma_{E_n}$ is the trace $\sigma$-algebra of $E_n$ in $\Sigma$.

Suppose that for every $m, n \in \N$, $f_m$ and $f_n$ satisfy:

$(1): \quad f_m \restriction_{E_m \cap E_n} = f_n \restriction_{E_m \cap E_n}$

that is, $f_m$ and $f_n$ coincide whenever both are defined; here $\restriction$ denotes restriction.

Define $f: X \to X'$ by:

$\ds \forall n \in \N, x \in E_n: \map f x := \map {f_n} x$

Then $f$ is a $\Sigma \, / \, \Sigma'$-measurable mapping.