Piecewise Combination of Measurable Mappings is Measurable/General Case

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Theorem

Let $\left({X, \Sigma}\right)$ and $\left({X', \Sigma'}\right)$ be measurable spaces.

Let $\left({E_n}\right)_{n \in \N} \in \Sigma, \displaystyle \bigcup_{n \mathop \in \N} E_n = X$ 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:

$\displaystyle \forall n \in \N, x \in E_n: f \left({x}\right) := f_n \left({x}\right)$


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


Proof

First, note that $f$ is well-defined, since if $x \in E_n$ and $x \in E_m$, we have that:

$f_n \left({x}\right) = f \left({x}\right) = f_m \left({x}\right)$

by $(1)$, since $x \in E_n \cap E_m$.


Let $E' \in \Sigma'$.

Then by definition of preimage, $f^{-1} \left({E'}\right) \subseteq X$, and hence:

\(\displaystyle f^{-1} \left({E'}\right)\) \(=\) \(\displaystyle X \cap h^{-1} \left({E'}\right)\) Intersection with Subset is Subset
\(\displaystyle \) \(=\) \(\displaystyle \left({\bigcup_{n \mathop \in \N} E_n}\right) \cap f^{-1} \left({E'}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \bigcup_{n \mathop \in \N} \left({E_n \cap f^{-1} \left({E'}\right)}\right)\) Intersection Distributes over Union: General Result
\(\displaystyle \) \(=\) \(\displaystyle \bigcup_{n \mathop \in \N} f_n^{-1} \left({E'}\right)\)

The last step we derive as follows.

Whenever $x \in E_n$, we have by definition of $f$ that $f \left({x}\right) = f_n \left({x}\right)$.

Hence, $x \in E_n \cap f^{-1} \left({E'}\right)$ implies $x \in f_n^{-1} \left({E'}\right)$.

Conversely, if $x \in f_n^{-1} \left({E'}\right)$, i.e. $f_n \left({x}\right) \in E'$, then $x \in E_n$ since $E_n$ is the domain of $f_n$.

By definition of $f$ then also $f \left({x}\right) = f_n \left({x}\right) \in E'$.

Hence we have deduced that $E_n \cap f^{-1} \left({E'}\right) = f_n^{-1} \left({E'}\right)$.


Since all $f_n$ are $\Sigma_{E_n} \, / \, \Sigma'$-measurable, it follows that for all $n \in \N$:

$f_n^{-1} \left({E'}\right) \in \Sigma$

By $\sigma$-algebra axiom $(3)$, it follows that:

$f^{-1} \left({E'}\right) = \displaystyle \bigcup_{n \mathop \in \N} f_n^{-1} \left({E'}\right) \in \Sigma$


Since $E' \in \Sigma'$ was arbitrary, it follows that $f$ is $\Sigma \, / \, \Sigma'$-measurable.

$\blacksquare$


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