Factorization Lemma/Extended Real-Valued Function

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Theorem

Let $X$ be a set.

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

Let $f: X \to Y$ be a mapping.


An extended real-valued function $g: X \to \overline \R$ is $\map \sigma f$-measurable if and only if:

There exists a $\Sigma$-measurable mapping $\tilde g: Y \to \overline \R$ such that $g = \tilde g \circ f$

where:

$\map \sigma f$ denotes the $\sigma$-algebra generated by $f$


Proof

Necessary Condition

Let $g$ be a $\map \sigma f \, / \, \overline \BB$-measurable function.

We need to construct a measurable $\tilde g$ such that $g = \tilde g \circ f$.


Let us proceed in the following fashion:

Establish the result for $g$ a characteristic function;
Establish the result for $g$ a simple function;
Establish the result for all $g$


So let $g = \chi_E$ be a characteristic function.

By Characteristic Function Measurable iff Set Measurable, it follows that $E$ is $\map \sigma f$-measurable.

Thus there exists some $A \in \Sigma$ such that $E = \map {f^{-1} } A$.

Again by Characteristic Function Measurable iff Set Measurable, we have $\chi_A: Y \to \overline \R$ is measurable.

It follows that $\chi_E = \chi_A \circ f$, and $\tilde g := \chi_A$ works.


Now let $g = \ds \sum_{i \mathop = 1}^n a_i \chi_{E_i}$ be a simple function.

Let $A_i$ be associated to $E_i$ as above. Then we have:

\(\ds \sum_{i \mathop = 1}^n a_i \chi_{E_i}\) \(=\) \(\ds \sum_{i \mathop = 1}^n a_i \paren {\chi_{A_i} \circ f}\) by the result for characteristic functions
\(\ds \) \(=\) \(\ds \paren {\sum_{i \mathop = 1}^n a_i \chi_{A_i} } \circ f\) Composition of Mappings is Linear

Now $\ds \sum_{i \mathop = 1}^n a_i \chi_{A_i}$ is a simple function, hence measurable by Simple Function is Measurable.

Therefore, it suffices as a choice for $\tilde g$.


Next, let $g \ge 0$ be a measurable function.

By Measurable Function is Pointwise Limit of Simple Functions, we find simple functions $g_j$ such that:

$\ds \lim_{j \mathop \to \infty} g_j = g$

Applying the previous step to each $g_j$, we find a sequence of $\tilde g_j$ satisfying:

$\ds \lim_{j \mathop \to \infty} \tilde g_j \circ f = g$

From Composition with Pointwise Limit it follows that we have, putting $\tilde g := \ds \lim_{j \mathop \to \infty} \tilde g_j$:

$\ds \lim_{j \mathop \to \infty} \tilde g_j \circ f = \tilde g \circ f$


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An application of Pointwise Limit of Measurable Functions is Measurable yields $\tilde g$ measurable.


Thus we have provided a suitable $\tilde g$ for every $g$, such that:

$g = \tilde g \circ f$

as desired.

$\blacksquare$


Sufficient Condition

Suppose that such a $\tilde g$ exists.

Note that $f$ is $\map \sigma f \,/\, \Sigma$-measurable by definition of $\map \sigma f$.


The result follows immediately from Composition of Measurable Mappings is Measurable.

$\blacksquare$


Sources

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