Factorization Lemma/Extended Real-Valued Function

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Theorem

Let $X$ be a set, and $\left({Y, \Sigma}\right)$ be a measurable space.

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


Then an extended real-valued function $g: X \to \overline{\R}$ is $\sigma \left({f}\right)$-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:

$\sigma \left({f}\right)$ denotes the $\sigma$-algebra generated by $f$


Proof

Necessary Condition

Let $g$ be a $\sigma \left({f}\right) \, / \, \overline{\mathcal B}$-measurable function.

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


Let us proceed in the following fashion:


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

By Characteristic Function Measurable iff Set Measurable, it follows that $E$ is $\sigma \left({f}\right)$-measurable.

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

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

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

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

Now $\displaystyle \sum_{i \mathop = 1}^n a_i \chi_{A_i}$ 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 Pointwise Limit of Simple Functions, we find simple functions $g_j$ such that:

$\displaystyle \lim_{j \to \infty} g_j = g$

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

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

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

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

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 $\sigma \left({f}\right) \,/\, \Sigma$-measurable by definition of $\sigma \left({f}\right)$.


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

$\blacksquare$


Sources