Pushforward Measure is Measure
Theorem
Let $\struct {X, \Sigma}$ and $\struct {X', \Sigma'}$ be measurable spaces.
Let $\mu$ be a measure on $\struct {X, \Sigma}$.
Let $f: X \to X'$ be a $\Sigma \, / \, \Sigma'$-measurable mapping.
Then the pushforward measure $f_* \mu: \Sigma' \to \overline \R$ is a measure.
Proof
To show that $f_* \mu$ is a measure, it will suffice to check the axioms $(1)$, $(2)$ and $(3')$ for a measure.
Axiom $(1)$
The statement of axiom $(1)$ for $f_* \mu$ is:
- $\forall E' \in \Sigma': \map {f_* \mu} {E'} \ge 0$
Now observe:
| \(\ds \map {f_* \mu} {E'}\) | \(=\) | \(\ds \map \mu {f^{-1} \sqbrk {E'} }\) | Definition of Pushforward Measure | |||||||||||
| \(\ds \) | \(\ge\) | \(\ds 0\) | $\mu$ is a Measure |
$\Box$
Axiom $(2)$
Let $\sequence {E'_n}_{n \mathop \in \N}$ be a sequence of pairwise disjoint sets in $\Sigma$.
The statement of axiom $(2)$ for $f_* \mu$ is:
- $\ds \map {f_* \mu} {\bigcup_{n \mathop \in \N} E'_n} = \sum_{n \mathop \in \N} \map {f_* \mu} {E'_n}$
Now compute:
| \(\ds \map {f_* \mu} {\bigcup_{n \mathop \in \N} E'_n}\) | \(=\) | \(\ds \map \mu {f^{-1} \sqbrk {\bigcup_{n \mathop \in \N} E'_n} }\) | Definition of Pushforward Measure | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map \mu {\bigcup_{n \mathop \in \N} f^{-1} \sqbrk {E'_n} }\) | Preimage of Union under Mapping: General Result | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{n \mathop \in \N} \map \mu {f^{-1} \sqbrk {E'_n} }\) | $\mu$ is a Measure | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{n \mathop \in \N} \map {f_* \mu} {E'_n}\) | Definition of Pushforward Measure |
Note that the second equality uses Preimage of Intersection under Mapping and Preimage of Empty Set is Empty to confirm that $\sequence { f^{-1} \sqbrk {E'_n} }_{n \mathop \in \N}$ is pairwise disjoint:
- $f^{-1} \sqbrk {E'_n} \cap f^{-1} \sqbrk {E'_m} = f^{-1} \sqbrk {E'_n \cap E'_m} = f^{-1} \sqbrk \O = \O$
$\Box$
Axiom $(3')$
The statement of axiom $(3')$ for $f_* \mu$ is:
- $\map {f_* \mu} \O = 0$
Now compute:
| \(\ds \map {f_* \mu} \O\) | \(=\) | \(\ds \map \mu {f^{-1} \sqbrk \O}\) | Definition of Pushforward Measure | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map \mu \O\) | Preimage of Empty Set is Empty | |||||||||||
| \(\ds \) | \(=\) | \(\ds 0\) | $\mu$ is a measure |
$\Box$
Thus $f_* \mu$, satisfying the axioms, is seen to be a measure.
$\blacksquare$
Sources
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $7.6$