Pushforward Measure is Measure

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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:

\(\ds \map {f_* \mu} {E'}\) \(=\) \(\ds \map \mu {\map {f^{-1} } {E'} }\) 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:

$\displaystyle \map {f_* \mu} {\bigcup_{n \mathop \in \N} E'_n} = \sum_{n \mathop \in \N} \map {f_* \mu} {E'_n}$


Now compute:

\(\ds \map {f_* \mu} {\bigcup_{n \mathop \in \N} E'_n}\) \(=\) \(\ds \map \mu {\map {f^{-1} } {\bigcup_{n \mathop \in \N} E'_n} }\) Definition of Pushforward Measure
\(\ds \) \(=\) \(\ds \map \mu {\bigcup_{n \mathop \in \N} \map {f^{-1} } {E'_n} }\) Preimage of Union under Mapping: General Result
\(\ds \) \(=\) \(\ds \sum_{n \mathop \in \N} \map \mu {\map {f^{-1} } {E'_n} }\) $\mu$ 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 $\sequence {\map {f^{-1} } {E'_n} }_{n \mathop \in \N}$ is pairwise disjoint:

$\map {f^{-1} } {E'_n} \cap \map {f^{-1} } {E'_m} = \map {f^{-1} } {E'_n \cap E'_m} = \map {f^{-1} } \O = \O$

$\Box$


Axiom $(3')$

The statement of axiom $(3')$ for $f_* \mu$ is:

$\map {f_* \mu} \O = 0$


Now compute:

\(\ds \map {f_* \mu} \O\) \(=\) \(\ds \map \mu {\map {f^{-1} } \O}\) 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.

$\blacksquare$


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