# 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:

 $\ds \map {f_* \mu} {E'}$ $=$ $\ds \map \mu {\map {f^{-1} } {E'} }$ 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.

$\blacksquare$