# Pushforward Measure is Measure

## Theorem

Let $\left({X, \Sigma}\right)$ and $\left({X', \Sigma'}\right)$ be measurable spaces.

Let $\mu$ be a measure on $\left({X, \Sigma}\right)$.

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:

$\forall E' \in \Sigma': f_* \mu \left({E'}\right) \ge 0$

Now observe:

 $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle f_* \mu \left({E'}\right)$$ $$=$$ $$\displaystyle$$ $$\displaystyle \mu \left({f^{-1} \left({E'}\right)}\right)$$ $$\displaystyle$$ $$\displaystyle$$ Definition of pushforward measure $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\ge$$ $$\displaystyle$$ $$\displaystyle 0$$ $$\displaystyle$$ $$\displaystyle$$ $\mu$ is a measure

$\Box$

### Axiom $(2)$

Let $\left({E'_n}\right)_{n \in \N}$ be a sequence of pairwise disjoint sets in $\Sigma$.

The statement of axiom $(2)$ for $f_* \mu$ is:

$\displaystyle f_* \mu \left({\bigcup_{n \mathop \in \N} E'_n}\right) = \sum_{n \mathop \in \N} f_* \mu \left({E'_n}\right)$

Now compute:

 $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle f_* \mu \left({\bigcup_{n \mathop \in \N} E'_n}\right)$$ $$=$$ $$\displaystyle$$ $$\displaystyle \mu \left({f^{-1} \left({\bigcup_{n \mathop \in \N} E'_n}\right)}\right)$$ $$\displaystyle$$ $$\displaystyle$$ Definition of pushforward measure $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \mu \left({\bigcup_{n \mathop \in \N} f^{-1} \left({E'_n}\right)}\right)$$ $$\displaystyle$$ $$\displaystyle$$ Mapping Preimage of Union: General Result $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \sum_{n \mathop \in \N} \mu \left({f^{-1} \left({E'_n}\right)}\right)$$ $$\displaystyle$$ $$\displaystyle$$ $\mu$ is a measure $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \sum_{n \mathop \in \N} f_* \mu \left({E'_n}\right)$$ $$\displaystyle$$ $$\displaystyle$$ Definition of pushforward measure

Note that the second equality uses Mapping Preimage of Intersection and Preimage of Empty Set is Empty to ensure that $\left({f^{-1} \left({E'_n}\right)}\right)_{n \in \N}$ is pairwise disjoint:

$f^{-1} \left({E'_n}\right) \cap f^{-1} \left({E'_m}\right) = f^{-1} \left({E'_n \cap E'_m}\right) = f^{-1} \left({\varnothing}\right) = \varnothing$

$\Box$

### Axiom $(3')$

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

$f_* \mu \left({\varnothing}\right) = 0$

Now compute:

 $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle f_* \mu \left({\varnothing}\right)$$ $$=$$ $$\displaystyle$$ $$\displaystyle \mu \left({f^{-1} \left({\varnothing}\right)}\right)$$ $$\displaystyle$$ $$\displaystyle$$ Definition of pushforward measure $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \mu \left({\varnothing}\right)$$ $$\displaystyle$$ $$\displaystyle$$ Preimage of Empty Set is Empty $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle 0$$ $$\displaystyle$$ $$\displaystyle$$ $\mu$ is a measure

$\Box$

Thus $f_* \mu$, satisfying the axioms, is seen to be a measure.

$\blacksquare$