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:

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:

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:


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

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


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

Now compute:

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