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:

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:

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:


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

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


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

Now compute:

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