Pushforward Measure is Measure

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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.

$\blacksquare$


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