Pushforward measure

Remarks

The set $f^{-1}(B)$ lies in $\Sigma$ because $f$ is measurable, so the formula makes sense. The map $f_* \mu$ is a measure: $f^{-1}(\varnothing) = \varnothing$, and the preimage of a disjoint union of sets $B_n$ is the disjoint union of the sets $f^{-1}(B_n)$. Other notations are $\mu \circ f^{-1}$ and $f_\# \mu$. Example: let $\mu$ be the measure on $\R$ with $\mu(A) =$ number of points of $A \cap \set{-1, 1}$, and $f(x) = x^2$. Then $f_* \mu$ gives mass $2$ to the point $1$, since $f^{-1}(\set{1}) = \set{-1, 1}$. Non-example: if $f$ is not measurable, some $f^{-1}(B)$ is outside $\Sigma$, and the formula does not define $f_* \mu$ on all of $\mathcal{T}$. For a measurable $g \colon Y \to [0, \infty]$ the change of variables formula holds: $$\int_Y g \, d(f_* \mu) = \int_X (g \circ f) \, d\mu.$$