Suppose $\mu$ and $\nu$ are probability measures on $\langle \Omega,\mathfrak{F}\rangle$ such that $$ (\forall A \in \mathfrak{F})\, \mu(A) \triangleq\int_{\Omega}1_A(x)f(x)d\nu(x), $$ for some measurable function $f$. Furthermore suppose that $\nu\sim\mu$ are equivalent probability measures.

If $F:\langle \Omega,\mathfrak{F}\rangle\rightarrow \langle \Omega,\mathfrak{F}\rangle$ is a bijective measurable function with measurable inverse then is the push-forward measure $F_{\star}\mu$ of the form $$ F_{\star}\mu(A) = \int_{\Omega} 1_AF\circ fd\nu? $$

  • $\begingroup$ What if in addition $\Omega$ was a manifold and $F$ were to be a diffeomorphism. $\endgroup$
    – user355356
    Oct 17, 2016 at 12:50
  • $\begingroup$ What do you mean by the notation $1_A F \circ f$? $\endgroup$
    – PhoemueX
    Oct 17, 2016 at 19:57

1 Answer 1


The change of variable formula implies the pushforward is $$ F_*\mu (A)=\int_\Omega\mathbb{1}_A\circ F(x)d\mu(x)=\int_\Omega\mathbb{1}_A\circ F(x)f(x)d\nu(x). $$ I don't see any reason that this integrand should equal yours, $1_AF\circ f,$ almost surely, as your conjecture suggests, even with the assumptions like bijectivity of $F$.

Concretely: let $\nu$ be lebesgue measure on the unit interval, $\mu$ its pushforward under $x\mapsto x^2$, and $F(x)=\sqrt{x}$, so that $F_*\mu$ is lebesgue measure $\nu$. From $$\mu((a,b))=\int_{F^{-1}(a,b)}d\nu=\int_\sqrt{a}^\sqrt{b}dx=\int_a^b1/(2\sqrt{x})dx, $$ we have $d\mu/d\nu= f(x)=1/(2\sqrt{x})$. Here $\mu,\nu$ are equivalent and $F$ is bijective, while your formula gives $$ (F_*\mu)((a,b))=\int_\Omega \mathbb{1}_{(a,b)}F\circ fd\nu=\int_a^b \sqrt{1/(2\sqrt{x})}dx\neq b-a. $$ You can see that the formula in the first display gives the correct result, $$ (F_*\mu)((a,b))=\int_\Omega\mathbb{1}_{(a,b)}\circ F(x)f(x)d\nu=\int_{a^2}^{b^2}f(x)dx=\sqrt{x}\Big|_{a^2}^{b^2}=b-a, $$ to match lebesgue measure on intervals, and therefore on the entire space.


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