Why is it called pushforward measure? I'm trying to get an intuitive sense of the pushforward measure.
Let $\Phi: \mathbb{R}^n \rightarrow \mathbb{R}^n$ be a vector field. We can define a corresponding pushforward measure: for an initial measure $\mu$, the pushforward induced by $\Phi$ would be $\nu$ such that
$$ \int_{\mathbb{R}^n} \nu(x) f(x) dx = \int_{\mathbb{R}^n} \mu(y) f(\Phi(x)) dx \;\;\;\; \text{ for any } f\in C^0_0(\mathbb{R}^n).$$
This says that $\mathbb{E}_\nu[f(x)] = \mathbb{E}_\mu [f(\Phi(x))]$.
I've also seen pushforward measure defined as: for any Borel set $B\in \mathbb{R}^n$,
$$ \nu(\Phi^{-1}(B)) = \mu(B).$$
This says that $\mu$ and $\nu$ are equivalent measures under a transformation of the state space by $\Phi$.

*

*Are these definitions equivalent? If we pick $f$ to be any continuous function with support on $B \in {\mathbb{R}^n}$, I can see how the first definition implies the second.

*Is my intuitive explanation that "$\mu$ and $\nu$ are equivalent under a transformation of the state space" correct?

*Why is it called pushforward measure? It seems more like a "pushbackward measure" since the original meausure of $B$ is $\mu(B)$ and the resulting measure $\nu$ takes $\Phi^{-1}(B)$ as its argument.

 A: Pushforward measures are even more generalizable than that. In general, consider a measure space $(M, \mu)$, a $\sigma$-algebra $N$, and a measurable map $f : M \to N$. Then we can define a measure on $N$ by $\nu(S) = \mu(f^{-1}(S))$.
The reason why this is known as the "push-forward" measure is that we're taking a measure on $M$ and "pushing it forward" through the map $f$ to get a measure on $N$.
Using the monotone convergence theorem, it's easy to show that for any measurable function $g : N \to \mathbb{R}_{\geq 0}$, $\int g d\nu = \int (f \circ g) d \mu$ (this can be done by taking an increasing sequence of simple functions $N \to \mathbb{R}_{\geq 0}$, the integrals of which converge to the integral of $g$). This can easily be extended to show $\int g d\nu = \int (f \circ g) d \mu$ for any measurable $g$, with the left side being defined iff the right side is.
So the two definitions are indeed equivalent. As for whether your intuitive explanation is correct, I don't think such a thing can be correct or incorrect, only useful or not useful.
