Radon-Nikodym derivative of pushforward of Lebesgue measure by differentiable function with respect to Lebesgue measure Here is the set up:

*

*$f:\mathbb{R}^n\to\mathbb{R}^m$ is a measurable function

*$\lambda^n$ is the $n$-dimensional Lebesgue measure

*$f_*\lambda^n$ is the pushforward of $\lambda^n$ by $f$

*$f_*\lambda^n \ll \lambda^m$

What do we know about the Radon-Nikodym derivative
$$
\frac{d f_*\lambda^n}{d\lambda^m}
$$
for different types of $f$?

Notice this Radon-Nikodym derivative exists as long as $f_*\lambda^n$ and $\lambda^m$ are sigma-finite measures on the same space, which is true in this setup.
Examples
For instance, using the Change of Variables formula and Integration by Substitution when $f$ is a diffeomorphism, the Radon-Nikodym derivative is the absolute determinant Jacobian (see Billingsley "Probability and Measure", Theorem 17.2)
$$
\frac{d f_*\lambda^n}{d\lambda^m} = |\det J_f|.
$$
I have an intuition, based on this question, (but I have never seen it proven or mentioned anywhere) that when $f$ is simply differentiable, then the Radon-Nikodym derivative is the multidimensional Jacobian defined in Federer's "Geometric Integration Theorem" (Lemma 5.1.4)
$$
\frac{d f_*\lambda^n}{d\lambda^m} = \begin{cases}
    \mathcal{J}_n f(x) = |\det J_f(x)| && \text{ if } n = m \\
    \mathcal{J}_n f(x) = \sqrt{\det J_f(x)^\top J_f(x)} && \text{ if } n \leq m \\
    \mathcal{J}_m f(x) = \sqrt{\det J_f(x) J_f(x)^\top} && \text{ if } n \geq m
\end{cases}
$$
However I have no idea how to go about proving this.
 A: Your intuition makes no sense - the Jacobian is at a point of the domain, and the Radon-Nikodym derivative is at a point in the target.
Notice this Radon-Nikodym derivative exists [...] in this setup.

Not really. As was pointed out, the constant map pushes forward to an "infinite-sized delta" at the value. This has no Radon-Nikodym derivative with respect to Lebesgue, both because the pushforward is not sigma-finite (hence "infinite-sized") and because it is not absolutely continuous relative to Lebesgue (hence "delta").
A simple case with a decent answer is when $f$ is a proper submersion. Then at $y\in \mathbb{R}^m$ we have $M=f^{-1}(y)$ a smooth compact submanifold of $\mathbb{R}^n$, and the derivative $\frac{d f_*\lambda^n}{d\lambda^m}(y)$ is the integral over $M$ of the multidimensional Jacobian you mention (note that this reduces to the correct answer when $f$ is a diffeo). The reason is that the multidimensional Jacobian is the infinitesimal volume stretching factor between tangent spaces on domain and target. In other words, a preimage of a small neighborhood of $y$ of volume $V$ is a thickening of $M$, fibered over $M$ by orthocomplements to the kernel of $f$ at every $x\in M$. The volume of each such orthocomplement is approximately $V\times Jf(x)$, so that the overall volume of the thickening is approximately $V\times\int_M (Jf(x)) d(vol M)$. The Radon-Nikodym derivative is the limit of the ratio of that to $V$  when $V$ becomes small.
A: If $U, V \subset \mathbb{R}^n$ are open sets and $f: U \to V$ is a smooth injective map with non-vanishing Jacobian determinant $J_f$, then the pushforward $\lambda \circ f^{-1}$ (denoted $f_*\lambda$ in the original question) of the Lebesgue measure $\lambda$ is absolutely continuous with respect to $\lambda$ (restricted to the $U$ and $V$, respectively) and it is given by:
$$
(\lambda \circ f^{-1})(\mathrm{d}y)
= \left[J_f(f^{-1}(y)) \right]^{-1} \lambda(\mathrm{d}y)
= J_{f^{-1}}(y) \lambda(\mathrm{d}y).
\tag{1}
$$
In other words, the term $J_{f^{-1}}(y)$ is the density of the pushforward measure $f_*\lambda$ with respect to the Lebesgue measure $\lambda$:
$$
\frac{(\lambda \circ f^{-1})(\mathrm{d}y)}{\lambda(\mathrm{d}y)} = J_{f^{-1}}(y).
\tag{2}
$$
More generally, if we have any measure $\mu(\mathrm{d}x) = h(x) \lambda(\mathrm{d}x)$ that is absolutely continuous with respect to $\lambda$, then
$$
\frac{(\mu \circ f^{-1})(\mathrm{d}y)}{\lambda(\mathrm{d}y)} = h(f^{-1}(y)) J_{f^{-1}}(y).
\tag{3}
$$
This equation corresponds to the usual rule that a derived random variable $Y = f(X)$ for 1-1 function $f$ has density
$$
p_Y(y) = p_X(f^{-1}(y))J_{f^{-1}}(y).
$$

Theorem 17.2 of Billingsley states a related property; namely, if $f: U \to V$ is defined as above and $\phi: V \to R$ is a Borel function, then
$$
\int_{V} \phi(x) \lambda(\mathrm{d}x)
= \int_{U} \phi(f(x)) \underbrace{J_f(x)\lambda(\mathrm{d}x)}_{:=\ 
 \nu(\mathrm{d}x)}
\tag{4}
$$
All that is left now is to do pattern matching on the definition of the pushforward measure. Namely, with $\nu$ as defined above, consider the pushforward measure $(\nu \circ f^{-1})$; we have:
$$
\int_{V}\phi(x) (\nu \circ f^{-1})(\mathrm{d}x) = \int_{U}\phi(f(x))\nu(\mathrm{d}x)
\tag{*}
$$
Therefore, the pushforward of the measure $\nu(dx) = J_f(x) \lambda(\mathrm{d}x)$ through $f$ (denoted $\nu \circ f^{-1})$ is exactly the Lebesgue measure $\lambda$.  This result is unsurprising, since applying $(3)$ we have
$$
\begin{align}
\frac{(\nu \circ f^{-1})(\mathrm{d}y)}{\lambda(\mathrm{d}y)}
&= J_f(f^{-1}(y)) J_{f^{-1}}(y)\\
&= J_f(f^{-1}(y)) \left[J_f(f^{-1}(y)) \right]^{-1} \\
&= 1.
\end{align}
$$

It should be noted that $(*)$ is a very special property of the Lebesgue measure, and it is the essential equation that is needed to obtain equations $(1)$ through $(3)$, and not the other way round, which is why Billingsley states it as the key theorem.
