# Radon-Nikodym derivative vs standard derivative. Multivariable case

For one dimensional case there is a nice connection of Radon-Nikodym derivative and "classical" derivative on real line. Is there some kind of analogy for higher dimensional cases?

## 1 Answer

Among other connections, the Radon-Nikodym derivative allows you to get the change of variable formula in $\mathbb{R}^n$.

Setup: Let $K\subset\mathbb{R}^n$ be compact and equal to the closure of its interior, let $U$ be an open neighborhood of $K$, and consider a $C^1$ map $T:U\to\mathbb{R}^n$ satisfying $|T(x)-T(y)|>\lambda|x-y|$ for all $x,y\in K$ and some $\lambda>0$. Then $T$ is one-to-one onto its image and $T^{-1}$ is Lipschitz with Lipschitz constant $\lambda^{-1}$.

Let $\mu = m\mid_K$ be the restriction of Lebesgue measure $m$ to $K$, i.e. $\mu(E) = m(E\cap K)$, and let $\nu = T\#\mu$ be the pullback measure $\nu(E) = \mu(T^{-1}(E))$. $\nu$ is absolutely continuous w.r.t. $m$, so $d\nu = fd\mu$ for some $f\in L^1(\mu)$. The Radon-Nikydym theorem and the general change of variable formula tell us that $$\int_U g\circ T~dm = \int_{T(U)}g~d(T\#\mu) = \int_{T(U)}gf~dm.$$ The general change of variable formula is hard to use in its usual form, but if we can obtain a formula for $f$ then we can get something much easier to work with.

In fact, under our current assumptions, we can get a formula for $f$. We then use the Lebesgue differentiation theorem to compute: $$f(x) = \lim_{r\to 0}\frac{1}{m(B_r(x))}\int_{B_r(x)}fdm = \lim_{r\to 0}\frac{\nu(B_r(x))}{\mu(B_r(x))} = \lim_{r\to 0}\frac{m(T^{-1}(B_r(x))\cap K)}{m(B_r(x))}.$$ The last limit essentially measures the volume distortion factor of $T^{-1}$, and therefore can be shown to be $|\det(DT^{-1})(x)| = |\det(DT)(x)|^{-1}$, where $DT$ is the Jacobian matrix. This is well defined because of the condition $|T(x)-T(y)|>\lambda|x-y|$, so $DT(x)$ is nonsingular for all $x$.

Consequently the change of variable formula is given by $$\int_U g\circ T~dm = \int_{T(U)}g(x)|\det(DT)(x)|^{-1}~dm(x).$$ Notice that we haven't invoked compactness or $T\in C^1$ yet. Since $T$ is $C^1$ and $K$ is compact (which is a very common scenario for integration, hence not a huge restriction), the derivative of $T$ is bounded on $K$ and hence $T$ is also Lipschitz. Write $S=T^{-1}$, $W=T(U)$; then our formula becomes $$\int_W g~dm = \int_W g\circ T\circ S~dm = \int_{S(W)}(g\circ T)|\det(DS)|^{-1}~dm = \int_{S(W)}(g\circ T)|\det(DT)|~dm,$$ that is, $$\int_{T(U)} g~dm = \int_U(g\circ T)|\det(DT)|~dm,$$ the familiar version of the change of variable formula from third semester calculus.