3
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.