Let $H^{s}$ be the $s$-dimensional Hausdorff measure, let $D$ be a nonsingular matrix. Consider the change of measure formula: $$ \int\limits_{A} f(Dx) \; \mathrm{d}H^{s}(x) = \int\limits_{ D A} f(y) \; \mathrm{d}D_*H^{s}(y)\ , $$ where $D_{*}H^{s}(M) = H^{s}(D^{-1}M)$ is the pushforward of the Hausdorff measure. Is it possible to find such a function $a(x)$ that $$ \int\limits_{ D A} f(y) \; \mathrm{d}D_{*}H^{s}(y) = \int\limits_{ D A} f(y) a(y) \; \mathrm{d}H^{s}(y)\ ? $$

I'm very interested in a usable general change of variables formula; does that exist?


First of all, let us make a few general considerations about the existence of such an $a(y)$. That amounts to asking whether the Radon-Nikódym derivative of $H^s\circ D$ with respect to $H^s$ exists. If $H^s$ and $H^s\circ D$ were $\sigma$-finite, by the Radon-Nikódym theorem this is the same as asking whether $H^s\circ D$ is absolutely continuous with respect to $H^s$: if $H^s(A)=0$, then $H^s(DA)=0$ for any $H^s$-measurable $A$. The problem is that if $s<n$, then $H^s$ is not $\sigma$-finite on $\mathbb{R}^n$, so we cannot appeal to the Radon-Nikódym theorem for existence unless $s=n$.

Now, if $D$ is an homothety, i.e. $D$ is of the form $D=\lambda U$ where $\lambda>0$ and $U$ is an isometry (i.e. an orthogonal matrix), then $H^s(DA)=\lambda^sH^s(A)$, hence in this case $a(y)=\lambda^{-s}$. If, however, $D$ is not an homothety - e.g. a symmetric, non-singular but non-orthogonal matrix with $\det D=1$ -, I do not know of any simple transformation formula for the $s$-dimensional Hausdorff measure, and I suspect that there may be none leading to a formula such as the one you are looking for.

The problem is that since such a linear map might expand certain directions while contracting others, the naive expectation that $H^s(DA)=|\det D|^{\frac{s}{n}}H^s(A)$ (where $n$ is the rank of $D$) coming from the case that $D$ is an homothety may fail short depending on how $A$ is oriented with respect to the principal axes of $D$, as shown here. Since one can change this relative orientation just by an isometry (which, as seen above, does not change Hausdorff measure), I do not see any chance of such a function $a(y)$ existing independently of the choice of $A$. More precisely, if we take $f\equiv 1$, then your last formula just becomes $$D_*H^s(DA)=H^s(A)=\int_{DA}a(y)\;\mathrm{d}H^s(y)$$ for all $A\subset\mathbb{R}^n$ $H^s$-measurable. If we replace $A$ with $UA$ where $U$ is a suitable isometry, we should have $$H^s(UA)=H^s(A)=\int_{D(UA)}a(y)\;\mathrm{d}H^s(y)\ ,$$ so $a(y)$ should have an "anisotropic" behavior which seems incompatible with the fact that it is a scalar function. I do not know if I was able to make my concerns sufficiently precise - if I can find a more explicit example, I will update my answer.

  • $\begingroup$ So to summarize, you're saying that $a$ is probably not a scalar function? That would mean that the choice of $A$ will have a fact on the "Jacobian" factor in the integral? If so the formula may be complex, unusable...is there any reason a formula might not exist? I'd still be interested in knowing the correct change-of-variables formula. $\endgroup$ – Zachary W. Robertson Jun 18 '16 at 18:00
  • $\begingroup$ If $s$ is an integer and $A$ is $s$-rectifiable, the change-of-variables formula is essentially provided by the area formula (encyclopediaofmath.org/index.php/Area_formula). However, if $s$ is not an integer and $A$ is $H^s$-measurable with $0\neq H^s(A)<+\infty$ (e.g. fractal sets), then $A$ is not $k$-rectifiable for any non-negative integer $k$, so the area formula cannot be applied. I don't know of any replacement to it in this case - see also the math.SE question math.stackexchange.com/questions/1058275/… $\endgroup$ – Pedro Lauridsen Ribeiro Jun 18 '16 at 20:13

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.