Change of variable within an integral of the Hausdorff measure

Question

Let $T \colon \mathbb{R}^n \to \mathbb{R}^n$ be a linear map, $H^{m}$ be a Hausdorff measure. Is it true that $$ \int\limits_{T(M)} f(x) H^{m}(dx) = |\det{T}| \int\limits_{M} f(T(x)) H^{m}(dx) $$ where $f(x)$ is some continuous function?

Answer 1

In general this formula doesn't hold.

Consider case $n=3$, and $M=\{(x,y,z):0\leq x\leq 1,\quad 0\leq y\leq 1\quad z=0\}$ is a square on $xy$-plane. Define linear transformation $T$ by matrix $$ T_k=\begin{vmatrix}1 && 0 && 0\\0 && 1 && 0\\0 && 0 && k \end{vmatrix} $$ Obviously $\operatorname{det}(T_k)=k$ and $T_k(M)=M$. Take $f:\mathbb{R}^3\to\mathbb{R}$ to be constant function, i.e. $f(x)=1$. Thus we see that $$ \int\limits_{T_k(M)}f(x)H^m(dx)=\iint\limits_{M}1dxdy=1 $$ $$ \operatorname{det}(T_k)\int\limits_{M}f(T_k(x))H^m(T_k(dx))=k\iint\limits_{M}1dxdy=k $$ Hence for $k\neq 1$ $$ \int\limits_{T_k(M)}f(x)H^m(dx)\neq\operatorname{det}(T_k)\int\limits_{M}f(T_k(x))H^m(T_k(dx)) $$ If we take $k=0$ we have counterexample for sigular matrices. If we take $k=2$ we have counterexample for non-singular matrices.

Answer 2

No; it must be slightly more interesting than that. Consider the Cantor set, which has $\frac{\log 2}{\log 3}$-dimensional measure 1. Expanding the set by a factor of 3 produces two identical copies of the original set, presumably doubling the $\frac{\log 2}{\log 3}$-dimensional measure.

One might suspect (as did I) that, instead, the formula should be $$ \int_{T(M)} f\,dH^\alpha = |\det T|^{\alpha/n} \int_M (f\circ T)\,dH^\alpha. $$

As a matter of fact, examining the definition of $\alpha$-dimensional Hausdorff measure, mere scaling does produce this behavior:

$$ \mu_\alpha(E) = \lim_{\delta \to 0} \inf\left\{\left.\sum_k(\mbox{diam }F_k)^\alpha\ \right|\ E \subset \bigcup_{k=1}^\infty F_k,\ \ \mbox{diam }F_k\leq\delta,\forall k\right\}, $$ so $$ \mu_\alpha(\lambda E) = \lim_{\delta \to 0} \inf\left\{\left.\sum_k(\mbox{diam }F_k)^\alpha\ \right|\ \lambda E \subset \bigcup_{k=1}^\infty F_k,\ \ \mbox{diam }F_k\leq\delta,\forall k\right\}\\ = \lim_{\delta \to 0} \inf\left\{\left.\sum_k(\mbox{diam }F_k)^\alpha\ \right|\ E \subset \bigcup_{k=1}^\infty \lambda^{-1}F_k,\ \ \mbox{diam }F_k\leq\delta,\forall k\right\}\\ =\lim_{\delta \to 0} \inf\left\{\left.\sum_k(\mbox{diam }\lambda F_k)^\alpha\ \right|\ E \subset \bigcup_{k=1}^\infty F_k,\ \ \mbox{diam }\lambda F_k\leq\delta,\forall k\right\}\\ = \lim_{\delta \to 0} \inf\left\{\left.\sum_k(\lambda\,\mbox{diam }F_k)^\alpha\ \right|\ E \subset \bigcup_{k=1}^\infty F_k,\ \ \lambda\,\mbox{diam }F_k\leq\delta,\forall k\right\}\\ =\lambda^\alpha \lim_{\delta \to 0} \inf\left\{\left.\sum_k(\mbox{diam } F_k)^\alpha\ \right|\ E \subset \bigcup_{k=1}^\infty F_k,\ \ \mbox{diam }F_k\leq\frac{\delta}{\lambda},\forall k\right\}\\ = \lambda^\alpha \lim_{\delta \to 0} \inf\left\{\left.\sum_k(\mbox{diam }F_k)^\alpha\ \right|\ E \subset \bigcup_{k=1}^\infty F_k,\ \ \mbox{diam }F_k\leq\delta,\forall k\right\}\\ =\lambda^\alpha \mu_\alpha(E). $$

However, again referring to the Cantor set, we can see that the formula I suspected is false by shearing the Cantor set vertcially; this does nothing whatever to the set, but the determinant of this shearing transformation is the factor by which we scale.