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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?

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You should look Measure thoery and fine properties of functions Lawrence C. Evans, Ronald F. Gariepy, Chapter 2 theorem 2. –  Norbert Feb 9 '12 at 12:41
    
Do you mean the (v)-th property? But it is valid only if $T$ is isometry. In an arbitrary case I think that $H(T(M))=| \det(T) | H(M)$ like it is in the case of Lebesgue measure. –  Nimza Feb 9 '12 at 13:24
    
I think that the main problem is to show that $\int\limits_{T(M)} f(x) H^m(dx) = \int\limits_{M} f(T(x)) H^m(T(dx))$ if $\det T \neq 0$. –  Nimza Feb 9 '12 at 13:43
    
Sorry, i have not saw this requirement. However, there are a lot of related theorems in section 3.2 –  Norbert Feb 9 '12 at 13:50
    
For $T(x) = \lambda x$ we can proove it using a standard procedure: at first we show this for indicators, then for weighed sums of indicators, then for positive functions and then for arbitraty measurable functions. To do this in general case we have to proove that $H(T(M)) = | \det(T) | H(M)$ (or to present a counterexample). –  Nimza Feb 9 '12 at 14:42
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In general this formula doesn't holds.

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.

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Thank you. And what about the most interesting case when $T$ is not singular? Is it true that $H^{s}(T(A)) = |\det{T}| H^{s}(A)$? –  Nimza Feb 9 '12 at 15:57
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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.

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Thank you, but why it doen't hold for $T(x) = \lambda x$, $f(x) = 1$: $\int\limits_{\lambda M} H^{m}(dx) = H^{m}(\lambda M) = \lambda^{m} H^{m}(M) = \lambda^{m} \int\limits_{M} H^{m}(dx)$, but $\det T = \lambda^{n}$, so $| \det T |^{\alpha} = \lambda^{mn}$? –  Nimza Feb 9 '12 at 16:35
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And what about $\mbox{diam }(TA) = | \det T |\mbox{ diam }(A)$? If we take $T = [1,m;0,1]$ (horizontal shear mapping) and $A = [0,1] \times [0,1]$ I think it won't work... –  Nimza Feb 9 '12 at 17:37
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I am a doofus! Yes, you're right. –  Robert Haraway Feb 9 '12 at 21:48
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