7
$\begingroup$

(Read bounty text for answering question)

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

What if we have a self similar object?

$A= D_1(A) \cup D_2(A)$

And transform $D_1(A) \rightarrow A$?

  • 0
    The symbol $k$ was lost in an edit. $k$ was the dimension of the space where the linear transformation $D$ is defined. So my idea of 2012 was: first do the case of arc-length in the plane.2016-06-18

1 Answers 1

5

Some remarks. The original question:

Does a function $a(y)$ exist?

is (by the Radon-Nikodym theorem) equivalent to the question

Is the image measure $D_*H^m$ absolutely continuous with respect to $H^m$?

(Let's suppose $A$ has sigma-finite measure.) And the answer is yes if $D$ is a non-singular matrix; the answer is sometimes no if $D$ is a singular matrix. But even knowing function $a(y)$ exists doesn't satisfy Zach, who wants an explicit formula.

Example (for the singular case)

Consider the Cantor singular function
picture
Our set $A$ is the graph of the function. It is a set with arc length exactly $2$ (that is, $1$-dimensional Hausdorff measure exactly $2$.)

Take $D$ to be the projection onto the $x$-axis. What is the image measure $D_*H^1$? Images of all the horizontal line-segments give us exactly Lebesgue measure in the real line. But images of the rest (a set of $H^1$-measure $1$) give us a singular measure concentrated on the Cantor set. The image measure is not absolutely continuous.

Or, alternatively, take $D$ to be the projection onto the $y$-axis. Now the line segments map to countably many point-masses (total mass $1$), while the rest of the graph maps exactly onto Lebesgue measure on the $y$-axis. But again, the image measure is not absolutely continuous.

There is a general theorem (due to Besicovitch) that generalizes this example. A set $E$ in Euclidean space with $0 < H^1(E) < \infty$ may be written as a disjoint union of two parts: one part is (up to measure zero) a countable union of rectifiable curves, the other part is "curve-free" and "dust-like". (See K. J. Falconer, Fractal Geometry, 3rd edition, section 5.2: "Structure of $1$-sets")

Remark (non-singular case: $D_*H^m \ll H^m$)

Suppose the matrix is non-singular. Then we have inequalities of the form $ \alpha \; \text{diam}\; A \le \text{diam}\; D(A) \le \beta \; \text{diam}\; A $ for all sets $A$. Here $\alpha, \beta$ are the smallest and largest singular values of $D$, and they are positive because $D$ is nonsingular.

Using that inequality we get $ \alpha^m H^m(A) \le H^m(D(A)) \le \beta^m H^m(A), \\ \alpha^m H^m(D^{-1}(A)) \le H^m(A) \le \beta^m H^m(D^{-1}(A)), $ so $D_*H^m \ll H^m$ as claimed.

  • 0
    Sorry if my edit came off a bit demanding. This is a great start! :)2016-06-18