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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 $D_1(A) \rightarrow A$?

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    What do you get when $k=2, m=1$?2012-04-06
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    I've commadiered your question for a bounty, hope you don't mind :)2016-06-18
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    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

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