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Let $U$ be an open set in $\mathbb{R}^n$, $T:U \rightarrow \mathbb{R}^n$ be a diffeomorphism defined on $U$, $\mu$ be the $d$-dimensional Hausdorff measure, $0<d<n$.

Is it true that we have the following formula:

For all borel measurable function $f:T(U)\rightarrow \mathbb{R}$, do we have $$ \int _{T(U)}f d\mu=\int_{U}f\circ T|det(DT)|^{d/n} d\mu $$ where $DT$ is the Jacobian of $T$?

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Can't it happen that $\mu(U) = \infty$ since $d<n$ and $U$ is an open subset of $\Bbb R^n$? – Ilya Jun 3 '13 at 15:35
Yes,$\mu(U)=\infty$ but $f$ can have very small support. – Ben Jun 4 '13 at 9:18
up vote 1 down vote accepted

No it isn't you cannot just "equally distribute" the determinant over the dimensions, as $T$s action will in general depend on the direction. Consider the following example: $U = \mathbb R^2$, $T = \binom{1\,0}{0\,2}$, $d=1$. Let $f = \chi_{[0,1]\times \{0\}}$. Then $$ \int_{\mathbb R^n} f \, d\mu = \mu([0,1]\times \{0\}) = 1 $$ but $$ \int_{\mathbb R^n} f \circ T (\det DT)^{1/2}\, d\mu = \sqrt 2 $$ as $T|_{[0,1]\times \{0\}} = \mathrm{id}$, $DT(x) = T$ for all $T$ and $\det T = 2$.

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Do you know is there any change of variable formula for non integral dimensional hausdorff measure? – Ben Jun 3 '13 at 13:58

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