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I posted this on MSE, but no answer is received, so I post this here.

I quote from wiki:

The coarea formula can be applied to Lipschitz functions $u$ defined in $Ω ⊂ \mathbb R^n$, taking on values in $\mathbb R^k$ where $k < n$. In this case, the following identity holds $$\int_\Omega g(x) |J_k u(x)|\, dx = \int_{\mathbb{R}^k} \left(\int_{u^{-1}(t)}g(x)\,dH_{n-k}(x)\right)\,dt$$

where $J_ku$ is the $k$-dimensional Jacobian of $u$.

I know that if $Du$ has full rank, $|J_k u(x)|=\sqrt{\det DuDu^T}$. But when $Du$ does not have full rank, how is $|J_k u(x)|$ defined?

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    $\begingroup$ Is it not just 0? $\endgroup$ Commented Feb 16, 2017 at 14:09
  • $\begingroup$ @ Steven Gubkin: so you mean the coarea formula fail in this case? $\endgroup$
    – JSCB
    Commented Feb 16, 2017 at 14:15
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    $\begingroup$ No. If I am interpreting your formula correctly, then both sides of the integral would be zero if this jacobian always had zero determinant. This makes sense: it says that the area of a curve is zero for instance. $\endgroup$ Commented Feb 16, 2017 at 14:18

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It is defined through the same formula! For $k \times n$ matrix $Df$, we have $Df \cdot Df^T$ a square $k \times k$ matrix, so its determinant is defined. There is a linear algebra lemma relating it to the minors of $Df$, some square root of squares of all $k$ by $k$ minors, i.e determinant of $k\times k$ submatrices of $Df$. This quantity measures the ratio of the $k$-dimensional volume, say, $k$-Hausdorff measure, of the image of a set in the domain (inside ambient $\mathbb{R}^n$) to the (Lebesgue) measure of the set.

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