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I am looking for a reference on the statement that real analytic sets (i.e. sets in the form $u^{-1}(0)$ where $0\not\equiv u:E\subset \mathbb{R}^n\to \mathbb{R}$ is analytic, or finite intersections of sets of this type for several analytic functions) are $(n-1)$-rectifiable (in the sense of https://en.wikipedia.org/wiki/Rectifiable_set).

I know this result is supposed to be found somewhere in Federer's Geometric measure theory, but the language of the book is too complex for me so I am having a hard time finding it and translating it in the simple terms of the above definitions.

EDIT: I believe the result is contained here (Federer's GMT, Theorem 3.4.8(13))

The main problem is understanding $\dim \alpha$, which is defined in a very complicated way in terms of ideals. If I could understand why $\dim \alpha\leq n-1$ when $\alpha$ is the germ of an analytic set, that would be enough.

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  • $\begingroup$ Do you mean $m$-rectifiable by rectifiable (and for what $m$)? $\endgroup$
    – Skeeve
    Commented Jun 6, 2019 at 10:09
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    $\begingroup$ I think the answer is essentially contained in Proposition 3 here. $\endgroup$
    – Skeeve
    Commented Jun 6, 2019 at 12:47
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    $\begingroup$ @Skeeve It says that the dimension is $\leq n-1$, but how does that imply that it is rectifiable? $\endgroup$
    – Lorenzo Q
    Commented Jun 6, 2019 at 12:54
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    $\begingroup$ In general finiteness of the Hausdorff measure of a set does not imply rectifiability, as far as I remember. But the cited note actually shows that $u^{-1}(0)$ can be covered by countably many graphs of $C^1$ functions from $\mathbb R^{n-1}$ to $\mathbb R^n$ (see the proofs of claims 2 and 1). $\endgroup$
    – Skeeve
    Commented Jun 6, 2019 at 15:43
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    $\begingroup$ Rectifiability of a set $F$ requires a covering up to a null set, i.e. ${F\setminus \bigcup _{i=0}^{\infty }f_{i}\left(\mathbb {R} ^{m}\right)}$ should be null (in particular it can be empty), not the symmetric difference ${F\mathop{\triangle} \bigcup _{i=0}^{\infty }f_{i}\left(\mathbb {R} ^{m}\right)}$. $\endgroup$
    – Skeeve
    Commented Jun 7, 2019 at 8:18

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