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In general no such stratification exists. Lipschitz functions coincide with $C^1$ functions on sets of positive measure:

Theorem. If $f$ is Lipschitz in $\mathbb{R}^n$, then for any $\epsilon>0$ there is $g\in C^1(\mathbb{R}^n)$ such that the Lebesgue measure of the set $\{ f\neq g\}$ is less than $\epsilon$.

You can find a proof of this result in almost any book on geometric measure theory, see for example Theorem 1 in Section 6.6.1 in L. C. Evans, R. F. Gariepy, Measure theory and fine properties of functions. Studies in Advanced Mathematics. CRC Press, Boca Raton, FL, 1992.

Therefore, you can decompose the graph of $f$ into countably many $C^1$ pieces (defined on measurable sets) and a set of measure zero. However, in general a Lipschitz function need not coincide with a $C^2$ function on a set of positive measure so there is no hope for any sort of $C^2$ stratification even for functions defined on a real line.

By the way, I checked the book by Vladimir Shikhman and there is no such result as the one formulated in the question. Instead, they consider Lipschitz functions $f : \mathbb{R}^n\to \mathbb{R}$ whose graphs admit a $C^\infty$- Whitney stratification, and that is a different story. No theorem, but assumption.

In general no such stratification exists. Lipschitz functions coincide with $C^1$ functions on sets of positive measure:

Theorem. If $f$ is Lipschitz in $\mathbb{R}^n$, then for any $\epsilon>0$ there is $g\in C^1(\mathbb{R}^n)$ such that the Lebesgue measure of the set $\{ f\neq g\}$ is less than $\epsilon$.

You can find a proof of this result in almost any book on geometric measure theory, see for example Theorem 1 in Section 6.6.1 in

L. C. Evans, R. F. Gariepy, Measure theory and fine properties of functions. Studies in Advanced Mathematics. CRC Press, Boca Raton, FL, 1992.

Therefore, you can decompose the graph of $f$ into countably many $C^1$ pieces (defined on measurable sets) and a set of measure zero. However, in general a Lipschitz function need not coincide with a $C^2$ function on a set of positive measure so there is no hope for any sort of $C^2$ stratification even for functions defined on a real line.

By the way, I checked the book by Vladimir Shikhman and there is no such result as the one formulated in the question. Instead, they consider Lipschitz functions $f : \mathbb{R}^n\to \mathbb{R}$ whose graphs admit a $C^\infty$- Whitney stratification, and that is a different story. No theorem, but assumption.

In general it is an impossible task. Lipschitz functions coincide with $C^1$ functions on sets of positive measure: If $f$ is Lipschitz in $\mathbb{R}^n$, then for any $\epsilon>0$ there is a $g\in C^1$ such that the measure of the set $\{ f\neq g\}$ is less than $\epsilon$ so you can decompose the graph of $f$ into countably many $C^1$ pieces (defined on measurable sets) and a set of measure zero. You can find a proof of this fact in almost any book on geometric measure theory. However, in general a Lipschitz function need not coincide with a $C^2$ function on a set of positive measure so there is no hope for any sort of $C^2$ stratification even for functions defined on a real line.

By the way I checked the book by Vladimir Shikhman and there is no such result as the one formulated in the question. Instead they consider Lipschitz functions $f : \mathbb{R}^n\to \mathbb{R}$ whose graphs admit a $C^\infty$- Whitney stratification and that is a different story. No theorem, but assumption.

In general no such stratification exists. Lipschitz functions coincide with $C^1$ functions on sets of positive measure:

Theorem. If $f$ is Lipschitz in $\mathbb{R}^n$, then for any $\epsilon>0$ there is $g\in C^1(\mathbb{R}^n)$ such that the Lebesgue measure of the set $\{ f\neq g\}$ is less than $\epsilon$.

You can find a proof of this result in almost any book on geometric measure theory, see for example Theorem 1 in Section 6.6.1 in L. C. Evans, R. F. Gariepy, Measure theory and fine properties of functions. Studies in Advanced Mathematics. CRC Press, Boca Raton, FL, 1992.

Therefore, you can decompose the graph of $f$ into countably many $C^1$ pieces (defined on measurable sets) and a set of measure zero. However, in general a Lipschitz function need not coincide with a $C^2$ function on a set of positive measure so there is no hope for any sort of $C^2$ stratification even for functions defined on a real line.

By the way, I checked the book by Vladimir Shikhman and there is no such result as the one formulated in the question. Instead, they consider Lipschitz functions $f : \mathbb{R}^n\to \mathbb{R}$ whose graphs admit a $C^\infty$- Whitney stratification, and that is a different story. No theorem, but assumption.

In general it is an impossible task. Lipschitz functions coincide with $C^1$ functions on sets of positive measure: If $f$ is Lipschitz in $\mathbb{R}^n$, then for any $\epsilon>0$ there is a $g\in C^1$ such that the measure of the set $\{ f\neq g\}$ is less than $\epsilon$ so you can decompose the graph of $f$ into countably many $C^1$ pieces (defined on measurable sets) and a set of measure zero. You can find a proof of this fact in almost any book on geometric measure theory. However, in general a Lipschitz function need not coincide with a $C^2$ function on a set of positive measure so there is no hope for any sort of $C^2$ stratification even for functions defined on a real line.

In general it is an impossible task. Lipschitz functions coincide with $C^1$ functions on sets of positive measure: If $f$ is Lipschitz in $\mathbb{R}^n$, then for any $\epsilon>0$ there is a $g\in C^1$ such that the measure of the set $\{ f\neq g\}$ is less than $\epsilon$ so you can decompose the graph of $f$ into countably many $C^1$ pieces (defined on measurable sets) and a set of measure zero. You can find a proof of this fact in almost any book on geometric measure theory. However, in general a Lipschitz function need not coincide with a $C^2$ function on a set of positive measure so there is no hope for any sort of $C^2$ stratification even for functions defined on a real line.

By the way I checked the book by Vladimir Shikhman and there is no such result as the one formulated in the question. Instead they consider Lipschitz functions $f : \mathbb{R}^n\to \mathbb{R}$ whose graphs admit a $C^\infty$- Whitney stratification and that is a different story. No theorem, but assumption.