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Piotr Hajlasz
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Answer to Q2 is yes.

If $\mathcal{H}^1(C_g)<\infty$, then the graph has finite length (it is known that for a one-to-one curve $\mathcal{H}^1$ coincides with the length). However that implies that $g$ has bounded variation (Theorem 97 in [H]) and functions of bounded variation are differentiable a.e.

[H] P. HajlaszHajłasz, Measure Theory

Answer to Q2 is yes.

If $\mathcal{H}^1(C_g)<\infty$, then the graph has finite length (it is known that for a one-to-one curve $\mathcal{H}^1$ coincides with the length). However that implies that $g$ has bounded variation (Theorem 97 in [H]) and functions of bounded variation are differentiable a.e.

P. Hajlasz, Measure Theory

Answer to Q2 is yes.

If $\mathcal{H}^1(C_g)<\infty$, then the graph has finite length (it is known that for a one-to-one curve $\mathcal{H}^1$ coincides with the length). However that implies that $g$ has bounded variation (Theorem 97 in [H]) and functions of bounded variation are differentiable a.e.

[H] P. Hajłasz, Measure Theory

Source Link
Piotr Hajlasz
  • 28k
  • 5
  • 85
  • 184

Answer to Q2 is yes.

If $\mathcal{H}^1(C_g)<\infty$, then the graph has finite length (it is known that for a one-to-one curve $\mathcal{H}^1$ coincides with the length). However that implies that $g$ has bounded variation (Theorem 97 in [H]) and functions of bounded variation are differentiable a.e.

P. Hajlasz, Measure Theory