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Feb 18, 2020 at 9:50 comment added Weekkola Thanks for you answer, Piotr! I think I should have chosen less confusing variable names for the dimensions because judging from your statement $T_x A = T_x M$ I think you assumed $\mathop{dim} M = d$. In any case, this shouldn't matter too much. If $A$ is $d$-rectifiable, it it is contained in countably many $d$-dimensional manifolds $N_i$ up to a set of $𝓗^d$-measure zero, and then for $𝓗^d$-a.e. $x \in A$ the approximate tangent space $T_x A$ agrees with the tangent space of one of the $N_i$ and this is obviously an invariant notion and independent of any embedding. Thanks again!
Feb 15, 2020 at 23:34 history answered Piotr Hajlasz CC BY-SA 4.0