Hausdorff and Lebesgue measure are equal (Proof)

We prove the equality of the Lebesgue measure with the Hausdorff measure (and with the Hausdorff content). We follow the amazing book "Measure theory and fine properties of functions" by Evans and Gariepy. One must use the isodiametric inequality.

[xxcheckerxx]

Hi,
Thanks for the video.
I was wondering in the "Corollary" you are stating that H^n_\infty (A) = 0 implies that H^n (A) = 0 if A is a null set, but do we really need this? In other words H^n(A) \leq C_n L^n(A) is not used, only H^n_\delta(A) \leq C_n L^n(A) for arbitrary \delta suffices?
Because later when adding and subtracting \cup B^i_j and using that the set (Q_i - \cup B^i_j) is a null set, we are using H^n_\delta and not H^n and then after pass to the limit in \delta.

I hope you can understand my point.

Best