Let K to be a Cantor middle-half set. Consider KXK.
figure 2 shows the 1-stage of this cantor set construction.
diam(B)=sup{|x-y|:x,ybelongs to B}
Want to show that
Ans from the article:
The lower bound can be derived by considering the probability measure μ on K^2 such that μ(K^2 ∩ Q) = 4^(−n) for each of the 4^n squares Q arising at stage n of the construction. This measure satisfies μ(B) ≤ 9diam(B) for any closed set B, since B can be covered by at most nine dyadic squares with side-length smaller than diam(B). Therefore, for any cover {Bi } of K^2 by disks, diam(Bi )>= μ(Bi )/9>=μ(K^2)/9 =1/9
Question
I do not understand why μ(B) ≤ 9diam(B). I do not understand why probability measure is related to the diameter.
Please help me!