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,$\displaystyle \sum_{i}$diam(Bi )>=$\displaystyle \sum_{i}$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!