PUMaC 2024 · 加试 · 第 4 题

Problem 4.2.2. Find a mass distribution on the Cantor set such that the hypothesis of the Mass Distribution Principle is not met. More specifically, that for all c, ϵ > 0, there is some U ⊂ C where | U | ≤ ϵ , but s μ ( U ) > c | U | The example for the Cantor set provides a way to come up with a natural measure on any attractor of an IFS that satisfies OSC. Let our IFS be ( f , ..., f ) with contraction 1 m ratios c , ..., c , and attractor E . Then let s be the dimension of E , as determined via the 1 m OSC. Lastly, denote f ( X ) to be f ( f ( ... ( f ( X )))). Then we can define i ,i ,...,i i i i 1 2 m 1 2 m s s s μ ( f ( E )) = c c ...c i ,i ,...i 1 2 m i i i 1 2 m Notice that μ ( E ) = 1. The crucial fact to verify is that ! m m X [ μ ( f ( E )) = μ f ( E ) = μ ( E ) k k k =1 k =1 And indeed this follows from the definition of the attractor and of μ . Lastly, we extend this to Borel subsets of E . The easiest way is indeed the right way to do it: if A ⊂ E , then let J = { f ( E ) : f ( E ) ⊂ A } . That is, J is all the level i ,i ,...i i ,i ,...i k k 1 2 k 1 2 k k copies of E that fit inside of A . X μ ( A ) = lim μ ( E ) k →∞ E ∈ J k This is fancy notation for saying in order to get μ ( A ), we simply add the measure of all the sets which we can pack into A that we’ve already defined the measure for. Theorem 4.2.4. For E the attractor of the IFS ( f , ...f ) satisfying the OSC, the natural 1 m measure described above is a mass distribution. Even though extension to Borel sets makes the natural measure an actual measure, notice that in example 4.2, we avoided having to use that part of the definition by passing immediately to the intervals using sub-additivity. It would be wise for you to do the same. s Theorem 4.2.5. If an attractor E satisfies the OSC, with dimension s , then H ( E ) > 0 . Proof. Too long.