monotone convergence theorem and calculaton of Hausdorff dimension

Question

I'm reading these notes, and I'm stuck on an application of the monotone convergence theorem. In computing the Hausdorff dimension of the Koch curve (page 13), the authors show that

1. $H^p(S^{i+1}(A)) > H^p(S^i(A))$ for all $i$
2. $H^p(K)$ is a finite limit of the sequence $H^p(S^{i}(A))$

where the $S^i(A)$ are the $i^{th}$ iteration of the recursive construction of the Koch curve, $K = \lim_{i \to \infty} S^i(A)$ is the Koch curve, and $p$ is the Hausdorff dimension of $K$. $H^p(X)$ is the Hausdorff measure of a set $X$.

The authors then use the monotone convergence theorem to conclude $$H^p(\lim_{i \to \infty} S^i(A)) = \lim_{i \to \infty} H^p(S^i(A)).$$ I'm not much of an analyst and really struggling to see how this is an application of the monotone convergence theorem. I'm used to this theorem being the statement that limits commute with integrals in the case of positive integrable functions in an increasing sequence. I don't see where the functions, or the integrals are in the above statement - could someone explain what the relevant measures, integrals, and functions are?

Edit: I've added a bounty because I still haven't gotten a satisfying answer here or found one any other way. I'm aware that the argument presented in the linked text might be incorrect/insufficiently rigorous, so I would also accept a rigorous computation using other methods, as long as they remain "at the level" of the paper (i.e. I've seen some methods that rely on Radon measures and some other measure theory background I'm not familiar with. I would accept an answer that is rigorous but doesn't use these methods).

Answer 1

There is no wonder you are suspicious of that article.

It is wrong that "$H^p(S^{i+1}(A)) > H^p(S^i(A))$ for all $i$"

In fact, $H^p(S^{i+1}(A))=H^p(S^{i}(A))=0$ for all $i$.

Here is why. Since $S^i(A)$ is a union of finitely many (straight) line segments, $H^1(S^{i}(A))$ is finite. By the theorem in section $3.1$ of that article, $H^p(S^i(A))=0$ since, as we know from elsewhere, $p=\frac{\log 4}{\log3}>1$.

By the way, that article reads "$p=\frac{\log 4}{\log(1/3)}\approx1.26$" on page $14$. That is an obvious typo, since $\log(1/3)<0$.

It is wrong that "$H^p(K)$ is a finite limit of the sequence $H^p(S^{i}(A))$"

As said above, $H^p(S^{i}(A))=0$ for all $i$. Their limit is $0$ while $H^p(K)$ is, as we know from elsewhere, nonzero.

A note on the definition of Hausdorff dimension

That article, Hausdorff Measure by Jimmy Briggs and Tim Tyree only defines Hausdorff dimension when the Haudorff measure is a finite, nonzero number.

Definition. We then see that if $H^p(A)$ is a finite nonzero value, then for all $k \in [0, +\infty)$ such that $k\not=p$, $H^k(A)$ is either infinite or zero. Then, we define the Hausdorff Dimension of set $A$, $\dim H(A) = p$ if and only if $H^p(A)$ is a finite, nonzero number.

However, such $p$ may not exist for some metric spaces, even if they are closed subsets of a finite-dimensional Euclidean space.

• There is a closed subset $M_{\frac12}$ of $\Bbb R$ such that
  • $H^p(M_\frac12)=\infty$ if $p<\frac12$ and
  • $H^p(M_\frac12)=0$ if $p\ge\frac12$.
• There is a closed subset $N_{\frac12}$ of $\Bbb R$ such that
  • $H^p(N_\frac12)=\infty$ if $p\le\frac12$ and
  • $H^p(N_\frac12)=0$ if $p>\frac12$.

The number $\frac12$ above could be replaced any number in $(0,1)$.

It turns out that article does not even prove the Koch curve does have a Hausdorff dimension according to its own definition!

Please check other definitions of Hausdorff dimension such as the one on Wikipedia.