Main Theme of the Question(s): Fractal Geometry by Kenneth Falconer describes the following method to construct a mass distribution $\mu$ (and later, a measure on $\mathbb R^n$) on a subset of $\mathbb R^n$. The author also says that we only concern ourselves with Borel sets. I need help understanding (i) the process of construction of this mass-distribution/measure, and (ii) the proof of a proposition that follows this discussion.
Slightly paraphrased from the book:
The following method is often used to construct a mass distribution on a subset of $ℝ^n$. It involves repeated subdivision of a mass between parts of a bounded Borel set $E$. Let $\mathcal{E}_0$ consist of the single set $E$. For $k = 1, 2, . . . $, we let $\mathcal{E}_k$ be a collection of disjoint Borel subsets of $E$ such that each set $U$ in $\mathcal{E}_k$ is contained in one of the sets of $\mathcal{E}_{k−1}$ and contains a finite number of the sets in $\mathcal{E}_{k+1}$. We assume that the maximum diameter of the sets in $\mathcal{E}_k$ tends to $0$ as $k \to \infty$. We define a mass distribution on $E$ by repeated subdivision. We let $\mu(E) \in (0,\infty)$. In general, we assign masses so that $$\sum_i\mu(U_i) = \mu(U)$$ for each set $U$ of $\mathcal{E}_k$, where $\{U_i\}$ are disjoint sets in $\mathcal{E}_{k+1}$ contained in $U$. For each $k$, let $E_k = \bigcup_{A\in \mathcal{E}_k} A$, and we define $\mu(A) = 0$ for all $A$ with $A\cap E_k = \varnothing$.
Now the author states a proposition, which enables us to determine $\mu(A)$ for any Borel set $A$.
Proposition $1.7.$ Let $$ be defined on a collection of sets $\mathcal E$ as above. Then the definition of $$ may be extended to all subsets of $ℝ^n$ so that $$ becomes a measure. The value of $(A)$ is uniquely determined if $A$ is a Borel set. The support of $$ is contained in $E_∞ = \bigcap_{k=1}^\infty \overline{E_k}$.
Note on Proof. If $A \subset \mathbb R^n$, let $$\mu(A) = \inf\left\{\sum_{i=1}^\infty \mu(U_i): A \cap E_\infty \subset \bigcup_{i=1}^\infty U_i \text{ and } U_i \in \mathcal{E}\right\}$$ It is not difficult to see that if $A\in\mathcal E$, then $\mu(A)$ above reduces to the mass $(A)$ specified in the construction. The complete proof that $$ satisfies all the conditions of a measure and that its values on the sets of $\mathcal E$ determine its values on the Borel sets is quite involved, and need not concern us here. As $(ℝ^n\setminus E_k) = 0$, we have $(A) = 0$ if $A$ is an open set that does not intersect $E_k$ for some $k$, so the support of $$ is in $E_k$ for all $k$.
I have six questions:
- In the proposition above, why is $\mu(A)$ defined the way it is (to extend $\mu$ to sets other than those in $\mathcal E$)?
- How do I show that $\mu$ as defined above is in fact a measure on $\mathbb R^n$? The author says the proof is involved, but I want to see it. I need to prove three things:
(a) $\mu(\varnothing) = 0$,
(b) $A \subset B \implies \mu(A) \le \mu(B)$,
(c) $\mu\left(\bigcup_{i=1}^\infty A_i\right) \le \sum_{i=1}^\infty \mu(A_i)$ with equality if $A_i$'s are disjoint Borel sets.
Here's what I've tried:
(a)$$\begin{align}
\mu(\varnothing) &= \inf\left\{\sum_{i=1}^\infty \mu(U_i): \varnothing \cap E_\infty \subset \bigcup_{i=1}^\infty U_i \text{ and } U_i \in \mathcal{E}\right\}\\
&= \inf\left\{\sum_{i=1}^\infty \mu(U_i): U_i \in \mathcal{E}\right\}
\end{align}$$
Why should $\inf\left\{\sum_{i=1}^\infty \mu(U_i): U_i \in \mathcal{E}\right\}$ be $0$? I'm not sure if $\varnothing\in\mathcal E$.
(b) Suppose $A\subset B$. Then, $A\cap E_\infty \subset B\cap E_\infty$. We have $$\mu(A) = \inf\left\{\sum_{i=1}^\infty \mu(U_i): A \cap E_\infty \subset \bigcup_{i=1}^\infty U_i \text{ and } U_i \in \mathcal{E}\right\}$$ and $$\mu(B) = \inf\left\{\sum_{i=1}^\infty \mu(U_i): B \cap E_\infty \subset \bigcup_{i=1}^\infty U_i \text{ and } U_i \in \mathcal{E}\right\}$$ and from what I understand - $A\cap E_\infty \subset B\cap E_\infty$ directly implies $\mu(A) \le \mu(B)$. This is due to the fact that any cover $\bigcup_{i=1}^\infty U_i$ of $B\cap E_\infty$ is also a cover of $A\cap E_\infty$, but the converse is not true in general. Since $A\cap E_\infty$ can possibly have more covers of the form $\bigcup_{i=1}^\infty U_i$ for $U_i\in\mathcal E$, the infimum of $\sum_{i=1}^\infty \mu(U_i)$ over covers of $A\cap E_\infty$ is in general $\le$ the infimum of $\sum_{i=1}^\infty \mu(U_i)$ over covers of $B\cap E_\infty$. Does this make sense?
(c)We have $$\mu(A_i) = \inf\left\{\sum_{i=1}^\infty \mu(U_i): A_i \cap E_\infty \subset \bigcup_{i=1}^\infty U_i \text{ and } U_i \in \mathcal{E}\right\}$$ and $$\mu\left(\bigcup_{i=1}^\infty A_i\right) = \inf\left\{\sum_{i=1}^\infty \mu(U_i): \bigcup_{i=1}^\infty A_i \cap E_\infty \subset \bigcup_{i=1}^\infty U_i \text{ and } U_i \in \mathcal{E}\right\}$$
Clearly, $A_i \subset \bigcup_{i=1}^\infty A_i$, and from the previous part, i.e. (b), we have $\mu(A_i) \le \mu\left(\bigcup_{i=1}^\infty A_i\right)$. I'm not sure how this helps, and I'm unable to proceed further.
- Why does $E_∞ = \bigcap_{k=1}^\infty \overline{E_k}$ contain $\operatorname{spt}\mu$?
$\color{blue}{\text{Answered.}}$ As $\mu(\mathbb R^n \setminus E_k) = 0$, by monotonicity, we have $\mu(\Bbb R^n\setminus \overline{E_k}) = 0$. By the definition of support, $\operatorname{spt}\mu \subset \overline{E_k}$ for all $k$. So, $\operatorname{spt}\mu \subset E_\infty = \bigcap_{k=1}^\infty \overline{E_k}$.
- If $A\in\mathcal E$, how does $\mu(A)$ (given by the wild expression in the proposition) reduce to the construction of $\mu(A)$ (i.e. the mass distribution earlier)?
- How do we know that the value of $\mu$ on sets of $\mathcal E$ determines $\mu$ uniquely on all Borel sets?
- What is the point of all this? What have we achieved? I need help understanding the bigger picture!
Thanks a lot for reading, and for your help (in advance)!
References:
Kenneth Falconer. Fractal Geometry.