Download PDF: Overview

Some notes on the taxonomy of spaces and some things to do for me.

Topological Spaces

Open sets are defined as sets where the $\varepsilon$-ball around each point in the set is contained in the set; i.e., all points are interior points. Closed sets are ones whose complements are open. Equivalently, they are sets that contain all of their limit points. This distinction in sets is useful because it allows to either bound quantities and ensure our set captures some $\varepsilon$-error, or ensure we do not miss any points a sequence converges to, respectively.

Topology also helps us define what is meant by a nice space. Consider the definition of a continuous function $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$ in topological terms: if, for every open set $V \subseteq \mathbb{R}^m$, the inverse image $f^{-1}(V) = \{x \in \mathbb{R}^n : f(x) \in V $ is an open subset of $\mathbb{R}^n$. A similar definition applies to measurable functions, $f: E \rightarrow \mathbb{R}^m$ in topological terms: if, for any interval $I \subseteq \mathbb{R}^m$, the inverse image $f^{-1}(I) = \{x \in \mathbb{R} : f(x) \in I \} \in \mathcal{M} $, where $E, \mathcal{M}$ are both measurable spaces. The definitions follow the same structure, namely: an operator has a certain property' if its image and inverse image have the property'.

The most natural question is whether the inverse image condition is necessary. After all, at first glance, we might use sets of the domain and their associated image. However, upon closer inspection, we find that the topological definition is merely a generalization of the analytic definitions. In analysis, we say: if you give me an $\varepsilon$, then I can find a $\delta$. The $\varepsilon$ applies to the range, while the $\delta$ must be found in the domain. Broadly speaking, the point of a function is that one has control over the inputs in order to realize a desired output. Thus, for a given output `property’, one attempts to find the corresponding conditions to place on the input. Hence, the inverse image defines the function.

Complete Spaces

Use my notes and introductory section in Real Analysis by Stein and Shakarchi. Already know definition in terms of metric spaces and how this can be generalized to function spaces. Check for a definition in terms of topological spaces.

Compact Spaces

Previously, we `constructed’ $\mathbb{R}$ with the axiom of completeness. However, sometimes this is too weak, and we must impose a stronger requirement on the spaces we work on. This is the aim of compactness. There are two definitions of this term, one for metric spaces and one for topological spaces. Fortunately, on $\mathbb{R}$ they are equivalent. Seeing that we defined completeness by metric spaces, we will start with the same for compactness.

Recall that completeness requires every Cauchy sequence to converge to a value in the space. Analogously, compactness requires that every sequence contain a convergent subsequence to a value in the space. At first glance, the subsequence requirement may seem arbitrary. After all, why not require every sequence to converge? The reason is that we wish to include sequences that have multiple accumulation points. Consider, for example, $\{ x_n \} = \{0,1,0,1,...\}$. The sequence is divergent, as successive terms always remain of constant distance apart, but also contains two accumulation points, namely, $0,1$. Thus, in order to include divergent sequences that have some accumulation point, we impose the criterion of convergent subsequences and not of convergent sequences. An important implication of this definition is the Bolzano-Weierstrass theorem, which states that every infinite bounded sequence in $\mathbb{R}$ has a convergent subsequence. The proof involves successively splitting the space into two intervals and noting that at least one should contain infinitely many points since the sequence is infinite and bounded. Thus, even as the length of the intervals tends to $0$, at least one will contain an accumulation point, and so a convergent subsequence can be found. The theorem explicitly gives the type of sequences we can expect to find in $\mathbb{R}$ when we require spaces to be compact.

The equivalent definition of compactness in topological spaces is as follows: every open cover of a set has a finite subcover. When an open cover is larger than the set, $A$, there always exists a finite subcover, as a subcover that is $A$, or even $\varepsilon$ larger, can just be used. Things become more tricky when we limit ourselves to open covers built by subsets of $A$. When $A$ is open, this is always possible; the trivial case is the open cover is $A$ itself. Other times it is a union of (in)finitely many joint subsets of $A$. A closed set $A$, on the other hand, does not have an open cover which can be constructed from only subsets of $A$, as the boundary point will always be missed. Thus, at the boundary of a closed set $A$, some overlap will inevitably occur with an open set that is not a subset of $A$. This is formalized in the Heine-Borel theorem, which states that a set is compact if and only if it is closed and bounded. Boundedness allows for any attempt at infinite subcovers to accumulate at a point, and closure ensures the accumulation is always within the set, and thus can be replaced by just one of the subcovers which contains the accumulation point, usually a subcover with elements not in $A$ if the accumulation point is a boundary point.

Measurable Spaces

A measurable space is defined by $(X, \mathcal{M})$, where $X$ is the underlying space and $\mathcal{M}$ is the $\sigma$-algebra. It can be constructed through topologies or through measures.

First, an introduction to the $\sigma$-algebra of a set. This means given a set, what are all possible sets that can be created using this set, where the only tools available are the complement and countable unions. To get the smallest possible $\sigma$-algebra, one takes the intersection of all possible $\sigma$-algebras of the generating set. It is perhaps easiest to understand measurable spaces by first understanding a finite $\sigma$-algebra, as they are a requirement of the space. Consider the finite set $X = \{1,2\}$. Then $\mathcal{P}(X) = \{\emptyset, \{1\}, \{2\}, \{1,2\}\}$. Note that $X \subseteq\mathcal{P}(X)$. Also note that $\forall A \subseteq \mathcal{P}(X), \: A^c \subseteq \mathcal{P}(X)$. For example, $\{1\}^c = \{2\}, \: \{1,2\}^c = \emptyset$ and so forth. Finally, any countable union of the subsets of $\mathcal{P}(X)$ will remain in $\mathcal{P}(X)$. Thus, $(X, \mathcal{P}(X))$ is a measurable space. This is always true for any finite set $X$. However, not all subsets of a measurable set are measurable, though usually only through contrived examples like the Vitali set.

To define a measurable space in $\mathbb{R}$ through topology requires the Borel-algebra. This is a generalization of finite $\sigma$-algebras, where the generating sets are now open sets and the $\sigma$-algebra is any set that can be `built’ by open sets. We cannot list all infinite Borel sets but do require the Borel-algebra to be the smallest $\sigma$-algebra of open sets. In practice then, to show a set is part of a measurable space defined by the Borel-algebra, all we need to do is show the set we are using is part of this $\sigma$-algebra of open sets. To do this, it is easiest to just construct our set from countable unions, countable intersections, and complements of open sets. Another way to do this is to show the $\sigma$-algebra generated by the set is equivalent to $\mathcal{B}$, though this is usually difficult. Notably, while the Borel algebra does include singleton sets, it does not include all null sets, and as such, is incomplete. Evidently, there are null sets which cannot be constructed by only open sets, namely the Vitali set. Thus, a complete measurable space generated by open sets requires a union of the Borel-algebra with all null sets.

To define a measurable space in $\mathbb{R}$ through measure requires the Caratheodory criterion. This states that a set is measurable if it can partition into two all other sets in the space. In practice, one can usually just choose an arbitrary representative of the space to test this against. The space of all sets which pass this criterion forms a measurable space. Naturally, such a space has a $\sigma$-algebra, for the outer measure is closed under complement and countable unions. In fact, over $\mathbb{R}$, the measurable space defined by the Caratheodory criterion is equivalent to the union of the measurable space defined by the Borel-algebra with all null sets. Notably, this makes the space of all sets which meet the Caratheodory criterion is complete, for all null sets have a defined outer measure, namely a vanishing one. Thus, the Caratheodory criterion is a slightly more robust and easier verification of whether a set is measurable.

Measure Spaces

Measure spaces are measurable spaces equipped with a measure, defined by $(X,\mathcal{M}, \mu)$. A function $\mu : X \rightarrow \mathbb{R}_{+}^{\infty}$ is considered a measure if it is non-negative, maps the empty set to zero, and satifies countable additivity: $\forall E_k \subseteq X, \: E_i \cap E_j = \emptyset, \: i \neq j \implies \mu \left( \bigcup_{k}^{\infty} E_k\right) = \sum_{k}^{\infty}E_k$

In using Caratheodory’s criterion to define a measurable space, it seemed a bit arbitrary to test a set against all other sets, even those which may not be measurable. Moreover, the test requires use of an outer measure. So it seems as if there is some circularity: a measure is being used to define a space that has no measure by taking into account how measurable a possibly non-measurable set is. In fact, this all just stems from obfuscatory names.

For one, the outer measure is not actually a measure, as it does not satisfy the requirement of additivity. However, since it does satisfy sub-additivity, the outer measure acts as a good litmus test for sets that might be measurable. This then leads to the next distinction: the outer measure is a function that can be applied to any set but does not alone determine whether a set is measurable. At best, it just tells us whether a set might be amenable to a measure. Again, this is because the outer measure is not a measure. Thus it is the quality of being a proto-measure combined with the criterion that allows us to define measurable spaces in terms of the outer measure. Essentially, it tells us whether the outer measure applied on a set yields a sensible result, only then can it be considered as a possible domain for a true measure; i.e., measurable. The remarkable consequence is that measurable spaces can be defined by this method or by the completion of Borel-algebras (on $\mathbb{R}$).

Difficult Proofs

\subsubsection*{Outer measure $=$ Length} Try original proof $\left| l(I) - m^*(I) \right| < \varepsilon$. The proof given in the book is sketched below.

Recall that $m^*(X)$ is the $\inf$ of all covers of a space. Since all intervals have an open cover, we can safely be sure all intervals are measurable and the outer measure preserves the notion of length. Since $m^*(A)$ is defined as the $\inf$ of all possible coverings of set $A$, it follows that $m^*(A) \leq \sum_{n}l\left( I_n \right)$ from definition, where $I_n$ is any open covering of $A$ and $l(I_n)$ is the length of interval $I_n$.

However, it is much harder to prove that $\exists I_n \: s.t. \sum_n{l\left( I_n \right)} \leq m^*(A)$. This is most commonly done by first observing that if $m^*(A)$ is the $\inf$ of all possible coverings of $A$, $\exists \varepsilon \in \mathbb{R}, \exists z \in Z \subseteq \mathbb{R} \: s.t. \inf Z \leq z \leq \inf Z + \varepsilon$. Thus, by replacing $z$ with a specifically constructed $\sum_n{l\left( I_n \right)}$ and $\inf Z$ with $m^*(A)$ and arguing the inequality holds as $\varepsilon \rightarrow 0$, we may achieve the desired result. This is usually done by constructing an open cover that has geometrically decreasing size for each successive interval, and is therefore only $r\cdot\varepsilon$ worse than $m^*(A)$, where $r$ is the constant determined by the geometric series.

Part 2 of Theorem 2.7 also uses these methods.

Caratheodory $\implies$ Borel

Sketch the proof. Theorem 2.8 essentially.

Subsets of Measurable Sets

Subsets of measurable sets are not necessarily measurable. Take the Vitali set in $\mathbb{R}$, for example. This is because the subset operation is not an operation used in a $\sigma$-algebra. However, this the exception and not the norm, and usually only contrived examples of subsets of measurable sets are not measurable.