Notes on Folland, Real Analysis, 2nd ed.

Author

Deepak Mirchandani

Prologue

The Language of Set Theory

Orderings

Cardinality

More about Well Ordered Sets

The Extended Real Number System

Metric Spaces

Measures

We often call Folland’s Lemma 1.1 by the more descriptive name minimality. If \(\mathcal{M}(\mathcal{E})\) then \(\mathcal{M}(\mathcal{E}) = \bigcap \{ \mathcal{A} : \mathcal{A} \text{ is a } \sigma\text{-algebra and } \mathcal{E}\subseteq \mathcal{A}\}\). Thus \(\mathcal{M}(\mathcal{E})\) is the smallest \(\sigma\)-algebra containing \(\mathcal{E}\). Here “smallest” means:

\[\text{if } \mathcal{A} \text{ is a } \sigma\text{-algebra and } \mathcal{E}\subseteq \mathcal{A}, \text{ then } \mathcal{M}(\mathcal{E})\subseteq \mathcal{A}.\]

This is the idea behind Folland’s Lemma 1.1:

\[\mathcal{E}\subseteq \mathcal{M}(\mathcal{F}) \implies \mathcal{M}(\mathcal{E})\subseteq \mathcal{M}(\mathcal{F}).\]

That is, \(\mathcal{M}(\mathcal{F})\) is itself a \(\sigma\)-algebra containing \(\mathcal{E}\), so by minimality the smallest \(\sigma\)-algebra containing \(\mathcal{E}\), namely \(\mathcal{M}(\mathcal{E})\), must be contained in \(\mathcal{M}(\mathcal{F})\).

Introduction

Vitali sets are not measurable

Consider the equivalence relation defined on \([0,1)\) such that \(x\sim y\) iff \(x-y\in \mathbb{Q}\). Construct \(N\subseteq [0,1)\) by adding precisely one member of each equivalence class. This construction depends on the Axiom of Choice, which we won’t go into.

Definition 1. For any \(x\in\mathbb{R}\), \(x\mod 1\) is the fractional part of \(x\).

Note first that all rationals in \([0,1)\) fall into one equivalence class, since for \(a,b\in \mathbb{Q}\), clearly \(a-b\in\mathbb{Q}\). Next, each equivalence class contains all numbers of the form \(x + q \mod 1 \mid q\in \mathbb{Q}\), and is therefore countable. The set N contains exactly one representative from each such class. In other words, \(N\) is uncountable, and each equivalence class contains a countable infinity of elements. Since we take just one representative from each equivalence class to add to \(N\), we have that \(N\) is a proper subset of \([0,1)\).

The construction of \(N_r\) is just saying, start with \(N\) and, as he says, “shift” the part that sticks out past 1 to the left. Then, \(N_r\), just like \(N\), is a proper subset of \([0,1)\).

Now consider \(\mu: \mathcal{P}(\mathbb{R}) \to [0, \infty]\): Since \(N_r\) is simply a translation of \(N\), we expect \(\mu(N) = \mu(N_r)\) for all \(r\in R\). By the argument Folland gives (“every \(x\in [0,1)\) belongs to...”), the \(N_r\) are disjoint. Also, it is impossible for there to exist any \(x\in [0,1)\) which is not in some \(N_r\). Therefore, \([0,1)\) is the disjoint union of the \(N_r\)’s, and we have \[\mu ([0,1)) = \sum_{r\in R} \mu(N_r) = \sum_{r\in R} \mu(N) = \aleph_0\times \mu(N).\]

But this \(\mu\) cannot exist, since the sum on the right is either 0 or \(\infty\), and not 1, as we would expect. This shows that the three desirable properties of \(\mu\) lead to a contradiction, and this (Vitali) set is not measurable.

Banach-Tarski

The setup is as follows: Let \(U\) and \(V\) be arbitrary bounded open sets in \(\mathbb{R}^n, n\geq 3\). There exist \(k\in \mathbb{N}\) and subsets \(E_1,\dots,E_k, F_1,\dots,F_k\) of \(\mathbb{R}^n\) such that

  • the \(E_j\)’s are disjoint and their union is \(U\);

  • the \(F_j\)’s are disjoint and their union is \(V\);

  • \(E_j\) is congruent to \(F_j\) for \(j = 1,\dots,k\)

The key to the construction, as one might imagine, is that the sets \(E_j\) and \(F_j\) are quite weird and based on the Vitali set above.

\(\sigma\)-algebras

Sequences of disjoint sets

For \(\lbrace E_j\rbrace_1^{\infty} \subseteq \mathcal{A}\), where \(\mathcal{A}\) is a \(\sigma\)-algebra over \(X\), we define \[F_k = E_k \backslash \left[\bigcup_{j=1}^{k-1} E_j\right] = E_k \cap \left[ \bigcup_{j=1}^{k-1} E_j\right]^c.\]

Folland claims that the \(F_k\)’s belong to \(\mathcal{A}\), are disjoint, and \(\cup_1^\infty E_j = \cup_1^\infty F_k\). To see this, we start building up the \(F_k\)’s, starting from \(k=1\). We have \[F_1 = E_1 \cap \left[\bigcup_{j=1}^0 E_j\right]^c = E_1\cap \emptyset^c = E_1\cap X = E_1.\]

Next, \[F_2 = E_2 \cap \left[ \bigcup_{j=1}^1 E_j \right]^c = E_2\cap E_1^c.\]

Now, if \(E_1\) and \(E_2\) are disjoint, then \(F_2\) will be just \(E_2\). The more interesting case is when they are not disjoint. In this case, we “take away” from \(E_2\) all points that are also in \(E_1\). In other words, each point in \(\cup_{j=1}^{\infty} E_j\) is assigned to the first set \(E_k\) in which it appears. We might see \(F_i\) as an impoverished version of \(E_i\); each point appears in exactly one \(F_k\), namely the earliest \(E_k\) containing it.

The rest of the construction proceeds similarly, by reallocating all intersections between \(E_k\)’s to just one \(F_k\), and this ensures that the \(F_k\)’s are disjoint (cf. ). Next, since each nonempty intersection between any \(E_j\)’s ends up in some \(F_k\), we have \(\cup_1^\infty E_j = \cup_1^\infty F_k\) (cf. ). Finally, since each \(F_k\) is the result of complementation and intersection operations on the \(E_j\in \mathcal{A}\), the \(F_k\)’s belong to \(\mathcal{A}\).

Proposition 1. \(F_m\cap F_n = \emptyset\) for \(m\neq n\).

Proof. Proof. Let \(m<n\). Then \[F_m = E_m \cap \left[ \bigcup_{j=1}^{m-1} E_j\right]^c = E_1^c\cap \dots \cap E_{m-1}^c\cap E_m\] which means that \(F_m\) consists of all points which are in \(E_m\) but in none of the preceding \(E_j\)’s. Similarly, \[F_n = E_n \cap \left[ \bigcup_{j=1}^{n-1} E_j\right]^c = E_1^c\cap \dots \cap E_{n-1}^c\cap E_n.\]

Since \(n>m\), no point in \(F_n\) can have occurred in any \(E_1,\dots,E_m\), which means that no point from \(F_n\) can be in \(F_m\). In other words, \(F_m\subseteq E_m\) and since \(F_n \subseteq \cap_{j=1}^{n-1} E_j^c\), we have \(F_n\subseteq E_m^c\), so \(F_m\cap F_n = \emptyset\). ◻

Proposition 2. \[\bigcup_{j=1}^\infty E_j = \bigcup_{j=1}^\infty F_j\]

Proof. Proof. The first direction is trivial: Since \(F_k\subseteq E_k\) for all \(k\), we have \(\cup_1^\infty F_j\subseteq \cup_1^\infty E_j\). For the other direction, let \(x \in \bigcup_{j=1}^\infty E_j\). Then \(x \in E_k\) for at least one \(k\). \(\mathbb{N}\) is well-ordered, so there exists a smallest such index \(k\). So, \(x \in E_k\) and \(x \notin E_j \text{ for all } j < k\). Therefore, \[x \notin \bigcup_{j=1}^{k-1} E_j \implies x \in E_k \setminus \bigcup_{j=1}^{k-1} E_j = F_k \implies \bigcup_{j=1}^\infty E_j \subseteq \bigcup_{k=1}^\infty F_k.\] ◻

Infinite unions (intersections) of closed (open) sets produce open (closed) sets

Proposition 3 (Intersections: \(G_\delta\)). \[[a,b] = \bigcap_{j=1}^\infty \left(a - \frac{1}{j}, b + \frac{1}{j}\right).\]

Proof. Proof. First, if \(x\in [a,b]\), then \(a\leq x\leq b\), which together with \(a - \frac{1}{j} < a\) and \(b < b + \frac{1}{j}\), implies that \(a - \frac{1}{j} < x < b + \frac{1}{j}\) for all \(j\), so \(x\in \cap_{j=1}^{\infty} (a - \frac{1}{j}, b + \frac{1}{j})\).

Next, let \(x\notin [a,b]\). Then either \(x<a\) or \(x>b\). Without loss of generality, let \(x < a\), and take \(j\) large enough such that \(a-x > \frac{1}{j}\)1. Then, \[a-x > \frac{1}{j} \implies x < a - \frac{1}{j}\] so \(x \notin (a - \frac{1}{j}, b + \frac{1}{j})\). But since \(x\) does not belong to at least one interval in the sequence, we have \(x \notin \cap_{j=1}^{\infty} (a - \frac{1}{j}, b + \frac{1}{j})\). ◻

Intuition 1. To get to \([a,b]\) we envision a countable infinity of open sets which are larger than the desired interval, and get ever closer to \([a,b]\). Then nothing past the end-points can exist in the resulting intersection.

Proposition 4 (Unions: \(F_\sigma\)). \[(a,b) = \bigcup_{j=1}^\infty \left[a + \frac{1}{j}, b - \frac{1}{j}\right].\]

Proof. Proof. Let \(x\in (a,b)\). Then \(a<x<b\). Let \(x-a=\epsilon\), and note that \(\epsilon > 0\). Using the Archimedean property again, take \(j\) large enough so that \(\frac{1}{j} < \epsilon\), which yields \[\frac{1}{j} < x-a \implies a < a + \frac{1}{j} < x \implies x \in \left[a + \frac{1}{j}, b - \frac{1}{j}\right]\] for some choice of \(j\), which implies that \(x \in \cup_{j=1}^\infty \left[a + \frac{1}{j}, b - \frac{1}{j}\right]\).

Conversely, if \(x \in \cup_{j=1}^\infty \left[a + \frac{1}{j}, b - \frac{1}{j}\right]\), there is some \(j\) such that \(x \in \left[a + \frac{1}{j}, b - \frac{1}{j}\right]\), which implies that \(a + \frac{1}{j} \leq x \leq b - \frac{1}{j}\). Now choose \(r\in\mathbb{R}\) such that \(r < \frac{1}{j}\). Then the open ball centered at \(x\) lies entirely within \((a,b)\), and so, \(x\in(a,b)\). ◻

Intuition 2. To generate \((a,b)\), we take a countable intersection of closed sets which are smaller than the desired set. By approaching the limits from the inside, we ensure that we can never actually get to \(a\) or \(b\), resulting in an open set.

Proposition 5. \(\mathcal{B}_\mathbb{R}\) is generated by the half-open intervals (this is exercise 2).

Proof. Proof. Following , we note that \[(a,b) = \bigcup_{j=1}^\infty \left( a, b - \frac{1}{j}\right]\] so \(\mathcal{B}_{\mathbb{R}} \subseteq \mathcal{M}(\mathcal{E}_3)\). A similar argument applies to the \(\mathcal{M}(\mathcal{E}_4)\) case. ◻

Proposition 6. \(\mathcal{B}_\mathbb{R}\) is generated by the open rays (this is exercise 2).

Proof. Proof. For \(\mathcal{E}_5\), let \(a<b\) and note that \[\left[ \left(a, \infty\right)^c \cup \left[\bigcap_{j=1}^\infty \left(b - \frac{1}{j}, \infty\right)\right] \right]^c = \left[ \left(-\infty, a\right] \cup \left[ b, \infty\right) \right]^c = (a,b).\]

Similar reasoning applies to \(\mathcal{E}_6\). These yield \(\mathcal{B}_{\mathbb{R}} \subseteq \mathcal{M}(\mathcal{E}_5)\) and \(\mathcal{B}_{\mathbb{R}} \subseteq \mathcal{M}(\mathcal{E}_6)\) respectively. ◻

The product \(\sigma\)-algebra

We first revisit Folland’s definition of Cartesian product. Recall from the Prologue that

If \(\{X_\alpha\}_{\alpha\in A}\) is an indexed family of sets, their Cartesian product \(\prod_{\alpha\in A} X_\alpha\) is the set of all maps \(f : A \to \cup_{\alpha\in A} X_\alpha\) such that \(f(\alpha)\in X_\alpha\) for every \(\alpha\in A\).

Usually, for sets \(X\) and \(Y\), we define the Cartesian product as \[X\times Y = \{(x, y), \dots \mid x \in X, y\in Y\}.\]

The current definition (call it \(D_2\)) is set-theoretically distinct from the usual one (\(D_1\)), but there is a bijection between them. The \(D_2\) product is \[X\times Y = \{f_1, f_2,\dots\}\] where each map \(f\) is nothing but an ordered set of elements from \(\bigcup_{\alpha\in A}\); the condition \(f(\alpha)\in X_\alpha\) ensures this. The bijection is clear. The map \(\pi_\alpha\) is then a map from \((x_1, x_2, \dots)\) (effectively) to \(x_i\); it simply picks out the \(\alpha\)-th component:

If \(X = \prod_{\alpha\in A} X_\alpha\) and \(\alpha \in A\), we define the \(\alpha\)th projection or coordinate map \(\pi_\alpha: X\to X_\alpha\) by \(\pi_\alpha (f) = f(\alpha)\).

And then, the product \(\sigma\)-algebra on \(X\) is the \(\sigma\)-algebra generated by \[\left\{\pi_\alpha^{-1} (E_\alpha) : E_\alpha \in \mathcal{M}_\alpha, \alpha \in A\right\},\] which we denote by \(\bigotimes_{\alpha\in A} \mathcal{M}_\alpha\). Folland suggests that the definition is not terribly intuitive, so we look at a concrete example to build intuition.

Intuition 3 (Product \(\sigma\)-algebras). Let \(X_1 = \{a_1, a_2\}, X_2 = \{b_1, b_2, b_3\}\). Then \(\prod_{\alpha\in A} X_\alpha = X_1\times X_2 = \{f_1, f_2, \dots, f_6\}\), where each \(f_i\) is a mapping from \(\{1,2\}\) to \(X_1\cup X_2\). For example, we might have \(f_1(1) = a_1, f_1(2) = b_1, f_2(1) = a_1, f_2(2) = b_2\), and so on. Then, the projection map takes functions to elements. For example, \(\pi_1\), being the projection map that picks out an element in \(X_1\), might look like this: \(\pi_1(f_1) = a_1, \pi_1(f_2) = a_1\), etc. In this toy example, \(\pi_1 (f_1) = \pi_1(f_2) = \pi_1(f_3) = a_1\). Now, consider the preimages2: We have \(\pi^{-1}_\alpha (\{a_1\}) = \{f_1, f_2, f_3\}\), or \(\{a_1\}\times X_2\), i.e. \(\{(a_1,b_1), (a_1, b_2), (a_1, b_3)\}\). In other words, for subsets \(E\subseteq X_1\) and \(F\subseteq X_2\), we have the “cylinder sets” \[\pi_1^{-1} (E) = E\times X_2 \quad \text{and} \quad \pi_2^{-1} (F) = X_1\times F.\]

Now, say we have \(E = \{a_1\}\) and \(F = \{b_2\}\). Then \(\pi_1^{-1}(\{a_1\}) = \{(a_1,b_1), (a_1, b_2), (a_1, b_3)\}\) and \(\pi_2^{-1} (\{b_2\}) = \{(a_1, b_2), (a_2, b_2)\}\). Note that their intersection is \(\{(a_1, b_2)\} = E\times F\). So, the product \(\sigma\)-algebra generated by all \(\pi_\alpha^{-1} (E_\alpha)\) contains all sets of the form \(E\times F\), where \(E\in \mathcal{M}_{X_1}\) and \(F\in \mathcal{M}_{X_2}\).

Intuitively (the rigorous statement is contained in Corollary 1.6, p. 23), we note that since the intersection of these preimages yields the measurable rectangle \(E\times F\), the \(\sigma\)-algebra generated by the coordinate projections contains all such rectangles. Conversely, because a preimage \(\pi_1^{-1} (E) = E\times X_2\) is itself a measurable rectangle, the \(\sigma\)-algebra generated by the rectangles contains all coordinate preimages. Therefore, defining the product \(\sigma\)-algebra via the projection maps (\(\pi_\alpha\)) is equivalent to defining it as the \(\sigma\)-algebra generated by all measurable rectangles.

The logic of Proposition 1.3

The claim is that \[G = \left\{\pi_\alpha^{-1} (E_\alpha) : E_\alpha \in \mathcal{M}_\alpha, \alpha \in A\right\} \quad \text{and} \quad H = \left\{\prod_{\alpha\in A} E_\alpha : E_\alpha \in \mathcal{M}_\alpha \right\}\] generate the same \(\sigma\)-algebras, if \(A\) is countable. That is, we want to show that \[\mathcal{M}(G) = \mathcal{M}(H)\] so we use Lemma 1.1 after establishing that \[G\subseteq \mathcal{M}(H) \quad \text{and} \quad H \subseteq \mathcal{M}(G).\]

A note on countability

David Williams says in Probability with Martingales:

[T]he measurable sets form a \(\sigma\)-algebra, a structure stable (or ‘closed’) under countably many set operations. Without the insistence that ‘only countably many operations are allowed’, measure theory would be self-contradictory - a point lost on certain philosophers of probability.

If we did allow arbitrary unions, then since singletons \(\{x\}\) are measurable, every subset of \(\mathbb{R}\) would be measurable. But as we saw in , this cannot be tolerated.

Proposition 1.4

Part 1

Suppose that \(\mathcal{M}_\alpha\) is generated by \(\mathcal{E}_\alpha , \alpha \in A\). Then \(\bigotimes_{\alpha\in A} \mathcal{M}_\alpha\) is generated by

\[\mathcal{F}_1 = \left\{\pi_\alpha^{-1} (E_\alpha) : E_\alpha \in \mathcal{E}_\alpha , \alpha\in A\right\}.\]

Proof. Proof. Let \[\mathcal{F}_0 = \left\{\pi_\alpha^{-1} (E_\alpha) : E_\alpha \in \mathcal{M}_\alpha , \alpha\in A\right\};\] By definition, we have \(\mathcal{M}(\mathcal{F}_0) = \bigotimes_{\alpha\in A} \mathcal{M}_\alpha\). Now, say that \(\mathcal{M}_\alpha\) is generated by \(\mathcal{E}_\alpha\), and define \(\mathcal{F}_1\) as above. Since \(\mathcal{F}_1 \subseteq \mathcal{F}_0\), we have that \(\mathcal{F}_1 \subseteq \mathcal{M}(\mathcal{F}_0)\), so by Lemma 1.1, \(\mathcal{M}(\mathcal{F}_1) \subseteq \mathcal{M}(\mathcal{F}_0)\). In other words, \(\mathcal{M}(\mathcal{F}_1)\subseteq \bigotimes_{\alpha\in A} \mathcal{M}_\alpha\). For the converse, since \(\mathcal{F}_1 \subseteq \mathcal{F}_0\), we have that \(\mathcal{F}_1 \subseteq \mathcal{M}(\mathcal{F}_0)\), so by Lemma 1.1, \(\mathcal{M}(\mathcal{F}_1) \subseteq \mathcal{M}(\mathcal{F}_0)\). In other words, \(\mathcal{M}(\mathcal{F}_1)\subseteq \bigotimes_{\alpha\in A} \mathcal{M}_\alpha\). For the converse direction, fix \(\alpha\in A\). Since \(\mathcal{M}_\alpha=\mathcal{M}(\mathcal{E}_\alpha)\), we have \[\pi_\alpha^{-1}(\mathcal{M}_\alpha) = \pi_\alpha^{-1}(\mathcal{M}(\mathcal{E}_\alpha)) = \mathcal{M}(\{\pi_\alpha^{-1}(E):E\in\mathcal{E}_\alpha\}) \subseteq \mathcal{M}(\mathcal{F}_1).\tag{The key step; cf \Cref{folland-ch-1-homomorphism}}\]

Thus every set of the form \(\pi_\alpha^{-1}(E_\alpha)\), with \(E_\alpha\in\mathcal{M}_\alpha\), belongs to \(\mathcal{M}(\mathcal{F}_1)\), which implies that \(\mathcal{F}_0 \subseteq \mathcal{M} (\mathcal{F}_1)\), and so using Lemma 1.1 again, we have \(\mathcal{M}(\mathcal{F}_0)\subseteq \mathcal{M}(\mathcal{F}_1)\).

Note that the key step above uses the fact that inverse images commute with unions, intersections, and complements; cf p. 4. This means that preimages (or inverse images, as Folland calls them) “pull back structure” and in particular, since \(\pi_\alpha^{-1} (E^c) = (\pi_\alpha^{-1}(E))^c\) and \(\pi_\alpha^{-1} \left(\bigcup_n E_n\right) = \bigcup_n \left(\pi_\alpha^{-1}(E_n)\right)\), it implies that if \(\mathcal{M}_\alpha = \mathcal{M}(\mathcal{E}_\alpha)\), then applying \(\pi_\alpha^{-1}\) to everything inside of \(\mathcal{M}_\alpha\) keeps you inside \(\mathcal{M}(\pi_\alpha^{-1} (\mathcal{E}_\alpha))\). ◻

Part 2

For the countable case, let \(\mathcal{F}_3 = \left\{\prod_{\alpha\in A} E_\alpha : E_\alpha \in \mathcal{M}_\alpha\right\}\) and \(\mathcal{F}_2 = \left\{\prod_{\alpha\in A} E_\alpha : E_\alpha \in \mathcal{E}_\alpha\right\}\), where \(\mathcal{M}_\alpha = \mathcal{M}(\mathcal{E}_\alpha)\), and \(\bigotimes_{\alpha\in A} \mathcal{M}_\alpha = \mathcal{M} (\mathcal{F}_3)\) (from Prop 1.3). Now, clearly \(\mathcal{F}_2 \subseteq \mathcal{M} (\mathcal{F}_3)\), so by minimality \(\mathcal{M}(\mathcal{F}_2) \subseteq \mathcal{M} (\mathcal{F}_3)\). Next, any element of \(\mathcal{F}_3\) is generated by operations on \(\mathcal{E}_\alpha\), so \(\mathcal{F}_3 \subseteq \mathcal{M}(\mathcal{F}_2)\), and using minimality again, we have \(\mathcal{M}(\mathcal{F}_3) \subseteq \mathcal{M}(\mathcal{F}_2)\), whence \(\bigotimes_{\alpha\in A} \mathcal{M}_\alpha = \mathcal{M} (\mathcal{F}_2) = \mathcal{M} (\mathcal{F}_3)\).

Proposition 1.5

Note that the open sets \(U_j\) generate the Borel \(\sigma\)-algebra \(\mathcal{B}_{X_j}\), so by Prop 1.4, \(\bigotimes_{j=1}^n \mathcal{B}_{X_j}\) is generated by \(\pi_j^{-1} (U_j) , 1\leq j \leq n\). Then by minimality, we have \(\pi_j^{-1} (U_j) \subseteq \mathcal{B}_X \implies \bigotimes_{j=1}^n \mathcal{B}_{X_j} \subseteq \mathcal{B}_X\).

Next, let \(\mathcal{F}_3 = \left\{\prod_{j=1}^n E_j : E_j \in \mathcal{E}_j\right\}\). By Prop 1.4, \(\mathcal{M}(\mathcal{F}_3) = \bigotimes_{j=1}^n \mathcal{B}_{X_j}\). The metric space argument shows that \(\mathcal{M}(\mathcal{F}_3) = \bigotimes_{j=1}^n \mathcal{B}_X\), which implies that \(\bigotimes_{j=1}^n \mathcal{B}_{X_j} = \mathcal{B}_X\).

The full statement is that \[\bigotimes_{j=1}^n \mathcal{B}_{X_j} = \mathcal{B}_{\prod_{j=1}^n X_j}\] which means (intuitively) that whether one starts with the Borel \(\sigma\)-algebras on each \(X_j\) and then takes the product, or one constructs the Borel \(\sigma\)-algebra on the (countable) product \(\prod_{j=1}^n X_j\) directly, it makes no difference.

Homomorphism

We observed in that preimages commute with unions, intersections, and complements, so \(\pi_\alpha^{-1}\) preserves the operations used to build \(\sigma\)-algebras. In other words, \(\pi_\alpha^{-1}\) is a \(\sigma\)-algebra homomorphism: \[\pi_\alpha^{-1} (\mathcal{M} (\mathcal{E}_\alpha)) = \mathcal{M} (\pi_\alpha^{-1} (\mathcal{E}_\alpha)).\]

Section summary

  • We start with the observation that there exist highly strange subsets of \(\mathbb{R}^n\), and these lead to paradoxes when we attempt to measure them, using quite reasonable criteria. The solution is to limit our measures to more reasonable subsets; we dare measure only in the domain of \(\sigma\)-algebras.

  • Algebras are closed under finite unions and complements, while \(\sigma\)-algebras are closed under countable unions and complements.

  • If \(\mathcal{E}\subseteq \mathcal{P}(X)\), then \(\mathcal{M}(\mathcal{E})\) is the smallest \(\sigma\)-algebra containing \(\mathcal{E}\). Since this is the smallest such algebra, it is the intersection of all \(\sigma\)-algebras containing the generating collection of sets \(\mathcal{E}\).

  • It follows from these observations that \(\mathcal{E}\subseteq \mathcal{M}(\mathcal{F}) \implies \mathcal{M}(\mathcal{E})\subseteq \mathcal{M}(\mathcal{F})\). This is minimality.

  • The product algebra \(\bigotimes_{\alpha\in A} \mathcal{M}_\alpha\) is generated by the cylinder sets \(\left\{\pi_\alpha^{-1} (E_\alpha) : E_\alpha \in \mathcal{M}_\alpha , \alpha\in A\right\}\). Note that these sets are of the form \(E\times X_i\) or \(X_i\times F\), so their intersection yields open subsets of the form \(E\times F\) of the product space.

  • \(\pi_\alpha^{-1}\) is a \(\sigma\)-algebra homomorphism: \(\pi_\alpha^{-1} (\mathcal{M} (\mathcal{E}_\alpha)) = \mathcal{M} (\pi_\alpha^{-1} (\mathcal{E}_\alpha))\).

  • We have that \(\bigotimes_{j=1}^n \mathcal{B}_{X_j} = \mathcal{B}_{\prod_{j=1}^n X_j}\), for separable metric spaces.

Exercises

Ex 2

Done above.

Ex 3

Prove that if \(\mathcal{M}\) is an infinite \(\sigma\)-algebra, then \(\mathcal{M}\) contains an infinite sequence of disjoint nonempty sets (see errata; the word “nonempty” is missing in the text).

Proof. Proof. Recall: Since \(\mathcal{M}\) is infinite, it contains sets \(\left\{E_j\right\}_{j=1}^\infty \subseteq \mathcal{M}\). And then, by the construction on p. 21, \(\mathcal{M}\) contains sets \[F_k = E_k\backslash \left[\bigcup_{j=1}^{k-1} E_j\right] = E_k \cap \left[ \bigcup_{j=1}^{k-1} E_j\right]^c\] and these are disjoint, so \(\mathcal{M}\) contains an infinite sequence of disjoint sets. However, this simple setup does not help us. Consider the case where the \(E_j\) satisfy \(E_1 \subseteq E_2 \subseteq\dots\) and arbitrarily many of the subset inclusions are equalities. In that case, the corresponding \(F_k\)’s would be empty. Therefore, the key here is to choose the \(E_j\)’s such that no \(F_k\) is empty.

Now3, either there exists an infinite sequence of sets such that \[E_1\subset E_2\subset \dots \tag{i.e. strict inclusion}\] or not.

If there does exist such a sequence, then each \(F_k\) is nonempty and disjoint, and we’re done. So assume no such sequence exists; then there exists a set \(E_1\) with no nonempty proper subsets (if \(E_1\) has a proper nonempty subset, call that one \(E_1\), and so on; this cannot go on forever by the assumption above). Then by the construction from p. 21, \[F_1 = E_1\backslash \emptyset = E_1.\]

Having found the first \(E_1\), construct \(X_1 = X\backslash E_1\) and find another such set. If we cannot find any such set then \(\mathcal{M}\) is finite and we have a contradiction. Proceed in this way, at each step \(n\) forming \[X_n = X\backslash \bigcup_{i=1}^n E_i\] and at each step constructing \(F_i\). This process can never complete else \(\mathcal{M}\) would be finite. Therefore, our sequence \(\left\{F_i\right\}_{i=1}^\infty\) is the desired infinite sequence of disjoint nonempty sets. ◻

Ex 4

An algebra \(\mathcal{A}\) is a \(\sigma\)-algebra iff \(\mathcal{A}\) is closed under countable increasing unions (i .e., if \(\left\{E_j\right\}_{j=1}^{\infty} \subseteq \mathcal{A}\) and \(E_1\subseteq E_2\subseteq \dots\), then \(\bigcup_{j=1}^\infty E_j\in \mathcal{A}\)).

Proof. Proof. Let \(\mathcal{A}\) be a \(\sigma\)-algebra. Then it contains all countable unions, so in particular, if \(\left\{E_j\right\}_{j=1}^{\infty} \subseteq \mathcal{A}\) then \(\bigcup_{j=1}^\infty E_j\in \mathcal{A}\). On the other hand, say that the algebra \(\mathcal{A}\) is closed under countable increasing unions. Let \(\left\{A_i\right\}_{i=1}^{\infty}\) be an arbitrary collection of sets in \(\mathcal{A}\); we will show that their union is in \(\mathcal{A}\). Define the sequence \(\left\{E_j\right\}\) as follows: \(E_1 = A_1, E_2 = A_1\cup A_2\), and in general, \[E_j = \bigcup_{i=1}^j A_i.\]

Now, by hypothesis, since each \(E_j\in\mathcal{A}\) and \(E_1\subseteq E_2\subseteq \dots\), we have that \(\bigcup_{j=1}^\infty E_j \in \mathcal{A}\), which, together with \(\bigcup_{j=1}^\infty E_j = \bigcup_{j=1}^\infty A_j\), implies that \(\bigcup_{i=1}^\infty A_i \in \mathcal{A}\), so \(\mathcal{A}\) is a \(\sigma\)-algebra. ◻

Contrapositive proof for the converse

Say that \(\mathcal{A}\) is an algebra but not a \(\sigma\)-algebra. Then there exists a collection of sets \(\left\{A_i\right\}_{i=1}^{\infty}\) for which \(\bigcup_{i=1}^\infty A_i \notin \mathcal{A}\). Using the construction above, we note that since \(\bigcup_{j=1}^\infty E_j = \bigcup_{j=1}^\infty A_j\), it follows that \(\bigcup_{j=1}^\infty E_j\) is not in \(\mathcal{A}\) either, so we have \(\left\{E_j\right\}_{j=1}^{\infty} \subseteq \mathcal{A}\) and \(E_1\subseteq E_2\subseteq \dots\), but not \(\bigcup_{j=1}^\infty E_j\in \mathcal{A}\).

Ex 5

If \(\mathcal{M} = \mathcal{M}(\mathcal{E})\) then \(\mathcal{M}\) is the union of the \(\sigma\)-algebras generated by \(\mathcal{F}\) as \(\mathcal{F}\) ranges over all countable subsets of \(\mathcal{E}\). That is, \[\mathcal M(\mathcal{E}) = \bigcup \{\mathcal M(\mathcal F) : \mathcal F \subseteq \mathcal E,\ \mathcal F \text{ countable}\}.\]

Proof. Proof. If \(B\in \bigcup \left\{\mathcal{M}(\mathcal{F})\right\}\) then \(B\in \mathcal{M}(\mathcal{F})\) for some \(\mathcal{F}\). But since \(\mathcal{F} \subseteq \mathcal{E}\), we also have \(B\in \mathcal{M}(\mathcal{E})\).

For the other direction, let \(\mathcal{U}\) be the union above. Then, if \(\mathcal{U}\) were a \(\sigma\)-algebra, since \(\mathcal{E}\subseteq \mathcal{U}\), by minimality we would have \(\mathcal{M}(\mathcal{E})\subseteq \mathcal{U}\). Therefore, all that remains to show is that \(\mathcal{U}\) is a \(\sigma\)-algebra.

Let \(E_1,E_2,\dots \in\mathcal{U}\). If the \(E_1,E_2,\dots\) are all in \(\mathcal{M}(\mathcal{F})\) for some \(\mathcal{F}\), then \(\bigcup_{j=1}^{\infty} E_j\in \mathcal{M}(\mathcal{F})\) and therefore in \(\mathcal{U}\). Otherwise, note that \[\tilde{\mathcal{F}} = \bigcup_{i=1}^{\infty} \mathcal{F}_i\] is countable, and must contain or generate all the \(E_1,E_2,\dots\), so \(\bigcup_{j=1}^\infty E_j \in \mathcal{M}(\tilde{\mathcal{F}})\), which in turn implies that \(\bigcup_{j=1}^\infty E_j \in \mathcal{U}\) as desired. For complements, if \(E\in \mathcal{U}\), then that \(E\) emanates from some \(\mathcal{M}(\mathcal{F})\), which implies that \(E^c \in \mathcal{M}(\mathcal{F})\) as well. Finally, we know the whole space \(X\) is contained in \(\mathcal{U}\) since each \(\mathcal{M}(\mathcal{F})\) contains it. Therefore, \(\mathcal{U}\) is a \(\sigma\)-algebra, and the conclusion follows. ◻

Measures

Section summary

Footnotes

  1. The Archimedean Property ensures that this is possible: Let \(a-x = \epsilon\) (this is positive since \(x<a\)), then since \(\mathbb{R}\) is an ordered field, we have \(\forall\epsilon\in\mathbb{R} \left(\epsilon > 0 \implies \exists j\in \mathbb{N} : \frac{1}{j} < \epsilon\right)\).↩︎

  2. Despite the notation, this is a preimage, not an inverse. Indeed, \(\pi_\alpha\) is not injective in general so will not usually have an inverse.↩︎

  3. Here we use the hint at Ari Stern’s Math 5051: “Hint: If \(\mathcal{M}\) contains an infinite sequence of strictly nested sets, then we’re done, so assume that no such sequence exists. Next, use this assumption to find a nonempty set in \(\mathcal{M}\) with no nonempty proper subsets in \(\mathcal{M}\). Finally, show that this can be done infinitely many times.”↩︎