[Measure Theory] Measures

A measure is a generalization of the notion of length (in 1D), area (in 2D), and volume (in 3D). Intuitively, if $A_1, A_2, \dots, A_n$ are pairwise disjoint sets, i.e. not overlapped we would expect the measure of $\bigcup_{i=1}^\infty A_i$ to be the sum of the measures of the $A_i$. If we extend this idea to countable unions (but not uncountable ones), we have the defining property of a measure.

Let $X$ be a set. And let $\mathcal{A}$ be a $\sigma$-algebra consisting of subsets of $X$.

$\color{blue}{\mathbf{Definition.}}$ A measure on $(X, \mathcal{A})$ is a function $\mu: \mathcal{A} \to [0, \infty]$ such that

1. $\mu(\emptyset) = 0$;
2. countable additivity ($\sigma$-additivity): if $A_i \in \mathcal{A}$, $i = 1, 2, \dots$, are pairwise disjoint, i.e. $A_i \cap A_j = \emptyset$ if $i \neq j$, then $$ \begin{aligned} \mu \left( \bigcup_{i=1}^\infty A_i \right) = \sum_{i=1}^\infty \mu(A_i). \end{aligned} $$

Then the triple $(X, \mathcal{A}, \mu)$ is referred to as measure space.

Note that by letting $A_i = \emptyset$ for all $i \geq n+1$, countable additivity implies finite additivity also; \(\mu \left( \bigcup_{i=1}^n A_i \right) = \sum_{i=1}^n \mu (A_i)\) whenever $A_1, \dots, A_n$ are in $\mathcal{A}$ and are pairwise disjoint.

But generally finite additivity doesn’t connect to countable additivity. And there are several assumptions that, together with finite additivity, imply countable additivity:

$\color{#bf1100}{\mathbf{Remark}}$. Let $(X, \mathcal{A})$ be a measurable space and $\mu$ is a non-negative set function that is finitely additive and such that $\mu(\emptyset) = 0$.

1. Suppose that whenever $A_i$ is an increasing sequence of sets in $\mathcal{A}$, then $\mu \left(\bigcup_i A_i \right) = \lim_{i \to \infty} \mu (A_i)$. Then, finitely additive $\mu$ is also countably additive, i.e. measure.
   $\mathbf{Proof.}$

   Let $A_1, A_2, \dots$ be pairwise disjoint. And due to finite additivity $\mu \left( \bigcup_{i=1}^n A_i \right) = \sum_{i=1}^n \mu (A_i)$.
   

   Since $\bigcup_{i=1}^n A_i \uparrow \bigcup_{i=1}^\infty A_i$,

   \[\begin{aligned} \mu \left( \bigcup_{n=1}^\infty \bigcup_{i=1}^n A_i \right) = \mu \left( \bigcup_{i=1}^\infty A_i \right) = \lim_{n \to \infty} \mu \left( \bigcup_{i=1}^n A_i \right) = \sum_{i=1}^\infty \mu (A_i) \end{aligned}\] \[\tag*{$\blacksquare$}\]
2. Let's further assume that $\mu(X) < \infty$. Suppose that whenever $A_i$ is a sequence of sets in $\mathcal{A}$ that decrease to $\emptyset$, then $\lim_{i \to \infty} \mu (A_i) = 0$. Then $\mu$ is countably additive, i.e. measure.
   $\mathbf{Proof.}$

   Let $A_1, A_2, \dots$ be pairwise disjoint. And due to finite additivity $\mu \left( \bigcup_{i=1}^n A_i \right) = \sum_{i=1}^n \mu (A_i)$.
   

   Let $B_n = \bigcup_{i=1}^\infty A_i - \bigcup_{i=1}^n A_i$. And due to finite additivity, it is clear that $\mu (B_n) = \mu \left( \bigcup_{i=1}^\infty A_i \right) - \mu \left( \bigcup_{i=1}^n A_i \right)$.

   Since $B_n \downarrow \emptyset$, $\lim_{n \to \infty} \mu (B_n) = 0$ which implies $\lim_{n \to \infty} \mu \left( \bigcup_{i=1}^n A_i \right) = \mu \left( \bigcup_{i=1}^\infty A_i \right)$.

   \[\tag*{$\blacksquare$}\]

Here are some examples.

click to expand

$\color{green}{\mathbf{Example\ 1.}}$ Let $X$ be any set, and let $\mathcal{A}$ be the collection of all subsets of $X$. Then the number of elements in set $A$, $\mu (A)$, is a measure and called counting measure.

$\color{green}{\mathbf{Example\ 2.}}$ Let $X = \mathbb{R}$, and let $\mathcal{A}$ be the collection of all subsets of $\mathbb{R}$. For $x_1, x_2, \dots \in \mathbb{R}$ and $a_1, a_2, \dots \geq 0$, set

\[\begin{aligned} \mu(A) = \sum_{\{ i \mid x_i \in A \}} a_i. \end{aligned}\]

Then $\mu$ is a measure.

$\color{green}{\mathbf{Example\ 3.}}$ Let $\delta_x (A) = 1$ if $x \in A$ and $0$ otherwise. This measure is called point mass at $x$.

And there are some useful properties that hold for any measure. The following hold:

$\color{#bf1100}{\mathbf{Proposition.}}$ Properties of measures

1. monotonicity: $A \subset B \in \mathcal{A} \rightarrow \mu(A) \leq \mu(B)$
2. countable monotonicity: $A_i \in \mathcal{A}$ and $A = \bigcup_{i=1}^\infty A_i \rightarrow \mu(A) \leq \sum_{i=1}^\infty \mu (A_i)$
3. continuity from below: Suppose $A_i \in \mathcal{A}$ and $A_i \uparrow A$. Then $\mu (A) = \text{lim}_{n \to \infty} \mu (A_n)$.
4. continuity from above: Suppose $A_i \in \mathcal{A}$ and $A_i \downarrow A$. If $\mu (A_1) < \infty$, then $\mu (A) = \text{lim}_{n \to \infty} \mu (A_n)$.

$\mathbf{Proof.}$

(1) Let $A_2 = B - A$ and $A_1 = A$. And let $A_3 = A_4 = \cdots = \emptyset$. Then $\mu(B) = \sum_{i=1}^\infty \mu(A_i) \geq \mu(A_1) = \mu(A)$.

(2) Let $B_1 = A_1$, $B_2 = A_2 - A_1$, $B_3 = A_3 - (A_2 \cup A_1)$, $B_4 = A_4 - (A_3 \cup A_2 \cup A_1)$, and in general $B_i = A_i - \bigcup_{j=1}^{i-1} A_j$.

Then, obviously $B_i$ are pairwise disjoint, $\bigcup_{i=1}^n B_i = \bigcup_{i=1}^n A_i$, and $B_i \uparrow A$.

\[\begin{aligned} \mu (A) & = \mu (\bigcup_{i=1}^n B_i) \\ & = \sum_{i=1}^\infty \mu (B_i) \quad \text{by definition of measure} \\ & \leq \sum_{i=1}^\infty \mu (A_i) \quad \because A_i \subset B_i \end{aligned}\]

(3) Define $B_i$ as (2). Then $\bigcup_{i=1}^\infty B_i = A = \bigcup_{i=1}^\infty A_i$

\[\begin{aligned} \mu (A) & = \mu \left(\bigcup_{i=1}^\infty A_i\right) = \mu \left(\bigcup_{i=1}^\infty B_i\right) = \sum_{i=1}^\infty \mu (B_i) \\ & = \lim_{n \to \infty} \sum_{i=1}^n \mu (B_i) = \lim_{n \to \infty} \mu \left(\bigcup_{i=1}^n B_i\right) = \lim_{n \to \infty} \mu \left(\bigcup_{i=1}^n A_i \right) \\ \end{aligned}\]

(4) Define $B_1 = \emptyset$ and $B_i = A_{1} - A_{i}$ for $i \geq 2$. Then $B_i \uparrow A_1 - A$. Thus

\[\begin{aligned} \mu(A_1 - A) = \mu (A_1) - \mu(A) = \lim_{n \to \infty} \mu (A_1 - A_n) = \lim_{n \to \infty} \left[ \mu (A_1) - \mu(A_n) \right]. \end{aligned}\] \[\tag*{$\blacksquare$}\]

In case of the last property, $\mu(A_1) < \infty$ is necessary. If not, let $X$ be the positive integers, $\mu$ counting measure, and \(A_i = \{i, i+1, \cdots \}\). Then the $A_i$ decrease, $\mu(A_i) = \infty$ for all $i$, but $\mu(\cap_i^\infty A_i) = \mu(\emptyset) = 0$.

$\color{#bf1100}{\mathbf{Proposition}}$. Let $(X, \mathcal{A})$ be a measurable space.

1. If $\mu_1, \mu_2, \cdots$ are measures on a measurable space, and each $a_n \in [0, \infty)$, then $\sum_{n=1}^\infty a_n \mu_n$ is also a measure.
   $\mathbf{Proof.}$

   We use the fact from undergraduate analysis that

   \[\begin{aligned} \sum_{i=1}^\infty \sum_{j=1}^\infty c_{ij} = \sum_{j=1}^\infty \sum_{i=1}^\infty c_{ij} \end{aligned}\]

   provided each of the $c_{ij}$ is a non-negative real or equal to $+\infty$. This will also be a consequence of the Fubini theorem discussed in the later post.

   First of all,

   \[\begin{aligned} \sum_{n=1}^\infty a_n \mu_n (\emptyset) = 0 \end{aligned}\]

   And, for any pairwise disjoint $A_1, A_2, \cdots$,

   \[\begin{aligned} \sum_{n=1}^\infty a_n \mu_n \left( \bigcup_{i=1}^\infty A_i \right) & = \sum_{n=1}^\infty a_n \sum_{i=1}^\infty \mu_n (A_i) \\ & = \sum_{i=1}^\infty \sum_{n=1}^\infty a_n \mu_n (A_i) \end{aligned}\] \[\tag*{$\blacksquare$}\]
2. Let $B \in \mathcal{A}$. Then, restriction $\nu (A) = \mu (A \cap B)$ is a measure.
   $\mathbf{Proof.}$

   One can easily prove this based on the definition of measure.

   \[\tag*{$\blacksquare$}\]
3. Let $B \in \mathcal{A}$. Then, restriction $\nu (A) = \mu (A \cap B)$ is a measure
   $\mathbf{Proof.}$

   One can easily prove this based on the definition of measure.

   \[\tag*{$\blacksquare$}\]

We end this post by introducing some terminologies related to measure.

$\color{blue}{\mathbf{Definition.}}$ Finite measure

A measure $\mu$ is a finite measure if $\mu(X) < \infty$.

If $\mu$ is a finite measure, then $(X, \mathcal{A}, \mu)$ is called a finite measure space.

$\color{blue}{\mathbf{Definition.}}$ $\sigma$-finite measure

A measure $\mu$ is a $\sigma$-finite measure or $\sigma$-finite if $X$ is a countable union of measurable sets each with finite measure, in other words, if there exist sets $E_i \in \mathcal{A}$ for $i = 1, 2, \dots$ such that $\mu (E_i) < \infty$ for each $i$ and $X = \bigcup_{i=1}^\infty E_i$.

If $\mu$ is a $\sigma$-finite measure, then $(X, \mathcal{A}, \mu)$ is called a $\sigma$-finite measure space.

$\color{#bf1100}{\mathbf{Remark.}}$

In $\sigma$-finite measure, there is no loss of generality even if we assume that $E_i$ is increasing.

Suppose $X$ is $\sigma$-finite so that $X = \bigcup_{i=1}^\infty E_i$ with $\mu (E_i) < \infty$ and $E_i \in \mathcal{A}$ for each $i$. If we set $F_n = \bigcup_{i=1}^n E_i$, then $\mu (F_n) < \infty$ and $F_n \in \mathcal{A}$ for each $i$, still satisfying $X = \bigcup_{i=1}^\infty E_i = \bigcup_{i=1}^\infty F_i$.

Let $(X, \mathcal{A}, \mu)$ be a measurable space.

$\color{blue}{\mathbf{Definition.}}$ complete measure space

A subset $A \subset X$ is a null set if there exists a set $B \in \mathcal{A}$ with $A \subset B$ and $\mu(B)=0$.

If $\mathcal{A}$ contains all the null sets, then $(X, \mathcal{A}, \mu)$ is referred to as complete measure space. Or, we just say $\mathcal{A}$ is complete or $\mu$ is complete if $(X, \mathcal{A}, \mu)$ is complete.

The completion of $\mathcal{A}$, $\bar{\mathcal{A}}$ is the smallest $\sigma$-algebra containing $\mathcal{A}$ such that $(X, \bar{\mathcal{A}}, \bar{\mu})$ is complete where $\bar{\mu}$ is a measure on $\bar{\mathcal{A}}$ that is an extension of $\mu$, i.e. $\bar{\mu}(B) = \mu(B)$ if $B \in \mathcal{A}$.

Note that a null set $A$ doesn’t need to be in $\mathcal{A}$.

We require measure spaces to be complete to treat sets of measure zero as negligible. For instance, if two functions $f$ and 𝑔 satisfy $f (x) = g (x)$ except $N$ and $N$ has measure zero, we aim to consider $f$ and $g$ as essentially equivalent. However, without completeness, it is not ensured that $N$ is measurable, therefore it’s possible that $f$ is measurable while $g$ is not. And the significance of completeness will become evident in the product measure that will be discussed in the later post.

As a final proposition, we end the post by showing that every measure space $(X, \mathcal{A}, \mu)$ has a completion $(X, \bar{\mathcal{A}}, \bar{\mu})$, which is the smallest complete measure space such that $\mathcal{A} \subset \bar{\mathcal{A}}$ and $\bar{\mu} |_{\mathcal{A}} = \mu$.

$\color{red}{\mathbf{Theorem}}$ Existence of completion
Let $(X, \mathcal{A}, \mu)$ be a measure space, and $\mathcal{N}$ be a collection of $\mu$-null set. Let $\mathcal{B} = \sigma (\mathcal{A} \cup \mathcal{N})$ and $\bar{\mu}$ be an extension of $\mu$ to $\mathcal{B}$. Then

1. $\mathcal{B} = \{ B | B = A \cup N \text{ for some } A \in \mathcal{A} \text{ and } N \in \mathcal{N} \}$;
2. $\bar{\mu}$ is well-defined and a measure on $\mathcal{B}$;
3. $(X, \mathcal{B}, \bar{\mu})$ is complete;
4. $\bar{\mu}$ is unique;

$\mathbf{Proof.}$

Denote

\[\begin{aligned} \mathcal{C} = \{ B | B = A \cup N \text{ for some } A \in \mathcal{A} \text{ and } N \in \mathcal{N} \} \end{aligned}\]

(1) Note that $\mathcal{B} = \sigma(\mathcal{A} \cup \mathcal{N}) \supset \mathcal{C}$. Hence it suffices to show that $\mathcal{B} = \sigma(\mathcal{C}) \subset \mathcal{C}$. And we will show that $\mathcal{C}$ is a $\sigma$-algebra containing $\mathcal{A} \cup \mathcal{N}$.

1. As $\emptyset$ is both in $\mathcal{A}$ and $\mathcal{N}$, $\emptyset \in \mathcal{C}$.
2. Suppose $B = A \cup N \in \mathcal{C}$. Then there exists $M \in \mathcal{A}$ such that $M \supset N$ and $\mu(M) = 0$. Then $$ \begin{aligned} A^c \cap N^c & = (A^c \cap N^c \cap M) \cup (A^c \cap N^c \cap M^c) \\ & = (A^c \cap N^c \cap M) \cup (A^c \cap M^c) \end{aligned} $$ Since $A^c \cap N^c \cap M \subset M$, it is a null set. And $A^c \cap M^c \in \mathcal{A}$.
3. Suppose any $B_n = A_n \cup N_n \in \mathcal{C}$ be given for each $n$. Then there exists $M_n \in \mathcal{A}$ such that $M_n \supset N_n$ and $\mu(M_n) = 0$ for each $n \in \mathbb{N}$. Then $$ \begin{aligned} \bigcup_{n=1}^\infty B_n = \left(\bigcup_{n=1}^\infty A_n\right) \cup \left(\bigcup_{n=1}^\infty N_n\right) \end{aligned} $$ Since $\bigcup_{n=1}^\infty N_n \in \bigcup_{n=1}^\infty M_n$ and $\mu \left(\bigcup_{n=1}^\infty M_n \right) = 0$, $\bigcup_{n=1}^\infty B_n \in \mathcal{C}$.

(2) First, we can easily show that $\bar{\mu}$ is well-defined. Suppose $B = A_1 \cup N_1 = A_2 \cup N_2$. Then there exists $N_n \subset M_n \in \mathcal{A}$ and $\mu(M_n)$ for $n = 1, 2$.

\[\begin{aligned} \bar{\mu} (B) = \mu (A_1) \leq \mu (A_1 \cup M_1) = \mu (A_2 \cup M_2) = \mu (A_2) \end{aligned}\]

therefore $\bar{\mu} (B) = \mu (A_1) = \mu (A_2)$. Also, one can verify that $\bar{\mu}$ is a measure on $\mathcal{B}$ following the definition of measure.

(3) Let any $\bar{\mu}$-null set $S$ be given. Then, there exists $T \in \mathcal{B}$ such that $T \supset S$ and $\bar{\mu}(T) = 0$. We should show that $S \in \mathcal{B}$.

As $T \in \mathcal{B}$, we may write $T = A \cup N$ where $M \supset N$, $M \in \mathcal{A}$ and $\mu(M) = 0$. Since $\bar{\mu}(T) = 0$, we have $\mu (A) = 0$ by definition of $\bar{\mu}$. Then, we claim that $S$ is a $\mu$-null set with $A \cup M \in \mathcal{A}$:

\[\begin{aligned} \mu (A \cup M) \leq \mu (A) + \mu (M) = 0 \text{ and } S \subset T = A \cup N \subset A \cup M \end{aligned}\]

Thus $S \in \mathcal{B}$.

(4) Finally, we show that $\bar{\mu}$ is a unique extension of $\mu$ to $\mathcal{B}$. Suppose there exists another extension $\nu$ of $\mu$ to $\mathcal{B}$. Then, for any $A \in \mathcal{A}$, $\nu (A) = \mu (A) = \bar{\mu}(A)$.

Suppose any $B = A \cup N \in \mathcal{B}$ be given and $M \supset N$, $M \in \mathcal{A}$, and $\mu(M) = 0$. Then

\[\begin{aligned} \bar{\mu} (B) = \mu (A) = \nu (A) \leq \nu (A \cup N) = \nu(B) \end{aligned}\]

And, as $A, M \in \mathcal{A}$,

\[\begin{aligned} \nu(B) \leq \nu (A \cup M) \leq \nu(A) + \nu(M) = \mu(A) + \mu(M) = \mu (A) = \bar{\mu} (B). \end{aligned}\] \[\tag*{$\blacksquare$}\]

Reference

[1] Richard F. Bass, Real Analysis for Graduate Students, Version 4.3
[2] Wikipedia, Measurable space