Measure spaces, completeness and outer measures

The lecture defines measures on $\sigma$-algebras, extends a measure space so that its negligible sets become measurable, and constructs measures from outer measures.

Measure spaces#

The main goal is a flexible theory of computing volumes, such as the volumes of the sets of $\mathbb{R}^n$ in Figure 1.

Figure 1. Two sets $E$ and $A$ of $\mathbb{R}^n,$ and a line that crosses $E.$

Definition 1 (Measure). Let $X$ be a set and $\mathcal{M}$ a $\sigma$-algebra on $X.$ A measure is a function ${\mu \colon \mathcal{M} \to [0, +\infty]}$ such that:
(i) ${\mu(\emptyset) = 0;}$

(ii) ($\sigma$-additivity) for every sequence ${(E_n) \subset \mathcal{M}}$ that is pairwise disjoint, that is, such that ${E_i \cap E_j = \emptyset}$ for every ${i \neq j,}$

$$\mu\Big( \bigcup_n E_n \Big) = \sum_{n=0}^{\infty} \mu(E_n).$$

Remark 2. It will turn out to be impossible to construct “interesting” measures defined on the whole power set $\mathcal{P}(X).$

Example 3 (Counting measure). The counting measure is the function ${\mu \colon \mathcal{P}(\mathbb{R}) \to [0, +\infty]}$ given by

$$\mu(E) = \begin{cases} \#E & \text{if } E \text{ is finite}, \\ +\infty & \text{if } E \text{ is infinite}. \end{cases}$$

Example 4 (Dirac delta measure). Let ${x_0 \in \mathbb{R}.}$ The Dirac delta measure at $x_0$ is the function ${\delta_{x_0} \colon \mathcal{P}(\mathbb{R}) \to [0, +\infty]}$ given by

$$\delta_{x_0}(E) = \begin{cases} 1 & \text{if } x_0 \in E, \\ 0 & \text{if } x_0 \notin E. \end{cases}$$

Figure 2 shows the two cases on the real line, with a set $E$ around 0 and a set $F$ around $x_0.$

x₀

3.00

Figure 2. An instance: $E$ and $F$ are intervals of $\mathbb{R},$ and the values of $\delta_{x_0}$ on them are written above. The slider moves $x_0.$

Remark 5 (Basic properties). A triple $(X, \mathcal{M}, \mu)$ of a set $X,$ a $\sigma$-algebra $\mathcal{M}$ on $X$ and a measure $\mu$ on $\mathcal{M}$ is called a measure space. Let $(X, \mathcal{M}, \mu)$ be a measure space.
(i) (Monotonicity) If ${E, F \in \mathcal{M}}$ and ${E \subset F,}$ then ${\mu(E) \le \mu(F).}$
(ii) (Excision) If ${E, F \in \mathcal{M},}$ ${E \subset F}$ and ${\mu(F) < +\infty,}$ then ${\mu(F \setminus E) = \mu(F) - \mu(E).}$

Proof (sketch). (i) For the sets $E \subset F$ of Figure 3,

$$\mu(E) \le \mu(E) + \mu(F \setminus E) = \mu\big( E \cup (F \setminus E) \big) = \mu(F).$$

Figure 3. A set $E$ inside a set $F,$ and the difference ${F \setminus E.}$

Details. The inequality holds because ${\mu(F \setminus E) \ge 0.}$ The set ${F \setminus E}$ belongs to $\mathcal{M}$ and is disjoint from $E,$ so $\sigma$-additivity applied to ${E, F \setminus E, \emptyset, \emptyset, \dots}$ gives the first equality, since ${\mu(\emptyset) = 0;}$ the second holds because ${E \cup (F \setminus E) = F}$ when ${E \subset F.}$ (ii) The same computation gives ${\mu(F) = \mu(E) + \mu(F \setminus E),}$ where ${\mu(E) \le \mu(F) < +\infty}$ by (i), and subtracting $\mu(E)$ gives the claim. $\blacksquare$

Proposition 6 (Continuity of measures). Let $(X, \mathcal{M}, \mu)$ be a measure space.
(i) Let ${(E_n) \subset \mathcal{M}}$ be such that ${E_n \subset E_{n+1}}$ for every $n,$ and let ${E = \bigcup_n E_n.}$ Then ${\mu(E) = \lim_{n \to +\infty} \mu(E_n).}$
(ii) Let ${(E_n) \subset \mathcal{M}}$ be such that ${E_{n+1} \subset E_n}$ for every $n,$ suppose ${\mu(E_0) < +\infty,}$ and let ${E = \bigcap_n E_n.}$ Then ${\mu(E) = \lim_{n \to +\infty} \mu(E_n).}$

Proof (complete). (i) Without loss of generality, ${\mu(E_n) < +\infty}$ for every $n.$ Indeed, if ${\mu(E_m) = +\infty}$ for some $m,$ then ${\mu(E_n) = +\infty}$ for every ${n \ge m}$ and ${\mu(E) = +\infty}$ by Remark 5 (i), so both sides equal $+\infty.$ Let ${F_0 = E_0}$ and ${F_{n+1} = E_{n+1} \setminus E_n}$ for every ${n \ge 0,}$ as in Figure 4. Then ${F_i \in \mathcal{M}}$ for every $i,$ ${F_i \cap F_j = \emptyset}$ for every ${i \neq j,}$ and ${E = \bigcup_n E_n = \bigcup_n F_n.}$ Hence

$$\begin{aligned} \mu(E) &= \mu\Big( \bigcup_n F_n \Big) = \sum_{n=0}^{\infty} \mu(F_n) \\ &= \lim_{N \to \infty} \sum_{n=0}^{N} \mu(F_n) = \lim_{N \to \infty} \mu(E_N), \end{aligned}$$

where the second equality is $\sigma$-additivity and the last one holds because the sum of the $\mu(F_n)$ with ${n \le N}$ equals ${\mu(E_N),}$ as $E_N$ is the union of the pairwise disjoint sets ${F_0, \dots, F_N}$ and $\sigma$-additivity applies as in the proof of Remark 5.

E₂ = F₀ ∪ … ∪ F₂

N

2

Figure 4. An instance of an increasing sequence: the sets ${E_0 \subset E_1 \subset \dots}$ and their union $E,$ dashed. The slider sets $N;$ the rings ${F_0, \dots, F_N}$ are shaded, and their union is $E_N.$

(ii) Let ${A_n := E_0 \setminus E_n}$ for every $n,$ as in Figure 5. Then ${A_n \subset A_{n+1}}$ and ${A_n \in \mathcal{M}}$ for every $n.$ Let

$$A = \bigcup_n A_n = \bigcup_n (E_0 \setminus E_n) = E_0 \setminus \bigcap_n E_n = E_0 \setminus E.$$

Then

$$\begin{aligned} \mu(E_0 \setminus E) &= \mu(E_0) - \mu(E) \\ &= \mu(A) = \mu\Big( \bigcup_n A_n \Big) \\ &= \lim_{n \to \infty} \mu(A_n) \\ &= \lim_{n \to \infty} \mu(E_0 \setminus E_n) \\ &= \lim_{n \to \infty} \big( \mu(E_0) - \mu(E_n) \big) \\ &= \mu(E_0) - \lim_{n \to \infty} \mu(E_n). \end{aligned}$$

Here the first and the fifth equality are Remark 5 (ii), which applies because $E$ and $E_n$ are contained in $E_0$ and ${\mu(E_0) < +\infty,}$ and the third is (i) applied to $(A_n).$ Comparing ${\mu(E_0) - \mu(E)}$ with the last line and cancelling the finite number $\mu(E_0)$ gives ${\mu(E) = \lim_{n \to \infty} \mu(E_n).}$ $\blacksquare$

A₁ = E₀ \ E₁

n

1

Figure 5. An instance of a decreasing sequence: the sets ${E_0 \supset E_1 \supset \dots}$ and their intersection $E,$ dashed. The slider sets $n;$ the shaded set is ${A_n = E_0 \setminus E_n.}$

Proposition 7 ($\sigma$-subadditivity). Let $(X, \mathcal{M}, \mu)$ be a measure space, ${(E_n) \subset \mathcal{M}}$ and ${E \in \mathcal{M}}$ with ${E \subset \bigcup_n E_n.}$ Then

$$\mu(E) \le \sum_{n=0}^{\infty} \mu(E_n).$$

Figure 6. A set $E$ contained in the union of sets $E_n.$

Definition 8 (Finite, probability and $\sigma$-finite measures). Let $(X, \mathcal{M}, \mu)$ be a measure space.
(i) The measure $\mu$ is called finite if ${\mu(X) < +\infty.}$
(ii) The measure $\mu$ is called a probability measure if ${\mu(X) = 1.}$
(iii) The measure $\mu$ is called $\sigma$-finite if there exists ${(E_n) \subset \mathcal{M}}$ such that ${\mu(E_n) < +\infty}$ for every $n$ and ${\bigcup_n E_n = X.}$

Complete measure spaces#

Consider the measure space ${(\mathbb{R}, \{\emptyset, \mathbb{R}\}, \mu)}$ with ${\mu(\emptyset) = 0}$ and ${\mu(\mathbb{R}) = 0.}$ An interval ${(a, b) \subset \mathbb{R}}$ is contained in a set of measure zero, but ${\mu((a, b)) = 0}$ cannot be written, because ${(a, b) \notin \{\emptyset, \mathbb{R}\}.}$

In a measure space $(X, \mathcal{M}, \mu),$ let ${A \in \mathcal{M},}$ let ${C \in \mathcal{M}}$ with ${\mu(C) = 0,}$ and let ${B \subset C,}$ as in Figure 7. One would like to write ${\mu(A) = \mu(A \cup B),}$ but $\mu(A \cup B)$ cannot be written, because ${A \cup B}$ may not belong to $\mathcal{M}.$

Figure 7. A set ${A \in \mathcal{M},}$ a set ${C \in \mathcal{M}}$ with ${\mu(C) = 0,}$ dashed, and a set ${B \subset C.}$

Definition 9 (Negligible set, complete measure space). Let $(X, \mathcal{M}, \mu)$ be a measure space.
(i) A set ${E \in \mathcal{M}}$ is called a set of zero measure if ${\mu(E) = 0.}$
(ii) A set ${F \subset X}$ is called negligible if there exists ${E \in \mathcal{M}}$ such that ${\mu(E) = 0}$ and ${F \subset E.}$
(iii) The measure space is called complete if every negligible set belongs to $\mathcal{M}.$

Theorem 10 (Completion of measure spaces). Let $(X, \mathcal{M}, \mu)$ be a measure space. There exists a measure space ${(X, \overline{\mathcal{M}}, \overline{\mu})}$ such that:
(i) ${(X, \overline{\mathcal{M}}, \overline{\mu})}$ is complete;
(ii) ${\mathcal{M} \subset \overline{\mathcal{M}};}$
(iii) ${\overline{\mu}(E) = \mu(E)}$ for every ${E \in \mathcal{M}.}$

Proof (sketch). One defines

$$\overline{\mathcal{M}} = \{ E \subset X : \text{there exist } F_1, F_2 \in \mathcal{M} \text{ such that } F_1 \subset E \subset F_2 \text{ and } \mu(F_2 \setminus F_1) = 0 \}.$$

For such sets $E,$ $F_1$ and $F_2,$ as in Figure 8, one defines

$$\overline{\mu}(E) := \mu(F_1) = \mu(F_2).$$

Figure 8. A set $E$ between two sets ${F_1 \subset E \subset F_2}$ of $\mathcal{M}$ with ${\mu(F_2 \setminus F_1) = 0.}$

Details. The two values agree, since ${\mu(F_2) = \mu(F_1) + \mu(F_2 \setminus F_1) = \mu(F_1)}$ as in the proof of Remark 5. They do not depend on the choice of $F_1$ and $F_2{:}$ if also ${G_1 \subset E \subset G_2}$ with ${G_1, G_2 \in \mathcal{M}}$ and ${\mu(G_2 \setminus G_1) = 0,}$ then ${\mu(F_1) \le \mu(G_2) = \mu(G_1) \le \mu(F_2) = \mu(F_1)}$ by Remark 5 (i). $\blacksquare$

Remark 11. Negligible sets in $(X, \mathcal{M}, \mu)$ become measurable in ${(X, \overline{\mathcal{M}}, \overline{\mu}).}$

Proof (complete). Let ${F \subset X}$ be negligible. Then there exists ${E \in \mathcal{M}}$ such that ${F \subset E}$ and ${\mu(E) = 0.}$ To check that ${F \in \overline{\mathcal{M}},}$ notice that ${\emptyset \subset F \subset E,}$ that ${\mu(\emptyset) = \mu(E) = 0}$ and that ${\mu(E \setminus \emptyset) = \mu(E) = 0.}$ Hence ${F \in \overline{\mathcal{M}}}$ and ${\overline{\mu}(F) = 0.}$ $\blacksquare$

Outer measures#

The plan starts from a function ${f \colon \mathcal{F} \to [0, +\infty]}$ on a family ${\mathcal{F} \subset \mathcal{P}(X)}$ and ends with a measure ${\mu \colon \mathcal{M} \to [0, +\infty]}$ with ${\mathcal{F} \subset \mathcal{M}.}$ As Figure 9 shows, the way from the start to the end passes through outer measures, not directly.

Figure 9. The plan: from a function $f$ on $\mathcal{F}$ to outer measures, and from outer measures to a measure $\mu$ on a $\sigma$-algebra ${\mathcal{M} \supset \mathcal{F}.}$ The direct way is crossed out.

Definition 12 (Outer measure). Let $X$ be a set. An outer measure is a function ${\mu^* \colon \mathcal{P}(X) \to [0, +\infty]}$ such that:
(i) ${\mu^*(\emptyset) = 0;}$

(ii) ($\sigma$-subadditivity) for every ${(E_n) \subset \mathcal{P}(X)}$ and every ${E \subset \bigcup_n E_n,}$

$$\mu^*(E) \le \sum_{n=0}^{\infty} \mu^*(E_n).$$

Remark 13 (Monotonicity). If $\mu^*$ is an outer measure and ${E \subset F,}$ then ${\mu^*(E) \le \mu^*(F).}$ Indeed, (ii) of Definition 12 applies with ${E_0 = F}$ and ${E_n = \emptyset}$ for ${n \ge 1,}$ and ${\mu^*(\emptyset) = 0.}$

Theorem 14 (Construction of outer measure). Let $X$ be a set, let ${\mathcal{F} \subset \mathcal{P}(X)}$ be such that ${\emptyset \in \mathcal{F},}$ and let ${f \colon \mathcal{F} \to [0, +\infty]}$ be such that ${f(\emptyset) = 0.}$ For every ${E \subset X}$ define

$$\mu^*(E) := \inf \Big\{ \sum_{n=0}^{\infty} f(A_n) : (A_n) \subset \mathcal{F}, \ E \subset \bigcup_n A_n \Big\}.$$

Then $\mu^*$ is an outer measure.

Figure 10. A set $E$ and a sequence of sets $A_n$ whose union contains $E,$ one of the sequences in the definition of $\mu^*(E).$

Remark 15. If there is no ${(A_n) \subset \mathcal{F}}$ such that ${E \subset \bigcup_n A_n,}$ then ${\mu^*(E) = +\infty.}$

Definition 16 ($\mu^*$-measurable set). Let $X$ be a set and $\mu^*$ an outer measure on $X.$ A set ${E \subset X}$ is called a $\mu^*$-measurable set if, for every ${Z \subset X,}$

$$\mu^*(Z) = \mu^*(Z \cap E) + \mu^*(Z \setminus E).$$

Figure 11. A set $E$ and a set $Z,$ split into ${Z \cap E}$ and ${Z \setminus E.}$

Remark 17. For every ${E, Z \subset X}$ it is always true that

$$\mu^*(Z) = \mu^*\big( (Z \cap E) \cup (Z \setminus E) \big) \le \mu^*(Z \cap E) + \mu^*(Z \setminus E).$$

Indeed, the inequality is (ii) of Definition 12 with ${E_0 = Z \cap E,}$ ${E_1 = Z \setminus E}$ and ${E_n = \emptyset}$ for ${n \ge 2.}$

Theorem 18 (Carathéodory’s construction of measures). Let $X$ be a set and $\mu^*$ an outer measure on $X.$ Let

$$\mathcal{M} := \{ E \subset X : E \text{ is } \mu^*\text{-measurable} \},$$

and let ${\mu = \mu^*|_{\mathcal{M}}.}$ Then:
(i) $\mathcal{M}$ is a $\sigma$-algebra;
(ii) $(X, \mathcal{M}, \mu)$ is a measure space;
(iii) $(X, \mathcal{M}, \mu)$ is complete.

Proof (sketch). Parts (ii) and (iii) are proved; part (i) is not.

(ii) Since ${\mu = \mu^*|_{\mathcal{M}},}$ what needs to be proved is that $\mu^*$ is $\sigma$-additive on $\mathcal{M},$ the family of the $\mu^*$-measurable sets.

Step 1. If ${E, F \in \mathcal{M}}$ and ${E \cap F = \emptyset,}$ then ${\mu^*(E \cup F) = \mu^*(E) + \mu^*(F).}$ Let ${Z = E \cup F,}$ as in Figure 12. Since ${E \in \mathcal{M},}$

$$\mu^*(E \cup F) = \mu^*(Z) = \mu^*(Z \cap E) + \mu^*(Z \setminus E) = \mu^*(E) + \mu^*(F).$$

Indeed, ${Z \cap E = E}$ and ${Z \setminus E = F}$ because ${E \cap F = \emptyset.}$

Figure 12. Two disjoint sets $E$ and $F$ and their union $Z.$

Step 2. Let ${(E_n) \subset \mathcal{M}}$ with ${E_i \cap E_j = \emptyset}$ for every ${i \neq j.}$ Then

$$\begin{aligned} \mu^*\Big( \bigcup_n E_n \Big) &\le \sum_{n=0}^{\infty} \mu^*(E_n) = \lim_{N \to \infty} \Big( \sum_{n=0}^{N} \mu^*(E_n) \Big) \\ &= \lim_{N \to \infty} \mu^*\Big( \bigcup_{n=0}^{N} E_n \Big) \le \mu^*\Big( \bigcup_{n=0}^{\infty} E_n \Big), \end{aligned}$$

where the first inequality is $\sigma$-subadditivity, the second equality is Step 1, and the last inequality is monotonicity, as ${\bigcup_{n=0}^{N} E_n \subset \bigcup_{n=0}^{\infty} E_n.}$ Indeed, Step 1 extends to ${N + 1}$ sets by induction on $N,$ because ${\bigcup_{n=0}^{N-1} E_n}$ belongs to $\mathcal{M}$ by (i) and is disjoint from $E_N.$ The chain starts and ends with the same number, hence

$$\mu^*\Big( \bigcup_{n=0}^{\infty} E_n \Big) = \sum_{n=0}^{\infty} \mu^*(E_n).$$

(iii) Step 1. If ${\mu^*(E) = 0,}$ then ${E \in \mathcal{M},}$ that is, $E$ is $\mu^*$-measurable. Let ${Z \subset X.}$ Then

$$\begin{aligned} \mu^*(Z \cap E) + \mu^*(Z \setminus E) &\le \mu^*(E) + \mu^*(Z) \\ &= \mu^*(Z) \\ &\le \mu^*(Z \cap E) + \mu^*(Z \setminus E), \end{aligned}$$

where the first inequality is monotonicity, as ${Z \cap E \subset E}$ and ${Z \setminus E \subset Z,}$ the equality uses ${\mu^*(E) = 0,}$ and the last inequality is the one of Remark 17. Hence equality holds throughout.

Step 2. Let $F$ be negligible in $(X, \mathcal{M}, \mu),$ that is, let there exist ${E \in \mathcal{M}}$ such that ${F \subset E}$ and ${\mu(E) = 0,}$ where ${\mu(E) = \mu^*(E).}$ By monotonicity, ${0 \le \mu^*(F) \le \mu^*(E) = 0,}$ so ${\mu^*(F) = 0}$ and ${F \in \mathcal{M}}$ by Step 1. $\blacksquare$

Exercises#

The first exercise computes a generated $\sigma$-algebra, and the second checks Definition 1 on a $\sigma$-algebra of subsets of $\mathbb{R}.$

A family of subsets of the integers#

Example 19. Let ${\mathcal{F} \subset \mathcal{P}(\mathbb{Z})}$ be

$$\begin{aligned} \mathcal{F} &:= \big\{ \emptyset, \{0\}, \{-1, 0, 1\}, \{-2, -1, 0, 1, 2\}, \dots \big\} \\ &= \{ \emptyset \} \cup \big\{ \{-k, -k+1, \dots, k-1, k\} : k \in \mathbb{N} \big\}. \end{aligned}$$

What is $\sigma_0(\mathcal{F})$? The guess is ${\sigma_0(\mathcal{F}) = S,}$ where

$$S := \{ A \subset \mathbb{Z} : x \in A \iff -x \in A \}.$$

The sets of $S$ are the symmetric subsets of $\mathbb{Z}.$

Step 1. The claim is ${S \subset \sigma_0(\mathcal{F}).}$ First, ${\{-k, k\} \in \sigma_0(\mathcal{F})}$ for every ${k \in \mathbb{N},}$ because, as in Figure 13,

$$\{-k, k\} = \{-k, \dots, k\} \setminus \{-k+1, \dots, k-1\}.$$

Indeed, for ${k \ge 1}$ both sets on the right belong to $\mathcal{F},$ for ${k = 0}$ the second set is empty and ${\{0\} \in \mathcal{F},}$ and a $\sigma$-algebra is closed under differences by Remark 24 of the previous post.

k

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Figure 13. The sets $\{-k, \dots, k\}$ and $\{-k+1, \dots, k-1\}$ of $\mathbb{Z},$ and their difference $\{-k, k\}.$ The slider sets $k.$

Now take ${E \in S,}$ as in Figure 14. The set $E$ is a countable union of sets of the form $\{-k, k\}$ with ${k \in \mathbb{N},}$ so ${E \in \sigma_0(\mathcal{F}),}$ and therefore ${S \subset \sigma_0(\mathcal{F}).}$ Indeed, since $E$ is symmetric, it is the union of the sets $\{-k, k\}$ over the ${k \in \mathbb{N}}$ that belong to $E.$

Figure 14. An instance of a set ${E \in S{:}}$ its points are filled, and each arc joins the two points of a set $\{-k, k\}.$

Step 2. The set $S$ is a $\sigma$-algebra. This concludes, as $\sigma_0(\mathcal{F})$ is the smallest $\sigma$-algebra containing $\mathcal{F}.$ Indeed, every set of $\mathcal{F}$ is symmetric, so ${\mathcal{F} \subset S}$ and ${\sigma_0(\mathcal{F}) \subset S}$ by Theorem 26 (ii) of the previous post. Now ${\emptyset \in S,}$ the complement ${\mathbb{Z} \setminus A}$ of a set ${A \in S}$ is also symmetric, as in Figure 15, and any union of symmetric sets is symmetric as well.

Figure 15. An instance of a set ${A \in S}$ and, below it, its complement ${\mathbb{Z} \setminus A,}$ which is symmetric too.

Countable sets and their complements#

Example 20. Let

$$\mathcal{M} := \{ A \subset \mathbb{R} : A \text{ is countable or } \mathbb{R} \setminus A \text{ is countable} \}.$$

(i) Is $\mathcal{M}$ a $\sigma$-algebra? The countable union property is checked: let ${(A_n) \subset \mathcal{M};}$ what needs to be proved is ${\bigcup_n A_n \in \mathcal{M}.}$ The other properties of a $\sigma$-algebra hold as well. Indeed, $\emptyset$ is countable, and the definition of $\mathcal{M}$ does not change when $A$ is replaced by ${\mathbb{R} \setminus A.}$

Case 1. Suppose that $A_n$ is countable for every $n.$ Then $\bigcup_n A_n$ is countable, hence ${\bigcup_n A_n \in \mathcal{M}.}$

Case 2. Suppose that there exists $n_0$ such that ${\mathbb{R} \setminus A_{n_0}}$ is countable. Then

$$\mathbb{R} \setminus \bigcup_n A_n = \bigcap_n (\mathbb{R} \setminus A_n) \subset \mathbb{R} \setminus A_{n_0}, \tag{1}$$

so ${\mathbb{R} \setminus \bigcup_n A_n}$ is countable, and ${\bigcup_n A_n \in \mathcal{M}.}$

(ii) Let ${\mu \colon \mathcal{M} \to [0, +\infty]}$ be

$$\mu(A) = \begin{cases} 0 & \text{if } A \text{ is countable}, \\ 1 & \text{if } \mathbb{R} \setminus A \text{ is countable}. \end{cases}$$

Is $\mu$ a measure? The property to check is $\sigma$-additivity: let ${(A_n) \subset \mathcal{M}}$ be such that ${A_i \cap A_j = \emptyset}$ for every ${i \neq j.}$ Indeed, $\mu$ is well defined, because $A$ and ${\mathbb{R} \setminus A}$ are not both countable, otherwise $\mathbb{R}$ would be countable, against Proposition 9 (i) of the previous post; and ${\mu(\emptyset) = 0.}$

Case 1. Suppose that $A_n$ is countable for every $n.$ Then $\bigcup_n A_n$ is countable, so both ${\mu(\bigcup_n A_n) = 0}$ and ${\mu(A_n) = 0}$ for every $n,$ and

$$\mu\Big( \bigcup_n A_n \Big) = 0 = \sum_{n=0}^{\infty} \mu(A_n).$$

Case 2. Suppose that there exists $n_0$ such that ${\mathbb{R} \setminus A_{n_0}}$ is countable. Also ${A_n \cap A_{n_0} = \emptyset}$ for every ${n \neq n_0,}$ so ${A_n \subset \mathbb{R} \setminus A_{n_0},}$ and $A_n$ is countable for every ${n \neq n_0.}$ Now

$$\sum_{n=0}^{\infty} \mu(A_n) = \mu(A_{n_0}) = 1 \qquad \text{and} \qquad \mu\Big( \bigcup_n A_n \Big) = 1,$$

where the last equality holds because, by (1), ${\mathbb{R} \setminus \bigcup_n A_n}$ is contained in the countable set ${\mathbb{R} \setminus A_{n_0}.}$