[Measure Theory] Signed Measure
Our next primary theme is the concept of differentiation: differentiating a measure $\nu$ with regard to another measure $\mu$ on the same $\sigma$-algebra. To do that, it is useful to extend the notion of measure so as to allow measures to have negative values. And signed measure forms a basic building block for steps required to deal with the differentiation, such as Radon-Nikodym theorem. It has the countable additivity property of measures, but are allowed to take negative as well as positive values.
There are two main results in this post.
-
Hahn decomposition theorem
If $\mu$ is a signed measure, then $X$ can be decomposed as the union of two disjoints sets $P$ and $N$ such that $\mu$ behaves like a positive measure (the measures we have been considering up until now) on $P$ and $-\mu$ behaves like a positive measure on $N$. -
Jordan decomposition theorem
A signed measure can be written as the difference of two positive measures.
Signed measures
Let $\mathcal{A}$ be a $\sigma$-algebra. A signed measure is a function $\mu: \mathcal{A} \to (-\infty, \infty]$ such that
- $\mu (\emptyset) = 0$
- 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 $$ \begin{aligned} \mu \left( \bigcup_{i=1}^\infty A_i \right) = \sum_{i=1}^\infty \mu(A_i). \end{aligned} $$ where the series converges absolutely if $\mu \left( \bigcup_{i=1}^\infty A_i \right)$ is finite.
Here, we need the absolute convergence to ensure that the order of summation doesn’t matter.
Intuitively, the next definition defines regions of $X$ that behaves as if $\mu$ is a positive measure (on positive set), and $-\mu$ is a positive measure (on negative set).
Let $\mu$ be a signed measure on $\mathcal{A}$. A set $A \in \mathcal{A}$ is called a positive set for $\mu$ if $\mu (B) \geq 0$ whenever $B \subset A$ and $B \in \mathcal{A}$. On the other hand, a set $A \in \mathcal{A}$ is called a negative set for $\mu$ if $\mu (B) \leq 0$ whenever $B \subset A$ and $B \in \mathcal{A}$. A null set $A \in \mathcal{A}$ is one where $\mu (B) = 0$ whenever $B \subset A$ and $B \in \mathcal{A}$.
Note that a null set for a positive measure is of course a null set in the above sense. Also, in the same way with positive measures, we can show continuity from below and above of signed measure by exploiting $\sigma$-additivity.
- continuity from below: Suppose $A_i \in \mathcal{A}$ and $A_i \uparrow A$. Then $\mu (A) = \text{lim}_{n \to \infty} \mu (A_n)$.
- 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 $B_1 = A_1$, $B_2 = A_2 - A_1$, $B_3 = A_3 - (A_2 \cup A_1)$, $B_4 = A_4 - (A_3 \cup A_2 \cup A_1)$, and in general $B_i = A_i - \bigcup_{j=1}^{i-1} A_j$.
Obviously $B_i$ are pairwise disjoint, $\bigcup_{i=1}^n B_i = \bigcup_{i=1}^n A_i$, and $B_i \uparrow A$. Then $\bigcup_{i=1}^\infty B_i = A = \bigcup_{i=1}^\infty A_i$, therefore
\[\begin{aligned} \mu (A) & = \mu \left(\bigcup_{i=1}^\infty A_i\right) = \mu \left(\bigcup_{i=1}^\infty B_i\right) = \sum_{i=1}^\infty \mu (B_i) \\ & = \lim_{n \to \infty} \sum_{i=1}^n \mu (B_i) = \lim_{n \to \infty} \mu \left(\bigcup_{i=1}^n B_i\right) = \lim_{n \to \infty} \mu \left(\bigcup_{i=1}^n A_i \right) \\ \end{aligned}\](2) Define $B_1 = \emptyset$ and $B_i = A_{1} - A_{i}$ for $i \geq 2$. Then $B_i \uparrow A_1 - A$. Thus
\[\begin{aligned} \mu(A_1 - A) = \mu (A_1) - \mu(A) = \lim_{n \to \infty} \mu (A_1 - A_n) = \lim_{n \to \infty} \left[ \mu (A_1) - \mu(A_n) \right]. \end{aligned}\] \[\tag*{$\blacksquare$}\]Here is an example of signed measure.
$\color{green}{\mathbf{Example.}}$
Suppose $m$ is Lebesgue measure and define
\[\begin{aligned} \mu (A) = \int_A f \; dm \end{aligned}\]for some integrable $f$. If $f$ takes both positive and negative values, then $\mu$ will be a signed measure.
If we let
\[\begin{aligned} P = \{ x : f(x) \geq 0\} \text { and } N = \{ x: f(x) < 0 \} \end{aligned}\]it is easy to show that $P$ and $N$ are positive and negative set, respectively.
The Hahn decomposition theorem provides a decomposition of $X$ into positive and negative sets $P$ and $N$. This decomposition is unique except that \(C = \{x : f (x) = 0 \}\) could be included in $N$ instead of $P$, or apportioned partially to $P$ and partially to $N$. Note that $C$ is a null set, however.
And the Jordan decomposition theorem provides a decomposition of $\mu$ into $\mu^+$ and $\mu^-$, where $\mu^+ (A) = \int_A f^+ \; dm$ and $\mu^- (A) = \int_A f^- \; dm$.
\[\tag*{$\blacksquare$}\]Let’s conclude the section with the important results that we can always extract a negative subset with negative measure from any measurable set with negative measure.
$\mathbf{Proof.}$
Step 1. If $E$ is a negative set, we are done. If not, there exists a measureable subset of $E$ with positive measure. Then let $n_1$ be the smallest integer such that there exists $E_1 \subset E$ satisfying $E_1 \in \mathcal{A}$ and $\mu (E_1) \geq \frac{1}{n_1}$.
Step 2. And by induction, we can define pairwise disjoint measurable sets $E_2, E_3, \cdots$. Let $k \geq 2$. $E_1, \dots E_{k-1}$ are pairwise disjoint measurable contained in $E$, satisfying $\mu (E_i) > 0$ for $i = 1, \cdots, k-1$.
Let
\[\begin{aligned} F_k = E - \bigcup_{n=1}^{k-1} E_n. \end{aligned}\]Then $\mu (E) = \mu (F_k) + \sum_{n=1}^{k-1} \mu (E_n)$, thus $\mu (F_k) \leq \mu (E) < 0$.
- If $F_k$ is a negative set, then $F_k$ is the desired $F$. We stop the construction of $E_{k + 1}$ and $F_{k + 1}$.
- If not, let $n_{k + 1} \leq n_{k}$ be the smallest positive integer such that there exists $E_{k + 1} \subset F_{k+1}$ satisfying $E_{k + 1} \in \mathcal{A}$ and $\mu (E_{k+1}) \geq \frac{1}{n_{k + 1}}$.
Step 3. But it is possible that we obtain infinitely many $F_k$. Then set
\[\begin{aligned} F = \bigcap_{k=1}^\infty F_k = E - \left( \bigcup_{k=1}^\infty E_k \right). \end{aligned}\]Note that we have $E = F \cup (\bigcup_{k=1}^\infty E_k )$, therefore
\[\begin{aligned} \mu (E) = \mu (F) + \sum_{k=1}^\infty \mu (E_k) \end{aligned}\]Since $\mu (E) < 0$ and $\mu (E_k) > 0$, $\mu (F) < \mu (E) < 0$. Note that from the definition of signed measure, $\mu (F) > -\infty$. Hence
\[\begin{aligned} \sum_{k=1}^\infty \mu (E_k) < \infty \end{aligned}\]As $\mu (E_k) \geq \frac{1}{n_k}$, $\sum_{k=1}^\infty \frac{1}{n_k} < \infty$ thus $n_k \to \infty$ as $k \to \infty$. With this fact, we now claim that $F$ is a negative set.
$\mathbf{Proof.}$
Suppose $G \subset F$ is measurable. If $\mu (G) > 0$, then $\mu (G) > \frac{1}{N}$ for some $N \in \mathbb{N}$. Then, for some $k$, $n_k > N$ as $n_k \to \infty$ as $k \to \infty$.
Then we would have the set $G$ instead of $E_k$ at stage $k$ and $n_k \leq N$. But this contradicts the construction as $F$ is constructed by subtracting all $E_k$ from $E$. Hence $\mu (G) \leq 0$ whenever $G \subset F$ and $G \in \mathcal{A}$.
\[\tag*{$\blacksquare$}\]Hahn decomposition theorem
Now we introduce Hahn decomposition theorem. This will be useful for proving the further theorems called Radon-Nikodym theorem that connects measure theory and differentiation.
Let $\mu$ be a signed measure taking values in $(- \infty, \infty]$.
- There exist disjoint measurable sets $E$ and $F$ such that $E \cup F = X$, $E$ is a negative set and $F$ is a positive set;
- If $E^\prime$ and $F^\prime$ are another such pair, then $E \Delta E^\prime = F \Delta F^\prime$ is a null set with regard to $\mu$;
- If $\mu$ is not a positive measure, then $\mu(E) < 0$. If $-\mu$ is not a positive measure, then $\mu(F) > 0$.
$\mathbf{Proof.}$
(1) Note that $\emptyset$ is a negative set. Let
\[\begin{aligned} L = \text{inf } \{ \mu (A) : A \text{ is a negative set } \}. \end{aligned}\]therefore $L \leq 0$. Then by definition of infimum we can choose negative sets $A_n$ such that $\mu (A_n) \to L$ as $n \to \infty$.
Let
\[\begin{aligned} E & = \bigcup_{n=1}^\infty A_n \\ B_n & = A_n - \left( \bigcup_{k=1}^{n-1} A_k \right) \end{aligned}\]Note that $B_n$ is a negative set and \(\{ B_n \}\) are pairwise disjoint, increasing to $E$. We claim that $E$ is a negative set.
$\mathbf{Proof.}$
Suppose $C \subset E$ is measurable. Then $C = \bigcup_{n=1}^\infty (C \cap B_n)$. We have
\[\begin{aligned} \mu (C) & = \lim_{n \to \infty} \mu \left( C \cap \left( \bigcup_{k=1}^n B_k \right) \right) \\ & = \lim_{n \to \infty} \sum_{k=1}^n \mu \left( C \cap B_k \right) \\ & \leq 0. \end{aligned}\] \[\tag*{$\blacksquare$}\]Hence $\mu (E) = \mu (A_n) + \mu (E - A_n) \leq \mu (A_n)$ for any $n \in \mathbb{N}$. Letting $n \to \infty, \mu (E) \leq L$. But since $E$ is negative set, we have $\mu (E) \geq L$ thus $\mu (E) = L > -\infty$.
Now, let $F = E^c$. We claim that $F$ is a positive set.
$\mathbf{Proof.}$
Suppose not. Then there exists a subset $B \subset F$ such that $\mu (B) < 0$. By the proposition above, there exists a measurable set $C \subset B$ with $\mu (C) < 0$. Then $E \cup C$ is a negative set and $\mu (E \cup C) = \mu (E) + \mu (C) < \mu (E) = L$, which is contradiction.
\[\tag*{$\blacksquare$}\](2) Let $A \subset E - E^\prime$ and $A \in \mathcal{A}$. Since $A \subset E$, $\mu (A) \leq 0$. But $A \subset E - E^\prime = F^\prime - F \subset F^\prime$, thus $\mu (A) \geq 0$. Hence $\mu (A) = 0$.
Similarly, we can prove that $\mu (A) = 0$ if $A \subset E^\prime - E$ and $A \in \mathcal{A}$. Therefore if $A \subset E \Delta E^\prime$ and $A \in \mathcal{A}$, $\mu (A) = 0$.
(3) Suppose $\mu$ is not a positive measure but $\mu (E) = 0$. Since $\mu$ is not a positive measure, there exists $A \in \mathcal{A}$ such that $\mu (A) < 0$.
And such $A \in \mathcal{A}$, we have
\[\begin{aligned} \mu (A) & = \mu (A \cap E) + \mu (A \cap F) \\ & \geq \mu (E) + \mu (A \cap F) \\ & \geq 0 + \mu (A \cap F) \geq 0 \end{aligned}\]since $F$ is a positive set, which is a contradiction.
\[\tag*{$\blacksquare$}\]Jordan decomposition theorem
The decomposition of $X$ in Hahn decomposition theorem is usually called a Hahn decomposition for $\mu$, and not unique in general. But, it leads to the Jordan decomposition theorem which states that the existence and uniqueness of a canonical representation of $\mu$ as the difference of two positive measures.
To state the theorem, we introduce a new concept called mutually singular: Two measures that are mutually singular live on disjoint sets;
Two positive measures $\mu$ and $\nu$ are called mutually singular and write $\mu \perp \nu$ if there exists $E, F \in \mathcal{A}$ with $$ \begin{aligned} E \cap F & = \emptyset \\ E \cup F & = X \end{aligned} \\ \begin{aligned} \mu (E) = \nu (F) = 0 \end{aligned} $$
For example, let $\mu$ be a Lebesgue measure. For $\nu$, consider Cantor-Lebesgue fuction $f$. Recall that it is increasing and continuous. Therefore, we can construct a Lebesgue-Stieltjes measure $\nu$ associated with $f$.
Then we claim that $\mu \perp \nu$ and $E = C$, $F = C^c$ where $C$ is Cantor set. Since $\mu (E) = 0$, we should show $\nu (F) = 0$. Note that Cantor set is closed, thus $F$ is open that implies $F$ can be written as disjoint countable union of open intervals.
Hence, it suffices to show that $\nu (I) = 0$ for every open intervals contained in $F$. And this will follow if we show $\nu (J) = 0$ for every interval of the form $J = (a, b]$ contained in $F$. But by construction of $f$, such $J$ must be satisfies $\nu (J) = f(b) - f(a) = 0$ as $b, a \in C^c$.
The following theorem, called Jordan decomposition theorem, is the second main result of this post. It can be derived with Hahn decomposition theorem.
If $\mu$ is a signed measure on a measure space $(X, \mathcal{A})$, there exist positive measures $\mu^+$ and $\mu^-$ such that $$ \begin{aligned} \mu = \mu^+ - \mu^- \end{aligned} $$ where $\mu^+$ and $\mu^-$ are mutually singular. And this decomposition is unique.
$\mathbf{Proof.}$
-
Existence
Let $X = E \cup F$ be a Hahn decomposition for $\mu$ where $E$ is a negative set and $F$ is a positive set. Define $$ \begin{aligned} \mu^+ (A) & := \mu (A \cap F) \\ \mu^- (A) & := - \mu (A \cap E) \end{aligned} $$ Then $\mu^+$ and $\mu^-$ satisfies the desired properties. -
Uniqueness
Let $\mu = \nu^+ - \nu^-$ be another decomposition with $\nu^+ \perp \nu^-$. Since $\nu^+ \perp \nu^-$, let $E^\prime$ and $F^\prime$ be measurable sets such that $$ \begin{aligned} E^\prime \cup F^\prime & = X \\ E^\prime \cap F^\prime & = \emptyset \end{aligned} $$ and $\nu^+ (E^\prime) = 0$ and $\nu^- (F^\prime) = 0$. We claim that $F^\prime$ and $E^\prime$ are a positive set and negative set for $\mu$, respectively:$\color{black}{\mathbf{Claim.}}$ $F^\prime$ and $E^\prime$ are a positive set and negative set for $\mu$, respectively.$\mathbf{Proof.}$
Let $A \subset F^\prime$ and $A \in \mathcal{A}$. Note that $\nu^-$ is a positive measure and $\nu^- (F^\prime) = 0$, so that $\nu^- (A^\prime) = 0$. Then
\[\begin{aligned} \mu (A) = \nu^+ (A) - \nu^- (A) = \nu^+ (A) \geq 0 \end{aligned}\]which shows $F^\prime$ is a positive set for $\mu$. Similarly, we can show that $E^\prime$ is a negative set for $\mu$.
\[\tag*{$\blacksquare$}\]
Thus, by Hahn decomposition theorem, $F \Delta F^\prime = E \Delta E^\prime$ are null sets for $\mu$. With this fact, we can show the uniqueness of decomposition:$\color{black}{\mathbf{Claim.}}$ $\nu^+ (A) = \mu^+ (A)$ for any $A \in \mathcal{A}$.$\mathbf{Proof.}$
Since $\nu^+ (E^\prime) = 0$ and $\nu^- (F^\prime) = 0$, we have
\[\begin{aligned} \nu^+ (A) & = \nu^+ (A \cap F^\prime) + \nu^+ (A \cap E^\prime) \\ & = \nu^+ (A \cap F^\prime) + 0 \\ & = \nu^+ (A \cap F^\prime) - \nu^- (A \cap F^\prime) \\ & = \mu (A \cap F^\prime) \\ & = \mu (A \cap F) - \mu (A \cap (F - F^\prime)) \\ & = \mu (A \cap F) \\ & := \mu^+ (A) \end{aligned}\]Note that $\mu (A \cap (F - F^\prime)) = 0$ as $F \Delta F^\prime = (F - F^\prime) \cup (F^\prime - F)$ is null set for $\mu$.
\[\tag*{$\blacksquare$}\]
Similarly, we can show that $\nu^- = \mu^-$.
In the proof, we constructed $\mu^+$ and $\mu^-$ as restrictions of $\mu$ to Hahn decomposition, but we can find the Jordan decomposition in another way:
$\mathbf{Proof.}$
Let any $B \in \mathcal{A}$ such that $B \subset A$ be given. Then
\[\begin{aligned} \mu (B) = \mu^+ (B) - \mu^- (B) \leq \mu^+ (B) \leq \mu^+ (A). \end{aligned}\]Also, let $E \cup F = X$ be a Hahn decomposition that defines $\mu^+ (A) = \mu (A \cap F)$. Since $A \cap F \in \mathcal{A}$ and $A \cap F \subset A$, we have
\[\begin{aligned} \mu^+ (A) = \mu (A \cap F) \leq \sup \; \{\mu(B): B \in \mathcal{A}, B \subset A\} \end{aligned}\]We can prove $\mu^-$ case in analogous way.
\[\tag*{$\blacksquare$}\]Also, one can easily show that for a signed measure $\lambda$, $\lambda^+$ and $\lambda^-$ in Jordan decomposition are the smallest of those positive measures $\mu$ and $\nu$ that satisfy $\lambda = \mu - \nu$. Specifically,
$\mathbf{Proof.}$
First, we have to prove that $\lambda$ is a signed measure. $\lambda(\emptyset) = 0$ obviously. Let \(\{A_i \}\) be given pairwise disjoint in $\mathcal{A}$. Since $\mu$, $\nu$ are finite, $\lambda$ is also finite, and we have
\[\begin{aligned} \lambda\left(\cup_{i=1}^{\infty} A_i\right) & =\mu\left(\cup_{i=1}^{\infty} A_i\right)-\nu\left(\cup_{i=1}^{\infty} A_i\right) \\ & =\sum_{i=1}^{\infty} \mu\left(A_i\right)-\sum_{i=1}^{\infty} \nu\left(A_i\right) \\ & =\sum_{i=1}^{\infty} \mu\left(A_i\right)-\nu\left(A_i\right) \\ & =\sum_{i=1}^{\infty} \lambda\left(A_i\right) . \end{aligned}\]Note that $\lambda = \mu − \nu$ is finite. And $\sum_{i=1}^\infty \lambda (A_i)$ is a difference of two absolutely convergence series, and thus it is absolutely converges. More precisely,
\[\begin{aligned} \sum_{i=1}^{\infty}\left|\lambda\left(A_i\right)\right| \leq \sum_{i=1}^{\infty}\left|\mu\left(A_i\right)\right|+\left|\nu\left(A_i\right)\right|<\infty \quad(\because \text { Def of finite signed measure }) \end{aligned}\]Thus, $\lambda$ is a finite signed measure on $(X, \mathcal{A})$.
Let $\lambda = \lambda^+ - \lambda^-$ be Jordan decomposition with positive set $F$ and negative set $E$ such that $X = E \cup F$, $E \cap F = \emptyset$, and $\lambda^+ (E) = \lambda^+(F) = 0$. We have
\[\begin{aligned} & \mu (A) \geq \mu (A \cap F) \geq \mu (A \cap F) - \nu (A \cap F) = \lambda^+ (A) \\ & \nu (A) \geq \nu (A \cap E) \geq \nu (A \cap E) - \mu (A \cap E) = \lambda^- (A) \\ \end{aligned}\] \[\tag*{$\blacksquare$}\]This proposition will be useful for further discussion. And we define total variation of measure that plays a similar role with modulus.
In Jordan decomposition of $\mu$, $\mu^+$ and $\mu^-$ are called the positive and negative variations of $\mu$. Furthermore, we define the total variation of $\mu$ to be the measure $| \mu |$ defined by $$ \begin{aligned} | \mu | = \mu^+ + \mu^-. \end{aligned} $$
Equivalently, total variation of measure can be defined in different way:
Let $\mu$ be a signed measure on $(X, \mathcal{A})$. Then $$ |\mu|(A)=\sup \left\{\sum_{j=1}^n\left|\mu(B_j)\right|: \operatorname{each} B_j \in \mathcal{A} \text{ the } B_j \text{ are disjoint, } \cup_{j=1}^n B_j=A \right\} $$
$\mathbf{Proof.}$
Let \(\{ B_j \}\) be any collection of pairwise disjoint measurable set, i.e. $B_j \in \mathcal{A}$, such that $\cup_{j=1}^n B_j = A$. Then
\[\begin{aligned} \sum_{j=1}^n | \mu (B_j) | \leq \sum_{j=1}^n | \mu | (B_j) = | \mu | (\cup_{j=1}^n B_j = A). \end{aligned}\]Let $B_1 = A \cap E$ and $B_2 = A \cap F$ where $\mu^+ (A) = \mu (A \cap F)$ and $\mu^- (A) = \mu (A \cap E)$. Then
\[\begin{aligned} |\mu(B_1)| + |\mu(B_2)| = \mu^+ (A) + \mu^- (A) = | \mu | (A). \end{aligned}\] \[\tag*{$\blacksquare$}\]Then, we have the following results:
$\mathbf{Proof.}$
For the first inequality, we should prove that $\mu + \nu$ is a finite signed measure and it is trivial. Let $\lambda = \mu + \nu$. Then $\lambda = (\mu^+ + \nu^+) - (\mu^- + \nu^-)$, where $\mu^+ + \nu^+$ and $\mu^- + \nu^-$ are finite positive measure. Thus we have
\[\begin{aligned} \lambda^+ (A) \leq \mu^+ (A) + \nu^+ (A) \\ \lambda^- (A) \leq \mu^- (A) + \nu^- (A) \end{aligned}\]For the second inequality, we should prove that $\mu - \nu$ is a finite signed measure and it is trivial. We will show that $\mu^+(A) + \nu^−(A) \geq \lambda^+(A)$ and $\mu^−(A) + \nu^+(A) \geq \lambda^– (A)$ for every $A \in \mathcal{A}$.
\[\begin{aligned} \lambda^{+}(A) & =\lambda(A \cap F)=\mu(A \cap F)-\nu(A \cap F) \\ & =\mu^{+}(A \cap F)-\mu^{-}(A \cap F)-\nu^{+}(A \cap F)+\nu^{-}(A \cap F) \\ & \leq \mu^{+}(A \cap F)+\nu^{-}(A \cap F) \leq \mu^{+}(A)+\nu^{-}(A) \end{aligned}\]Likewise,
\[\begin{aligned} \lambda^{-}(A) & =-\lambda(A \cap E)=-\mu(A \cap E)+\nu(A \cap E) \\ & =-\mu^{+}(A \cap E)+\mu^{-}(A \cap E)+\nu^{+}(A \cap E)-\nu^{-}(A \cap E) \\ & \leq \nu^{+}(A \cap E)+\mu^{-}(A \cap E) \leq \mu^{-}(A)+\nu^{+}(A) \end{aligned}\]Thus we conclude that
\[\begin{aligned} | \lambda | (A) = | \mu - \nu | (A) \leq | \mu (A) | + | \nu (A) |. \end{aligned}\] \[\tag*{$\blacksquare$}\]There are some useful properties related to $\mu$ and $| \mu |$. Here are examples.
$\mathbf{Proof.}$
Let $\mu^+ (B) = \mu (B \cap F)$ and $\mu^- (B) = \mu (B \cap E)$ as in the proof of Jordan decomposition theorem.
Suppose $A$ is a null set. Then $\mu (A \cap F) = \mu^+ (A) = 0$ and $\mu (A \cap E) = \mu^- (A) = 0$, thus $| \mu | (A) = 0$.
Suppose $| \mu | (A) = 0$. In other words, $\mu^+ (A) = 0$ and $\mu^- (A) = 0$. For any $B \subset A$ and $B \in \mathcal{A}$, note that $\mu^+(B) \leq \mu^+ (A) = 0$, and $\mu^-(B) \leq \mu^- (A) = 0$. Hence
\[\begin{aligned} \mu (B) = \mu^+ (B) - \mu^- (B) = 0. \end{aligned}\] \[\tag*{$\blacksquare$}\]Reference
[1] Richard F. Bass, Real Analysis for Graduate Students, Version 4.3
[2] Folland, Gerald B. Real analysis: modern techniques and their applications. Vol. 40. John Wiley & Sons, 1999.
[3] W. Rudin. Real and Complex Analysis, 3rd ed. McGraw-Hill, New York, 1987.
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