Signed Measures

Let $(X, \mathcal{S})$ be a measurable space. A function $\nu: \mathcal{S}\rightarrow [-\infty, \infty]$ is a signed measure if

• $\nu(\emptyset) = 0$
• $\{ -\infty, \infty \} \cap$ (range $\nu$ ) is a singleton set or empty.
• if $\{ E_i \}_{i=1}^\infty$ are a pairwise disjoint collection of measurable sets, then
  $$\nu(\bigcup_{i=1}^\infty E_i) = \sum_{i=1}^\infty \nu(E_i)$$
  where the sum converges absolutely if $\sum_1^\infty \nu(E_i) < \infty$ .

Example: Let $g \in L^1(X, \mathcal{S}, \mu)$ (where $\mu$ ) is a measure and define

$$\nu(E) = \int_E g d\mu = \int_E g^+ d\mu - \int_E g^- d\mu$$

Definition: A set $E \in \mathcal{S}$ is a positive set for the signed measure $\nu$ if $\nu(F) \geq 0$ for all $F \in \mathcal{S}$ with $F \subset E$ . Similarly, we can define a negative set.

Lemma: Suppose $\nu$ is a signed measure and $E$ is a positive set. If $F \subset E$ is measurable, then $F$ is a positive set. Moreover, if a countable family $E_i \in \mathcal{S}$ are all positive, so is $\bigcup_1^\infty E_i$ .
Proof:
The first assertion follows immediately from the definition.
Now suppose that $\{E_i\}_1^\infty$ are positive sets, and write $\bigcup_1^\infty E_i$ as a disjoint union, $\bigcup_1^\infty F_i$ so $F_i \subset E_i$ . Moreover, if $B \subset \bigcup_1^\infty E_i = \bigcup_1^\infty F_i$ , then

$$B = \bigcup_1^\infty (B \cap F_i)$$

Then,

$$\nu(B) = \sum_{i=1}^\infty \nu(B \cap F_i) \geq 0$$

as for all $i$ , $B \cap F_i \subset E_i$ .

Lemma: (Pac-Man Lemma) Let $(X, \mathcal{S})$ be a measurable space, $\nu$ a signed measure. Suppose we have a subset $E \in \mathcal{S}$ and $0 < \nu(E) < \infty$ . Then, there exists a positive set $P \subset E$ such that $\nu(P) > 0$ .
Proof: If $E$ is positive, take $E = P$ and we are done.
So assume there exists a subset $N \subset E$ such that $\nu(N) < 0$ . Let $\mathcal{N}(E) = \{ N \in \mathcal{S}: N \subset E, \nu(N) < 0 \}$ . As $E$ is not positive, $\mathcal{N}$ is nonempty.
Let $n \in \mathbb{N}$ be the smallest natural number such that there exists a $N_1 \in \mathcal{N}(E)$ with $\nu(N_1) < -\frac{1}{n_1}$ . If $E \setminus N_1$ is positive, we are done.
If not, proceed inductively. Let $n_2$ be the smallest positive integer such that there exists a $N_2 \in \mathcal{N}(E)$ such that $\nu(N_2) < -\frac{1}{n_2}$ . Repeat to obtain $N_1, \ldots, N_k$ such that $\nu(N_j) < -\frac{1}{n_j}$ , $1 \leq j \leq k$ and $n_{k+1}$ is chosen to be the smallest positive integer such that there exists $N_{k+1} \in \mathcal{N}(E \setminus \bigcup_1^k N_j)$ such that $\nu(N_{k+1}) < -\frac{1}{n_{k+1}}$ .
Notice that $n_{k+1} \geq n_k$ and $\{ N_k \}$ is a disjoint family of sets. Let

$$P = E \setminus \left( \bigcup_{k=1}^\infty N_k \right)$$

We have

$$E = P \cup \left(\bigcup_{k=1}^\infty N_k\right)$$

So that

$$\nu(E) = \nu(P) + \sum_1^\infty \nu(N_k)$$

As $\nu(E) < \infty$ , we get that the sum on the right converges absolutely. So,

$$\sum_{k=1}^\infty \frac{1}{n_k} < \sum_{k=1}^\infty \vert \nu(N_k) \vert < \infty$$

In particular, $n_k \rightarrow \infty$ as $k \rightarrow \infty$ . We have

$$\nu( \bigcup_{k=1}^\infty N_k ) < 0$$

and

$$\nu(E) > 0$$

so $\nu(P) > 0$ . Let $\epsilon > 0$ , and choose $K$ so large that

$$\frac{1}{n_{k+1} - 1} < \epsilon , \forall k \geq K$$

We know that

$$P \subset E \setminus \left( \bigcup_{k=1}^K N_k \right),$$

so by construction, $P$ contains no measurable set $F$ with

$$\nu(F) < \frac{-1}{n_{k+1} - 1}.$$

Hence, $P$ contains no subset $F \in \mathcal{S}$ with $\nu(F) < -\epsilon$ . But this is true for all $\epsilon > 0$ , so $P$ must be positive. $\square$

Theorem: Hahn Decomposition
Let $\nu$ be a signed measure on $(X, \mathcal{S})$ . Then, there exists a positive set $P \subset X$ and a negative set $N \subset X$ such that $X = P \cup N$ and $P \cap N = \emptyset$ .
Proof: $\nu$ omits at least one of $+\infty$ , $-\infty$ . Assume $\nu(E) = \neq +\infty$ .
Let $\alpha = \sup\{\nu(A) : A \in S$ and $A$ is a positive set $\}$ . Let $\alpha_n \in \mathbb{R}$ and $\alpha_n < \alpha$ and $\alpha_n \rightarrow \alpha$ . Let $P_n \in \mathcal{S}$ be positive sets such that $\alpha_n < \nu(P_n) \leq \alpha$ for all $n$ (by our previous lemma, we can do this). Let

$$P = \bigcup_{n=1}^\infty P_n.$$

Then $P$ is a positive set so $\nu(P) \leq \alpha$ . But, $P \setminus P_n \subset P$ , so

$$0 \geq \nu(P \setminus P_n)$$

and

$$\alpha \geq \nu(P) = \nu(P \setminus P_n) + \nu(P_n)$$

$$> \nu(P \setminus P_n) + \alpha_n \geq \alpha_n$$

Hence, by taking $n \rightarrow \infty$ , $\nu(P) = \alpha$ . Because $\nu(P) \neq +\infty$ , $\alpha < \infty$ .

Wednesday, 12-8-2004

Let $N = X \setminus P$ .
Suppose $E \subset N$ , $E \in \mathcal{S}$ and $\nu(E) > 0$ . By the ``Pac-Man" lemma, there exists a positive set $E' \subset E$ , $E' \in \mathcal{S}$ . Then, $E' \cup P$ is a positive set and $E' \cap P = \emptyset$ (from the definition of $N$ ). Therefore, $\alpha \leq \nu(E' \cup P) = \nu(E') + \nu(P) > \alpha$ . Because $\alpha < \infty$ , this says $\nu(E') = 0$ , a contradiction. $\square$

The decomposition of $X$ into $P$ and $N$ is called a Hahn-Decomposition of $X$ . In general, this decomposition is not unique. This is due to null sets.

Suppose $\{ P, N \}$ is a Hahn decomposition. Let $\nu^+(E) = \nu(E \cap P)$ and $\nu^-(E) = - \nu(E \cap N)$ . Then, $\nu^+$ and $\nu^-$ are measures on $X$ and $\nu(E) = \nu^+(E) - \nu^-(E)$ .

Definition: To measures $\mu_1$ and $\mu_2$ on $(X, \mathcal{S})$ are mutually singular (written $\mu_1 \perp \mu_2$ ) if there exists disjoint measurable sets $A$ and $B$ such that $X = A \cup B$ . and $\mu_1(B) = \mu_2(A) = 0$ .

Example: $\nu^+$ and $\nu^-$ are mutually singular.

Two signed measures $\mu$ and $\sigma$ are mutually singular if $(\mu^+ + \mu^-) \perp (\sigma^+ + \sigma^-)$ .

Examples :

• Lebesgue measure on Borel sets and the point-mass measure.
• Lebesgue measure on Borel sets and the counting measure.
• Lebesgue measure and a measure derived from the Cantor set.
  

Definition: If $\nu$ is a signed measure, then $\vert\nu\vert =$ total variation = $\nu^+ + \nu^-$ .

Theorem: Jordan Decomposition Theorem
Let $\nu$ be a signed measure on $(X, \mathcal{S})$ . Then there exists two measures $\nu^+$ and $\nu^-$ on $(X, \mathcal{S})$ such that

• $\nu = \nu^+ + \nu^-$ and
• $\nu^+ \perp \nu^-$ .

Moreover, if $\mu^+$ and $\mu^-$ on $(X, \mathcal{S})$ such that

• $\nu = \mu^+ - \mu^-$ , and
• $\mu^+ \perp \mu^-$ ,

Then $\mu^+ = \nu^+$ and $\nu^- = \mu^-$ .
Proof: Existence has already been done.
Suppose $\mu^+$ , $\mu^-$ satisfy the last two conditions. Then, there exists measurable sets $A, B$ such that $X = A \cup B$ and $A \cap B = \emptyset$ . Further, $\mu^+(B) = \mu^-(A) = 0$ .
Let $P, N$ be the corresponding sets for $\nu^+$ and $\nu^-$ . Then,

• $A$ is positive for $\nu$ ,
• $P$ is positive for $\nu$ ,
• $B$ is negative for $\nu$ , and
• $N$ is negative for $\nu$ ,

from the last two properties.
Notice,

$$P \setminus A \subset P$$    and $$P \setminus A \subset B$$

So,

$$\nu( P \setminus A) = \mu^+(P \setminus A) - \mu^-( P \setminus A)$$

$$= 0 - \mu^-(P \setminus A) \leq 0$$

Therefore, we get that $\mu^-(P \setminus A) = 0$ and $\nu( P \setminus A) = 0$ . Similarly, $\nu(A \setminus P) = 0$ .
For $E \in \mathcal{S}$ , we find

$$\mu^+(E) = \mu^+(E \cap A) = \nu(E \cap A) = \nu(E \cap (P \cap A)) + \nu(E \cap (P \setminus A))$$

$$= \nu(E \cap P \cap A) = \nu((E \cap A) \cap P) = \nu^+(E \cap A) = \nu^+(E)$$

Similarly, we get $\mu^-(E) = \nu^-(E)$ for all $E \subset \mathcal{S}$ . $\square$

Definition $\nu^+$ and $\nu^-$ are called the positive and negative variation of $\nu$ . $\nu^+ + \nu^-$ is called the total variation of $\nu$ , $\vert\nu\vert$ . From this last theorem, we get this is well-defined.

Remark: $E \subset \mathcal{S}$ is null for a signed measure $\nu$ if $\nu(F) = 0$ for all $F \subset E$ , $F \in \mathcal{S}$ . This is true if and only if $\vert v\vert(E) = 0$ .

Recall that for signed measures $\nu_1, \nu_2$ , we say $\nu_1 \perp \nu_2$ if and only if $\vert \nu_1 \vert \perp \vert \nu_2 \vert$ .

Definition: For a signed measure $\nu$ and a measurable function $f$ , we say that $f$ is integrable with respect to $\nu$ if $f$ is integrable with respect to $\nu^+$ and $f$ is integrable with respect to $\nu^-$ . Moreover, we define

$$\int_X f d\nu = \int_X f d\nu^+ - \int_X f d\nu^-$$

On the right hand side, we can equivalently take out the integration over $X$ and replace it with integration over $P$ and $N$ , respectively.

Recall: If $\nu$ is a signed measure and $\mu$ is a (honest) measure on a measure space $(X, \mathcal{S})$ (same space for both measures), we say $\nu$ is absolutely continuous with respect to $\mu$ if whenever $E \in \mathcal{S}$ and $\mu(E) = 0$ , we have $\nu(E)$ .
Write $\nu « \mu$ whn $\nu$ is absolutely continuous.

Note: $\nu « \mu \Leftrightarrow \vert \nu \vert « \mu$ .

Remark: ``Absolutely continuous" and ``mutually singular" are in some sense opposites. If $\nu$ and $\mu$ are measures such that both are $\mu \perp \nu$ and $\nu « \mu$ . Then, $\nu \equiv 0$ . The same holds for $\nu$ a signed measure.
Proof:
To say that $\nu \perp \mu$ means that $X = A \cup B$ where $A, B \in \mathcal{S}$ , $A \cap B = \emptyset$ , and $\nu(B) = \mu(A) = 0$ . Then since $\nu « \mu$ , we get $\nu(A) = \nu(B) = 0$ , and we get that $\nu(X) = \nu(A \cup B) = \nu(A) + \nu(B) = 0$ . If we wanted to do this for signed measures, we simply replace $\nu$ and $\mu$ in the statement by the total variance, $\vert\nu\vert$ and $\vert\mu\vert$ .

Proposition: Suppose $\nu$ is a finite signed measure (i.e. $\vert\nu\vert(X) < \infty$ ) and $\mu$ is a positive measure, both on the measure space $(X, \mathcal{S})$ . Then, $\nu « \mu$ if and only if for all $\epsilon > 0$ , there exists a $\delta > 0$ such that $\vert v(E)\vert < \epsilon$ whenever $\mu(E) < \delta$ .
Proof: We have $\nu « \mu \Leftrightarrow \vert \nu \vert « \mu$ , so without loss of generality assume that $\nu$ is a measure (i.e., $\nu \geq 0$ ). If the $\epsilon -\delta$ condition holds, if $\mu(E) = 0$ then $\mu(E) < \delta$ for all $\delta > 0$ . In particular, we get $\nu(E) < \epsilon$ for all $\epsilon > 0$ . So, $\nu(E) = 0$ . So, $\nu « \mu$ .
Conversely, suppose the $\epsilon -\delta$ condition fails. Hence there exists $\epsilon > 0$ with no $\delta$ . Fix such an $\epsilon > 0$ . Hence, $\forall \delta > 0$ , there exists a $E_\delta \in \mathcal{S}$ with $\mu(E_\delta) < \delta$ and $\nu(E_\delta) \geq \epsilon$ . Let $E_n \in \mathcal{S}$ satisfy $\mu(E_n) < \frac{1}{2^n}$ and $\nu(E_n) \geq \epsilon$ . Put $F_k = \bigcup_{n=k}^\infty E_n$ . Then, $\mu(F_k) \leq \frac{1}{2^{k-1}}$ and $\nu(F_k) \geq \epsilon$ . Using the continuity theorem, we get

$$\mu( \bigcap_{k=1}^\infty F_k ) = 0,$$

yet,

$$\nu(\bigcap_{k=1}^\infty F_k) = \lim_{k \rightarrow \infty} v(F_k) \geq \epsilon$$

Hence, $\nu$ is not absolutely continuous with respect to $\mu$ . $\square$

Corollary: If $f \in L^1(X, \mu)$ ($\mu$ a measure), then for all $\epsilon > 0$ , there exists a $\delta > 0$ such that $\vert \int_E f d\mu \vert < \epsilon$ whenever $\mu(E) < \delta$ .
Proof: Let $\nu(E) = \int_E f d\mu$ . Then $\nu \ll \mu$ and apply the proposition.

Theorem: (Radon-Nikodym): Let $\nu$ be a $\sigma$ -finite signed measure on $(X, \mathcal{S})$ and let $\mu$ be a $\sigma$ -finite positive measure on $(X, \mathcal{S})$ . If $\nu \ll \mu$ , then there exists an extended-real valued $\mu$ -integrable function $f: X \rightarrow \bar \mathbb{R}$ such that $\nu(E) = \int_E f d\mu$ . If $f_1: X \rightarrow \bar \mathbb{R}$ is $\mu$ -integrable and satisfies $\nu(E) = \int_E f_1 d\mu$ , then $f = f_1$ almost everywhere.

Theorem: (Lebesgue-Decomposition) Let $\nu$ be a $\sigma$ -finite signed measure on $(X, \mathcal{S})$ and $\mu$ be a $\sigma$ -finite positive measure on $(X, \mathcal{S})$ . Then there exists unique $\sigma$ -finite signed measures $\nu_0$ and $\nu_1$ on $(X, \mathcal{S})$ such that $\nu = \nu_0 + \nu_1$ , $\nu_0 \perp \mu$ and $\nu_1 \ll \mu$ .