Differentiating Measures on $\mathbb{R}^n$

Let $m$ be Lebesgue measurable on $\mathbb{R}^k$ . An integrable $f$ is with respect to $m$ .

Definition 3.1   Shrinks Nicely:
Fix $x \in \mathbb{R}^k$ . A sequence of Borel sets $\{ E_i \}_{i=1}^\infty$ is said to shrink nicely to $x$ if $\exists \alpha > 0$ and $\exists r_i > 0$ such that

(a)

$E_i \subset B(x, r_i)$

(b)

$m(E_i) > \alpha m(B(x, r_i))$

(c)

$r_i \rightarrow 0$

Remarks:

1. Notice that we do not require $x \in E_i$ .
2. Some authors use an index $r \in (0, \infty)$ and define $E_r$ shrunking nicely to $x$ .
   

Definition 3.2   Let $\mu$ be a complex Borel measure on $\mathbb{R}^k$ and let $x \in \mathbb{R}^k$ . If $\lambda \in \mathcal{C}$ and

$$\lim_{i \rightarrow \infty} \frac{\mu(E_i)}{m(E_i)} = \lambda$$

for every sequence $\{ E_i \}_{i=1}^\infty$ of Borel sets shrinking nicely to $x$ , we say $\lambda$ is the derivative of $\mu$ at $x$ and write $(D \mu )(x) = \lambda$ .

Lemma 3.1   Covering Lemma:
Let $\mathcal{C}$ be a nonempty collection of open balls in $\mathbb{R}^k$ and let $U = \displaystyle \bigcup_{B \in \mathcal{C}} B$ . If $c < m(U)$ , there exists disjoint balls $B_1, \ldots B_n \in \mathcal{C}$ such that $\sum_{i=1}^n m(B_i) > 3^{-k} c$ .

Proof. First recall that the Lebesgue measure is regular in the sense that given a measurable set $A$ ,

$$m(A) = \sup \{ m(K) : K$$    is compact $$, K \subset A \}.$$

Let $K \subset U$ be a compact set such that $m(K) > c$ . Finitely many of the balls in $\mathcal{C}$ cover $K$ , say $A_1, \ldots, A_p \in \mathcal{C}$ and $\bigcup_1^p A_i \supset K$ . We may assume that $radius(A_1) \geq \cdots \geq radius(A_p)$ .
Let $B_1 = A_1$ . Discard all the $A_i$ with $i > 1$ such that $A_i \cap B_1 \neq \emptyset$ . The first remaining $A_{i_2}$ is $B_2$ . Define $B_1, \ldots, B_n$ in this manner. It is obvious from this construction that all the $B_i$ 's are disjoint.
Let $A_j$ be one of the discarded balls. Then $A_j$ meets at least one of the $B_1, \ldots B_n$ . Let $i$ be the smallest integer such that $B_i \cap A_j \neq \emptyset$ . We know $radius(B_i) \geq radius(A_j)$ , so $A_j \subset \tilde B_i$ where $\tilde B_i$ is concentric with $B_i$ and $radius(\tilde B_i) = 3 radius(B_i)$ . Hence, $K \subset \bigcup_1^p A_j \subset \bigcup_{j=1}^n \tilde B_j$ , and

$$c \leq \sum_{j=1}^n m(\tilde B_j) = 3^k \sum_1^n m(B_j)$$

Rearranging, we get

$$\sum_{i=1}^n m(B_i) = 3^{-k} c.$$

$\qedsymbol$

Definition 3.3   Locally Integrable: If $f : \mathbb{R}^k \rightarrow \mathcal{C}$ is measurable, we say $f$ is locally integrable if for all bounded measurable sets $E \subset \mathbb{R}^k$ , $\int_E \vert f(x) \vert dm < \infty$ . Write $L^1_{loc}(\mathbb{R}^k)$ for the collection of locally integrable functions

Definition 3.4   For $f \in L^1_{loc}(\mathbb{R}^k)$ and $x \in \mathbb{R}^k$ let

$$(A_r f)(x) := \frac1{m(B(x,r))} \int_{B(x,r)} f dm.$$

Lemma 3.2   If $f \in L^1_{loc}(\mathbb{R}^k)$ and $x \in \mathbb{R}^k$ , then the map $G(x,r) = (A_r f)(x)$ is jointly continuous in $x$ and $r$ . (i.e., $G:\mathbb{R}^k \times (0,\infty) \rightarrow \mathbb{C}$ is continuous).

Proof. Let $S(x,r)$ be the sphere of radius $r$ centered at $x$ (the edge of the ball $B(x,r)$ . Now, $m(B(x,r)) = c m(B(0,1))$ for some $c$ and $m(S(x,r)) = 0$ . Fix $r_0 \in (0, \infty)$ and $x_0 \in \mathbb{R}^k$ . As $r \rightarrow r_0$ and $x \rightarrow x_0$ ,

$$\chi_{B(x,r)} \rightarrow \chi_{B(x_0, r_0)}$$

pointwise on $\mathbb{R}^k \setminus S(x_0, r_0)$ . So,

$$\chi_{B(x,r)} \rightarrow \chi_{B(x_0, r_0)}$$    almost everywhere $$$$

and

$$\vert \chi_{B(x,r)} \vert \leq \chi_{B(x_0, r_0 + 1)}$$

if $r < r_0 + \frac12$ and $\vert x - x_0\vert < \frac12$ . If $r_n \rightarrow r_0$ and $x_n \rightarrow x_0$ , the Dominated Convergence Theorem says

$$\int_{B(x_n,r_n)} f dm \rightarrow \int_{B(x_0, r_0)} f dm$$

Hence the map

$$(x,r) \rightarrow \int_{B(x,r)} f dm$$

is continuous. Then so is

$$(A_r f)(x) = \frac{1}{c(x,r) m(B(0,1))} \int_{B(x,r)} f dm$$

Therefore, $(A_r f)(x) = G(x,r)$ is continuous. $\qedsymbol$

Definition 3.5   Hardy-Littlewood maximal function:
For $f \in L_{loc}(\mathbb{R}^k)$ , the Hardy-Littlewood maximal function for $f$ is

$$(Hf)(x) = \sup_{r > 0} (A_r \vert f\vert)(x)$$

Claim: $Hf$ is measurable.
By the above lemma, for $r > 0$ ,

$$(A_r \vert f\vert)^{-1}(a, \infty)$$

is an open set for $a \in \mathbb{R}^k$ , as $A$ is continuous. Therefore,

$$\bigcup_{r > 0} (A_r \vert f\vert)^{-1}(a,\infty)$$

is open (an arbitrary union of open sets is open). But,

$$(Hf)^{-1}(a, \infty) = \bigcup_{r > 0}( A_r\vert f\vert)^{-1}(a,\infty)$$

If $x \in \bigcup_{r > 0} (A_r \vert f\vert)^{-1}(a,\infty)$ , then for some $r > 0$ ,

$$(A_r\vert f\vert)(x) \in (a, \infty)$$

So, $\sup_{r > 0} (A_r\vert f\vert)(x) > a$ . Thus, $x \in (Hf)^{-1}(a, \infty)$ . If $(Hf)(x) \in (a,\infty)$ , then for some $r > 0$ , $A_r\vert f\vert(x) > a$ , so $x \in \bigcup_{r > 0} (A_r \vert f\vert)^{-1}(a,\infty)$ . We conclude $(Hf)^{-1}(a,\infty)$ is open for all $a \in \mathbb{R}^k$ , and $Hf$ is measurable.

Theorem 6   Hardy-Littlewood Maximal Theorem:
There exists a constant $c > 0$ such that for all $f \in L^1_{loc}(\mathbb{R}^k)$ and all $\alpha > 0$ ,

$$m(\{ x : (Hf)(x) > \alpha \}) < \frac{C}{\alpha} \int_{\mathbb{R}^k} \vert f \vert dm$$

In fact, $C \leq 3^k$ .

Proof. Given $\alpha > 0$ and $f \in L_{loc}^1$ , let $E_\alpha = \{ x \in \mathbb{R}^k : (Hf)(x) > \alpha \}$ . For each $x \in E_\alpha$ , let $r_x > 0$ such that $A_{r_x} \vert f\vert(x) > \alpha$ . Then,

$$\bigcup_{x \in E_\alpha} B(x,r_x) \supset E_\alpha,$$

so by the covering lemma, if $c < m(E_\alpha)$ , then there exists $x_1, \ldots, x_n \in E_\alpha$ such that $B_j := B(x_j, r_{x_j})$ are a disjoint family and

$$\sum_{j=1}^n m(B_j) > 3^{-k} c$$

So,

$$c < 3^k \sum_{j=1}^n m(B_j) \leq 3^k \sum_{j=1}^n m(B_n) \frac{A_... ...j)}{\alpha} \leq \frac{3^k}{\alpha} \sum_{j=1}^n \int_{B_j} \vert f(y)\vert dy$$

$$\leq \frac{3^k}{\alpha} \int_{\mathbb{R}^k} \vert f(y)\vert dy.$$

By taking supremums over $c$ , we get

$$m(E_\alpha) \leq \frac{3^k}{\alpha} \int_{\mathbb{R}^k} \vert f\vert dm.$$

$\qedsymbol$

Recall: If $f : \mathbb{R}^k \rightarrow \mathbb{R}$ , then

$$\limsup_{y \rightarrow x} f(x) = \inf_{\delta > 0} \sup_{0 < \vert x - y \vert < \delta} f(y)$$

Fact: $$\lim_{y \rightarrow x} f(y) = c \equiv \limsup_{y \rightarrow x} \vert f(y) - c\vert = 0$$ .

Theorem 7   For $f \in L^1_{loc}(\mathbb{R}^k)$ ,

$$\lim_{r \rightarrow 0} (A_r f)(x) = f(x)$$

for almost every $x \in \mathbb{R}^k$ .

Proof. For any $N \in \mathbb{N}$ , we'll show that

$$\lim_{r \rightarrow 0} (A_r f)(x) = f(x)$$

for almost every $x \in \mathbb{R}^k$ with $\vert x\vert \in N$ .
Note that for $\vert x\vert < N$ and $r \leq 1$ , the values of $(A_r f)(x)$ only depend on the values of $f$ in $B(0, N+1)$ . So, we can replace $f$ with $f \chi_{B(0,N+1)}$ as necessary. Hence, we may assume that $f \in L^1(\mathbb{R}^k)$ .
So, WLOG, take $f \in L^1(\mathbb{R}^k)$ . Let $\epsilon > 0$ . Then, there exists a continuous function $g \in L^1(\mathbb{R}^k)$ such that

$$\int_{\mathbb{R}^k}\vert f(y) - g(y)\vert dm(y) < \epsilon$$

Then,

$$(A_r g)(x) - g(x) = \frac{1}{m(B(x,r))}\int_{B(x,r)} g(y) - g(x) dm(y)$$

Since $g$ is continuous, given $\delta > 0$ , there exists some $r > 0$ such that

$$\vert g(y) - g(x) \vert < \delta$$

if $y \in B(x,r)$ . So,

$$\vert (A_r g)(x) - g(x) \vert < \delta$$

Thus,

$$\lim_{ r \rightarrow 0} (A_r g)(x) = g(x)$$

We have

$$\limsup_{r \rightarrow 0} \vert (A_r f)(x) - f(x) \vert = \limsup... ...ightarrow 0} \vert(A_r( f - g))(x) + [ (A_r g)(x) - g(x) ] + g(x) - f(x) \vert$$

$$\leq \limsup_{r \rightarrow 0} \vert A_r (f - g)(x) \vert+ \limsu... ...rt (A_r g)(x) - g(x) \vert + \limsup_{r \rightarrow 0} \vert g(x) - f(x) \vert$$

$$\leq (H(f-g))(x) + 0 + \limsup_{r \rightarrow 0} \vert g(x) - f(x)\vert$$

Thus, we have the following inequality

$$\limsup_{r \rightarrow 0} \vert (A_r f)(x) - f(x) \vert \leq H(f-g)(x) + \limsup_{r \rightarrow 0} \vert g(x) - f(x)\vert$$ (1)

For $\alpha > 0$ , let

$$E_\alpha := \{ x : \limsup_{r \rightarrow 0} \vert (A_r f)(x) - f(x) \vert > \alpha \}$$

$$F_\alpha := \{ x : \vert f(x) - g(x) \vert > \alpha \}$$

Then, we claim $E_\alpha = F_{\frac\alpha2} \cup \{ x : H(f - g)(x) > \frac{\alpha}{2} \}$ . This follows directly from the above inequality (1). So,

$$\epsilon > \int_{\mathbb{R}^k} \vert f - g \vert dm \geq \int_{F_{\frac\alpha2}} \vert f - g\vert dm \geq \frac{\alpha}{2} m(F_{\frac{\alpha}{2}})$$

$$m(F_{\frac\alpha2}) < \frac{2 \epsilon }{\alpha}.$$

By the Hardy-Littlewood Maximal Theorem,

$$m ( \{ x : H(f - g)(x) > \frac{\alpha}{2} \}) < \frac{2C}{\alpha} \int_{\mathbb{R}^k} \vert f - g \vert dm$$

So,

$$m(E_\alpha) < \frac{ 2 \epsilon }{\alpha} + \frac{2C}{\alpha} \in... ...} \vert f - g\vert dm = \frac{2\epsilon }{\alpha} + \frac{2C\epsilon }{\alpha}$$

As $\epsilon$ is arbitrary, $m(E_\alpha) = 0$ . This is true for all $\alpha$ , so

$$m( \bigcup_{n=1}^\infty E_{\frac{1}{n}}) = 0$$

Hence, if $x \notin \bigcup_{n=1}^\infty E_{\frac{1}{n}}$ , then

$$\limsup_{r \rightarrow 0} \vert (A_r f)(x) - f(x) \vert = 0$$

Hence,

$$\lim_{r \rightarrow 0} (A_r f)(x) = f(x)$$

for almost all $x \in \mathbb{R}^k$ . $\qedsymbol$

Remark: The conclusion of the theorem is equivalent to

$$\lim_{r \rightarrow 0} \frac{1}{m(B(x,r))} \int_{B(x,r)} f(y) - f(x) dy = 0.$$

In fact, we can do better than this; the conclusion is still valid if we use absolute values under the integrand.

Definition 3.6   Lebesgue Set:
Given $f \in L^1_{loc}(\mathbb{R}^k)$ , the Lebesgue set for $f$ is the set

$$L_f := \{ x \in \mathbb{R}^k : \lim_{r \rightarrow 0} \frac{1}{m(B(x,r))} \int_{B(x,r)} \vert f(y) - f(x)\vert dy= 0 \}$$

Note that for $x \in L_f$ , $\lim_{r \rightarrow 0} (A_r f)(x) = f(x)$ .

Theorem 8   For $f \in L_{loc}^1 (\mathbb{R}^k)$ , $m(\mathbb{R}^k \setminus L_f) = 0$ .

Proof. For $\lambda \in \mathcal{C}$ , put $g_\lambda(x) = \vert f(x) - \lambda\vert$ . Notice that $g_\lambda \in L^1_{loc}$ , so the previous theorem applies to $g_\lambda$ , i.e., the set

$$E_\lambda := \{ x \in \mathbb{R}^k : \lim_{r \rightarrow 0} (A_r g_\lambda)(x) \neq g_\lambda(x) \}$$

has measure 0. Let $\{ \lambda_i \}_{i=1}^\infty$ be a countable dense set in $\mathcal{C}$ and put

$$E = \bigcup_{i=1}^\infty E_{\lambda_i}$$

It is clear that $m(E) = 0$ .

Friday, 1-28-2005

Let $\epsilon > 0$ , and $x \notin E$ . Pick $i \in \mathbb{N}$ such that $\vert \lambda_i - f(x) \vert < \epsilon$ . For any $y \in \mathbb{R}^k$ ,

$$\vert f(y) - f(x) \vert < \vert f(y) - \lambda_i \vert + \epsilon .$$

Then,

$$\limsup_{r \rightarrow 0} \frac{1}{m(B_r(x))} \int_{B_r(x)} \vert... ...0} \frac{1}{m(B_r(x))} \int_{B_r(x)} \vert f(y) - \lambda_i\vert dy + \epsilon$$

As $x \notin E$ ,

$$(A_r g_{\lambda_i})(x) \rightarrow g_{\lambda_i}(x)$$

I.e.,

$$\lim_{r \rightarrow 0} \frac{1}{m(B_r(x))} \int_{B_r(x)} \vert f(y) - \lambda_i\vert dy = \vert f(x) - \lambda_i \vert$$

Therefore,

$$\limsup_{r \rightarrow 0} \frac{1}{m(B_r(x))} \int_{B_r(x)} \vert f(y) - f(x)\vert dy < \epsilon + \epsilon$$

Thus, for $x \in E$ ,

$$\lim_{r \rightarrow 0} \frac{1}{B_r(x)} \int_{B_r(x)} \vert f(y) - f(x) \vert dy = 0$$

Hence, $x \notin E$ implies $x \notin L_f$ . Turning this expression around,

$$m(\mathbb{R}^k \setminus L_f) \leq m(E)$$

$\qedsymbol$

Theorem 9   Lebesgue Differentiation Theorem:
Suppose $f \in L^1_{loc}$ . Then for every $x \in L_f$ , we have

$$\lim_{r \rightarrow 0} \frac{1}{m(E_r)} \int_{E_r} \vert f(y) - f(x)\vert dy = 0$$

for every family $\{ E_r \}_{r \geq 0}$ which shrinks nicely to $x$ .

Proof. For some $\alpha > 0$ , we know that $E_r \subset B_r(x)$ and $m(E_r) > \alpha m(B_r(x))$ . So,

$$\frac{1}{m(E_r)} \int_{E_r} \vert f(y) - f(x)\vert dy \leq \frac{1}{m(E_r)} \int_{B_r(x)} \vert f(y) - f(x)\vert dy$$

$$\leq \frac{1}{\alpha m(B_r(x))} \int_{B_r(x)} \vert f(y) - f(x)\vert dy.$$

Now, apply the previous theorem. So, if $x \in L_f$ , the limit is 0.
$\qedsymbol$

Theorem 10   Suppose $\mu$ is a complex Borel measure on $\mathbb{R}^k$ and $\mu \ll m$ . Then, $(D \mu)(x)$ exists for almost all $x \in \mathbb{R}^k$ and if $f = \left[ \frac{d\mu}{dm} \right]$ , then $(D \mu)(x) = f(x)$ almost everywhere.

Proof. We know that $f \in L^1_{loc}(m)$ , so by the the Lebesgue Differentiation Theorem, if $x \in L_f$ ,

$$\lim_{r \rightarrow 0} \frac{1}{m(Er)} \int_{E_r} f(y) dm(y) = f(x)$$

i.e.,

$$\frac{\mu(E_r)}{m(E_r)} \rightarrow f(x)$$    for almost all $$x$$

$\qedsymbol$

Pseudo-Example
Suppose that for $f \in L^1(R)$ , we can write

$$F(x) = \int_{-\infty}^x f(t) dt.$$

By the theorem, $F'(x) = f(x)$ for almost all $x$ . Note that we have no need for continuity of $f$ . Also, $F$ is continuous almost everywhere.
For $f \in L^1_{loc}$ , define

$$F(x) = \int_0^x f(t) dt$$

and we can get

$$F'(x) = f(x),$$    almost everywhere $$$$

From a previous homework, we've seen the function

$$f(x) = \int_1^\infty \frac{g(x - r_i)}{2^{r_i}}$$

$$g(x) = \frac{1}{\sqrt{x}} \chi_{(0,1)}$$

where $\{r_i\}$ is a countable dense subset of the real line. Then, we get that the integral of $f$ is continuous almost everywhere.

Suppose we have

$$d\nu = d\lambda + fdm$$

where $\lambda \perp m$ (we have used the Lebesgue Decomposition) - how would we start with a general measure and find these? We'll show that $(D \mu)$ exists almost everywhere and is equal to $f$ .

Definition 3.7   A positive Borel measure $\mu$ on $\mathbb{R}^k$ is regular if

(a)

For all compact sets $K \subset \mathbb{R}^k$ , $m(K) < \infty$ .

(b)

For all $E \subset \mathbb{R}^k$ , $E$ Borel,

$$\mu(E) = \inf \{ \mu(U) : U$$    is open $$, U \supset E \}.$$

A complex or signed measure is regular if its total variation is.

Observation: If $f \geq 0$ is measurable, then $d\mu = f dm$ is regular if and only if $f \in L_{loc}^1 (\mathbb{R}^k)$ .

Proof. If $\mu$ is regular, then for any bounded set $E$ ,

$$\int_E f dm \leq \int_{\bar E} f dm = \mu(\bar E) < \infty$$

Thus, $f \in L^1_{loc}(\mathbb{R}^k)$ .
If $f \in L^1_{loc}$ , then (a) in the definition of regular is obvious. We leave part (b) as an exercise.

Monday, 1-31-2005

Let $E \subset \mathbb{R}$ be Borel and bounded. Let $\epsilon > 0$ , let $B \subset R$ be a bounded open set. Since $f \in L^1_{loc}$ , there exists a $\delta > 0$ such that whenever $F \subset B$ and $m(F) < \delta$ then

$$\int_F f dm < \epsilon .$$

Find an open set $U$ such that $U \supset E$ and $m(U \setminus E) < \delta$ . We can do this because last semester we showed that the Lebesgue measure was regular. Then,

$$m(B \cap U) = \int_{U \cap B} f dm = \int_E f dm + \int _{(U \cap B) \setminus E} f dm$$

$$< \mu(E) + \epsilon$$

Therefore, (b) holds for $E$ bounded.
If $E$ isn't bounded ($E$ Borel), cover it with bounded sets, and approximate each of these bounded sets by $2^{-n} \epsilon$ (use standard techniques). $\qedsymbol$

Example: Suppose $\mu$ is a regular (signed / complex) Borel measure on a $\sigma$ -finite measure space. (Note that $\sigma$ -finite is also implied by the regularity conditions.)
Let

$$d\mu = d\lambda + f dm$$

be the Lebesgue Decomposition of $\mu$ with respect to $m$ , where $\lambda \perp m$ .

Claim: $d\lambda$ and $f dm$ are regular.
To see this, we first observe

$$d\vert\mu\vert = d\vert\lambda\vert + \vert f\vert dm$$

Proof. $\lambda \perp m$ , so $\vert\lambda\vert \perp m$ . Find Borel sets $A, B \subset \mathbb{R}^k$ such that $A \cup B = \mathbb{R}^k$ , $\vert\lambda\vert(B) = m(A) = 0$ , and $A \cap B = \emptyset$ . Let $g$ be such that $d \lambda = g d\vert\lambda\vert$ , where $\vert g\vert = 1$ .
For $E \subset \mathbb{R}^k$ , $E$ Borel,

$$\mu(E) = \int_E g d\vert\lambda\vert + \int f dm$$

$$= \int_E g \chi_A d\vert\lambda\vert + \int_E f \chi_B dm$$

$$= \int_E (g \chi_A + f \chi_B) d\vert\lambda\vert + \int_E (g \chi_A + f\chi_B) dm$$

$$= \int_E (g \chi_A + f \chi_B)d(\vert\lambda\vert + m) = \mu(E)$$

Also, $\mu \ll (\vert\lambda\vert + m)$ . Hence,

$$\vert\mu\vert(E) = \int_E \vert g \chi_A + f \chi_B \vert d(\vert\lambda\vert + m)$$

As $A \cap B = \emptyset$ ,

$$= \chi_E \chi_A + \vert f\vert \chi_B d(\vert\lambda\vert + m)$$

$$= \int_E d\vert\lambda\vert + \int_E \vert f\vert dm$$

so, $d\vert\mu\vert = d\vert\lambda\vert + \vert f\vert dm$ .
Since $\vert\mu\vert$ is regular, $\mu$ is regular. So, if $K$ is compact, then

$$\int_K \vert f\vert dm \leq \vert\mu\vert (K) < \infty$$

So, $f \in L^1_{loc}$ . Hence, $f dm$ is regular by our previous work.

To see $\vert\lambda\vert$ is regular, let $E$ be Borel and find $U$ such that $U \supset E$ is open and $\vert\mu\vert(U \setminus E) < \epsilon$ . Then, $\vert\lambda\vert(U \setminus E) < \epsilon$ , so

$$\lambda\vert(U) = \vert\lambda\vert(E) + \vert\lambda\vert(U \setminus E) < \vert\lambda\vert(E) + \epsilon$$

$\qedsymbol$

Theorem 11   Let $\lambda$ be a regular complex or signed Borel measure such that $\lambda \perp m$ . Then for $m$ -almost all $x \in \mathbb{R}^k$ ,

$$\lim_{r \rightarrow 0} \frac{\lambda(E_r)}{m(E_r)} = 0$$

for every family $\{ E_r \}_{r \geq 0}$ of Borel sets which shrink nicely to $x$ .

Proof. Without loss of generality, we may assume that $\lambda$ is a positive regular Borel measure and that $E_r = B(r,x)$ .
Let $\lambda \perp m$ . Then, there exists a Borel set $A \subset \mathbb{R}^k$ such that $\lambda(A) = m(A^c) = 0$ . Put

$$F_n = \{ x \in A : \limsup_{r \rightarrow 0} \frac{\lambda(B(r,x))}{m(B(r,x))} > \frac1n \}$$

We'll prove that $m(F_n) = 0$ for each $n$ . Hence the theorem follows: for if

$$x \notin \bigcup_{n=1}^\infty F_n \cup A^c$$

then $$\limsup_{r \rightarrow 0} \frac{\lambda(B(r,x)}{m(B(r,x)} = 0$$ , as desired.
Fix $n \in \mathbb{N}$ . By the regularity of $\lambda$ , given $\epsilon > 0$ , there exists an open set $U_\epsilon$ such that $U_\epsilon \supset A$ and $\lambda(U_\epsilon ) < \epsilon$ . By the definition of $F_n$ , given $x \in F_n$ , there exists a $r_x > 0$ such that $B(r_x,x) \subset U_\epsilon$ and

$$\frac{\lambda(B(r,x))}{m(B(r,x))} > \frac1n$$

Let $V_\epsilon = \bigcup_{x \in F_n} B(r_x, x)$ . Let $0 < c < m(V_\epsilon$ . By the covering lemma, there exists $x_1, \ldots, x_p \in F_n$ such that $B(r_{x_1},x_1), \ldots, B(r_{x_n},x_n)$ are disjoint and

$$c < 3^k \sum_{j=1}^p m(B(r_{x_j},x_j)) < 3^k n \sum_{j=1}^p \lambda(B(r_{x_j}, x_j)).$$

Because the balls are disjoint and $\lambda$ is a measure,

$$= 3^k n \lambda (\bigcup_{j=1}^p B(r_{x_j}, x_j) ) \leq 3^k n \lambda(V_\epsilon )$$

By our clever construction, we get

$$\leq 3^k n \lambda(U_\epsilon ) < 3^k n \epsilon .$$

Hence,

$$m(V_\epsilon ) \leq 3^k n \epsilon .$$

$$m(F_n) \leq 3^k n \epsilon$$

As $\epsilon$ is arbitrary, we get $m(F_n) = 0$ . $\qedsymbol$

Combining this result with the Lebesgue Differentiation Theorem, we immediately see the following:

Theorem 12   Let $\mu$ be a regular signed or complex Borel measure on $\mathbb{R}^k$ , and let $d\mu = d\lambda + fdm$ be its Lebesgue decomposition with respect to m. Then, for $m$ -almost every $x \in \mathbb{R}^k$ , $$\lim_{r \rightarrow 0} \frac{\mu(E_r)}{m(E)} = f(x)$$ for every family $\{ E_r \}_{r \geq 0}$ shrinking nicely to $x$ .

Example:
Let $F(x)$ be the Cantor function $F:[0,1] \rightarrow [0,1]$ . We can extend $F$ to a function $F: \mathbb{R}\rightarrow [0,1]$ .
Define $\mu([a,b]) = F(b) - F(a)$ . Then this determines a unique Borel measure on $\mathbb{R}$ such that $F(x) = \int_0^x d\mu$ . Since $F'(x) = 0$ for almost every $x$ , $\mu \perp m$ .

Application:

Theorem 13   Let $F: \mathbb{R}\rightarrow \mathbb{R}$ be non-decreasing and let

$$G(x) = F(x^+)$$

Then, $F$ is continuous, except at countably many points. Further, for almost all $x \in \mathbb{R}$ , $F$ and $G$ are differentiable and their derivatives are equal almost everywhere.

Proof. Since $F$ is nondecreasing, for all $x \in \mathbb{R}$

$$F(x^-) \leq F(x) \leq F(x^+)$$

Hence the open intervals, $(F(x^-), F(x^+))$ are disjoint and if $\vert x\vert < N \in \mathbb{N}$ , we have $(F(x^-), F(x^+)) \subset (F(-N), F(N))$ . So, the sum

$$\sum_{\vert x\vert < N} (F(x^+) - F(x^-)) \leq F(N) - F(-N).$$

Hence except for at most countably many $x$ , $F(x^+) = F(x^-)$ ; i.e., $F$ is continuous on $(-N, N)$ except at countably many points. This gives the first statement.

We have $G(x) = F(x)$ whenever $F$ is continuous at $x$ . Observe that there exists a regular Borel measure $\mu_G$ such that $\mu_G((a,b]) = G(b) - G(a)$ .

$$G(x + h) - G(x) = \begin{cases}\mu_G( (x, x+h] ), & h > 0 \ \mu_G( (x+h, x]), & h < 0 \end{cases}$$

Intervals such as $(x - r, x]$ and $(x, x+r]$ shrink nicely to $x$ . Since $\mu_G$ is regular, $D\mu_G$ exists for almost every $x \in \mathbb{R}$ . Hence

$$\lim_{h ra 0^+} \frac{\mu_G(x,x+h]}{h}$$

and

$$\lim_{h \rightarrow 0^-} \frac{\mu_G(x+h, x]}{h}$$

exist for almost every $x$ . Hence, $G$ is differentiable for almost all $x$ . Next, we need to prove that $F'(x) = G'(x)$ almost everywhere. Define

$$H(x) = G(x) - F(x);$$

from the definition of $G$ , $H$ is nonnegative. We want to show $H$ is diff. almost everywhere and $H'(x) = 0$ almost everywhere. We have

$$F(x^-) \leq F(x) \leq F(x^+) = G(x)$$

Therefore,

$$H(x) \leq F(x^+) - F(x^-).$$

We get, for $N \in \mathbb{N}$ ,

$$\sum_{\vert x\vert < N} H(x) \leq \sum_{\vert x\vert < N} F(x^+) - F(x^-) < F(N) - F(-N) < \infty$$

So, $H(x)$ has to be zero except on countably may points. Let $\{ x_j \}_{j=1}^\infty$ be an enumeration of $B = \{ x : H(x) \neq 0 \}$ .
Let $\delta_{x_j}$ be the point mass at $x_j$ and put

$$\mu = \sum_{j=1}^\infty H(x_j) \delta_{x_j}.$$

Then, $\mu(K) < \infty$ if $K \subset \mathbb{R}$ is compact (by the previous argument, as $K$ has to be bounded). Also if $E$ is Borel, it isn't hard to prove that $\mu(E) = \inf\{ \mu(U) : U$ is open $\}$ . Hence, $\mu$ is regular.
Note that

$$\mu(B^c) = 0 = m(B)$$

and $\mu \perp m$ . Hence, $(D\mu)(x) = 0$ for almost all $x$ . Note,

$$\left\vert \frac{H(x + h) - H(x)}{h} \right\vert \leq \frac{H(x + h) + H(x)}{\vert h\vert}$$

$$\leq \frac{\mu(x - 2h, x+2h)}{\vert h\vert} = 4\frac{\mu(x - 2h, x+2h)}{4\vert h\vert}$$

Hence, since $(D\mu)(x) = 0$ for almost all $x$ , we get

$$H'(x) = \lim_{h \rightarrow 0} 4\frac{\mu(x - 2h, x+2h)}{4\vert h\vert} = 0$$

for almost all $x$ .
$\qedsymbol$

Problem: Which functions are of the form

$$F(x) = \int_{-\infty}^x d\mu$$

for a Borel complex measure?
We already know this answer, in part. For every increasing real valued function that is continuous on the right, we can get a Baire measure (look at last semester's notes).

If $F$ is continuously differentiable (and was nice enough) and

$$F(x) = \int_{-\infty}^x F'(t) dt,$$

we'd expect

$$\int_{-\infty}^x \vert F'(t)\vert dt < \infty$$

for all $x$ .

Given a function $F: \mathbb{R}\rightarrow \mathbb{C}$ , we define the total variation of $F$ over $\mathbb{R}$ by

$$\sup \{ \sum_{j=0}^{n+1} \vert F(x_{j+1}) - F(x_{j})\vert : x_0 \leq x_1 \leq \cdots \leq n$$    is a finite partition of $\mathbb{R}$ $$\}$$

More generally, if $[a,b]$ is an interval, then

$$T_a^b(f) = \sup\{ \sum_{j=1}^n \vert F(x_j) - F(x_{j-1})\vert : \{x_j\}$$    partion $$[a,b] \}$$

is the total variation of $F$ from $a$ to $b$ .

If $a = - \infty$ , write $V_f(x) = T_{-\infty}^x(f)$ . $V_f(x)$ is the total variation function of $f$

Definition 3.8   Bounded Variation:
If $V_f(+\infty) < \infty$ , then we say $f$ is of bounded variation on $\mathbb{R}$ . If $T_a^b(f) < \infty$ , we say $f$ is of bounded variation on $[a,b]$ .

Monday, 2-7-2005

Definition 3.9   $BV[a,b]$ :
$BV[a,b]$ is the set of all functions of bounded variation on $[a,b]$ .

Some Examples:

• A bounded increasing function has finite total variation.
  
• The set of all functions of bounded variation is a linear space.
  
• A function with bounded derivative is of bounded variation on a finite interval.
• $\sin x$ is an example of a function with bounded variation on a finite interval, but not of bounded variation on the real line.

• $$f(x) = \begin{cases}x \sin \frac1x, & x \neq 0 \ x = 0, & x = 0. \end{cases}$$
  If $0 \notin [a,b]$ , then $f \in BV[a,b]$ . If $0 \in [a,b]$ , $f \notin BV[a,b]$ .
  

Remark:
Notice that if $f$ has bounded variation , then (for $x > y$ )

$$V_f(x) - V_f(y) = T_y^x(f).$$

Proposition 3.1   Suppose $F: \mathbb{R}\rightarrow \mathbb{R}$ and $F \in BV$ . Then $V_F + F$ and $V_F - F$ are both increasing functions. Thus in particular,

$$F = \frac{(V_F + F)}2 - \frac{(V_F - F)}2.$$

In other words, all functions of bounded variation is the difference of two increasing functions.

Remark: The proposition is valid if we use $F:[a,b] \rightarrow \mathbb{R}$ instead of $F: \mathbb{R}\rightarrow \mathbb{R}$ .

Proof. Let $x < y$ and $\epsilon > 0$ . Find a partition $\{ x_0, \ldots, x_n \}$ such that

$$\sum_{j=1}^n \vert F(x_j) - F(x_{j-1})\vert \geq V_F(x) - \epsilon$$

Then,

$$(\sum_{j=1}^n\vert F(x_j) - F(x_{j-1})\vert) + \vert F(y) - F(x)\vert$$

approximates $V_F(y)$ and $F(y) = (F(y) - F(x)) + F(x)$ .
So,

$$V_F(y) + F(y) \geq \sum_1^n \vert F(x_j) - F(x_{j-1})\vert + \vert F(y) - F(x)\vert + [ F(y) - F(x) ] + F(x).$$

$$\geq V_F(x) - \epsilon + F(x)$$

So,

$$V_F(y) + F(y) \geq V_F(x) + F(x),$$

and $V_F + F$ is an increasing function. Changing the appropriate plus signs to minus signs, we get that $V_F - F$ is also increasing.
$\qedsymbol$

Combining this with our work on increasing functions leads to this theorem:

Theorem 14   The following are true:

1. $F \in BV \Leftrightarrow Re(F) \in BV$ and $Im(F) \in BV$
2. If $F: \mathbb{R}\rightarrow \mathbb{R}$ , then $F \in BV \Leftrightarrow F$ is a difference of two bounded increasing functions.
3. If $F \in BV$ , then the following limits exist for every $x \in \mathbb{R}$ :
   1. $$\lim_{y \rightarrow x^+} F(y)$$
   2. $$\lim_{y \rightarrow x^-} F(y)$$
   3. $$\lim_{y \rightarrow \pm \infty} F(y)$$
4. If $F \in BV$ , the set of discontinuities of $F$ is countable.
5. For $F \in BV$ , let $G(x) = F(x^+)$ . Then $F', G'$ exists almost everywhere and $F' = G'$ almost everywhere.

Proof. .

1. Use
   $$Re(F) = \frac{F + \bar F}2, Im(F) = \frac{F - \bar F}{2i}$$

2. We've done $\implies$ , and need to go the other way. However, this is obvious from the definition of bounded variation and the triangle inequality.
   
3. - (5) have been proved previously

$\qedsymbol$

Definition 3.10   If $F: \mathbb{R}\rightarrow \mathbb{R}$ is of bounded variation, we call the decomposition

$$F = \frac{V_F + F}2 - \frac{V_F - F}2$$

the Jordan decomposition of $F$ .

Definition 3.11   A function $F \in NBV$ (normalized bounded variation) if $F \in BV$ , $F(-\infty) = 0$ , and $F$ is right continuous.

We can always make a function of $BV$ into a function of $NBV$ . If $F \in BV$ , then $G(x) := F(x^+) - F(-\infty) \in NBV$ . Moreover, $F'(x) = G'(x)$ for almost all $x$ .

Lemma 3.3   If $F \in BV$ , then $V_F(-\infty) = 0$ and if $F$ is right continuous, then so is $V_F$ .

Proof. Wednesday, 2-9-2005

Let $\epsilon > 0$ , $x \in \mathbb{R}$ . Let $x_0 < x_1 < \cdots < x_n = x$ satisfy

$$\sum_{i=1}^n \vert F(x_j) - F(x_{j-1})\vert \geq V_F(x) - \epsilon$$

We have then,

$$T_{x_0}^x(F) = V_F(x) - V_F(x_0) = \sup\{ \sum_{i=1}^n \vert F(y_i) - F(y_{i-1})\vert : x_0 = y_0 < \cdots < y_n = x \}$$

Therefore,

$$V_F(x) - V_F(x_0) \geq \sum_{j=1}^n \vert F(x_j) - F(x_{j-1}) \vert \geq V_F(x) - \epsilon$$

and $V_F(x-0) < \epsilon$ . Hence, if $y < x_0$ , we get $V_F(y) < \epsilon$ , so $V_F(-\infty) = 0.$

Suppose that $F$ is right continuous. Fix $x \in \mathbb{R}$ , $\epsilon > 0$ , and put $\alpha = V_F(x^+) - V_F(x)$ . Find $\delta > 0$ such that $\vert F(x+h) - F(x)\vert < \epsilon$ and $V_F(x + h) - V_F(x^+) < \epsilon$ if $0 < h < \delta$ .

For $0 < h < \delta$ , find a partition $x = x_0 < \cdots < x_n = x + h$ such that

$$\sum_{j=1}^n \vert F(x_j) - F(x_{j-1})\vert \geq \frac34(V_F(x+h) - V_F(x)).$$

$$\geq \frac34 \alpha,$$    as $V_F(x + h) > V_F(x^+)$  $$$$

Now, partition the interval $x_0 = t_0 < \cdots < t_m = x_1$ . such that

$$\sum_{j=1}^m \vert F(t_j) - F(t_{j-1}) \vert \geq \frac34 \alpha.$$

We have

$$\alpha + \epsilon > V_F(x + h) - V_F(x) \geq \sum_{j=1}^m \vert F(t_j) - F(t_{j-1})\vert + \sum_{i=2}^n \vert F(x_i) - F(x_{i-1})\vert$$

$$\geq \frac34 \alpha - \frac34 \alpha - \vert F(x_1) - F(x)\vert \geq \frac32 - \epsilon$$

Manipulating this, we get

$$\alpha < 4 \epsilon$$

for all $\epsilon$ - and hence, $\alpha = 0$ . $\qedsymbol$

Theorem 15   If $\mu$ is a complex, regular Borel measure on $\mathbb{R}$ and if $F(x) = \mu(-\infty, x]$ , then $F \in NBV$ . Conversely, if $F \in NBV$ , then there exists a unique regular complex Borel measure $\mu_F$ such that $F(x) = \mu(-\infty, x]$ .

Proof. If $\mu$ is a regular Borel complex measure, write

$$\mu = \mu_1^+ - \mu_1^-) + i (\mu_2^+ - \mu_2^-)$$

where $\mu_1 = Re \mu$ , $\mu_2 = Im \mu$ , and $\mu_i^\pm$ are positive (finite) measures. For $j=1,2$ , let $F_j^\pm(x) = \mu_j^\pm (-\infty, x]$ . Then $F_j^\pm$ are increasing functions and the continuity theorem shows they are continuous on the right and $F_j^\pm(-\infty) = 0$ . We conclude that $F_j^\pm \in NBV.$ Hence, $F = F_1^+ - F_1^- + i(F_2^+ - F_2^-) \in NBV$ .

Conversely, write

$$F = Re F + i Im F$$

Since $Re F$ and $Im F$ are real-valued functions of bounded variation, we can write the Jordan Decomposition:

$$= \frac{V_{F_1} + F_1}2 - \frac{V_{F_1} - F_1}{2} + i \left( \frac{V_{F_2} + F_2}{2} - \frac{V_{F_2} - F_2}2 \right)$$

By the Lemma,

$$\frac{V_{F_1} \pm F_1}{2}, \frac{V_{F_2} \pm F_2}{2} \in NBV.$$

By last semester's work, there are unique regular and Borel measures $\mu_i^\pm$ such that

$$\mu_i^\pm(-\infty, x] = \frac{V_{F_i} \pm F_i}{2}(x).$$

So, if $\mu = (\mu_1^+ - \mu_1^-) + i(\mu_2^+ - \mu_2^-)$ , we have $\mu$ is regular, Borel, and $\mu(-\infty, x] = F(x)$ . $\qedsymbol$

Remark: It can be shown that for $F \in NBV$ , $\vert\mu_F\vert = \mu_{V_f}$ , and if $F$ is real, then the Jordan Decomposition of $\mu_F$ corresponds to $$\mu_{\frac12 (V_F \pm F)}$$ . This naturally leads to the question: Which $F$ satisfy $\mu_F \ll m$ ?

Proposition 3.2   If $F \in NBV$ , then $F' \in L^1(m)$ . Moreover, $\mu_F \perp m \Leftrightarrow F' = 0$ (Lebesgue) almost-everywhere. Finally, $\mu_F \ll m$ if and only if

$$F(x) = \int_{(-\infty,x]} F'(t) dt.$$

Proof. Observe that

$$F'(x) = \lim_{r \rightarrow 0} = \frac{\mu_F(E_r)}{m(E_r)}$$

where $E_r = (x,x+r]$ or $E_r = (x-r,x]$ . Write $\mu = \lambda + \nu$ where $\lambda \perp m$ and $\nu \ll m$ ; then, for almost every $x$ ,

$$D_{\mu_F}(x) = F'(x) = \left[ \frac{d\nu}{dm}\right](x)$$

Hence, $\mu_F \perp m \Leftrightarrow \nu = 0$ almost everwhere $\Leftrightarrow F'(x) = 0$ almost everywhere. On the other hand,

$$\mu_F \ll m \Leftrightarrow F(x) = \mu_F(-\infty,x] = \int_{-\infty}^x d\mu_F$$

$$= \int_{(-\infty,x)} F'(x) dm(x).$$

Finally, $F' \in L^1(m)$ because the Radon-Nikodym derivative is, and $F' = \left[ \frac{d\nu}{dm}\right]$ almost everywhere.
$\qedsymbol$

This proposition isn't very satisfactory because we must compute both the derivative and the integral of the derivative.

Definition 3.12   We say that $F: \mathbb{R}\rightarrow \mathbb{C}$ is an absolutely continuous function if for all $\epsilon > 0$ , there exists a $\delta > 0$ such that for every finite family $(a_1,b_1), \ldots, (a_n,b_n)$ of disjoint intervals with

$$\sum_{j=1}^n (b_j - a_j)< \delta$$

we have

$$\sum_{j=1}^n \vert F(b_j) - F(a_j)\vert < \epsilon .$$

We say that $F$ is absolutely continuous on the interval $[a,b]$ if the same condition holds for all disjoint intervals contained in $[a,b]$ .

Remarks:

1. If $F$ is absolutely continuous, then $F$ is uniformly continuous.
2. If $F$ is everywhere differentiable and $F'$ is bounded, then $F$ is absolutely continuous.

Proposition 3.3   If $F \in NBV$ , then $F$ is absolutely continuous if and only if $\mu_F \ll m$ .

Proof. Suppose $\mu_F \ll m$ . Then $F' \in L^1$ and

$$F(x) = \int_{-\infty}^x F'(t) dt$$

by the previous proposition. Given $\epsilon > 0$ , there exists a $\delta > 0$ such that

$$\int_E \vert F'(t)\vert dm(t) < \epsilon$$

whenever $m(E) < \delta$ . Taking $E$ to be a collection of disjoint intervals, we find that $F$ is absolutely continuous.

Conversely, suppose that $F$ is absolutely continuous. Let $E$ be a Borel set such that $m(E) = 0$ . Given $\epsilon > 0$ , we may find $\delta > 0$ which satisfies the absolutely continuous condition for $F$ .

Let $\{U_j\}_{j=1}^\infty$ be a sequence of open sets such that $U_1 \supset U_2 \supset \cdots \supset E$ , $m(U_j) < \delta$ , and

$$\lim_{j \rightarrow \infty} \mu_F(U_j) = \mu_F(E).$$

Every $U_j$ is a disjoint union of open intervals, so write

$$U_j = \bigcup_{k=1}^\infty (a^k_j, b_j^k).$$

Then for any $N \in \mathbb{N}$ ,

$$\sum_{k=1}^N (b_j^k - a_j^k) < \delta.$$

Hence,

$$\sum_{k=1}^N \vert F(b_j^k) - F(a_j^k) \vert < \epsilon .$$

In particular,

$$\sum_{k=1}^\infty \vert F(b_j^k) - F(a_j^k) \vert \leq \epsilon .$$

Therefore,

$$\left\vert \sum_{k=1}^\infty F(b_j^k) - F(a_j^k) \right\vert = \vert \mu_F(U_j) \vert \leq \epsilon .$$

Taking $j \rightarrow \infty$ , we get $\mu_F(E) = 0$ so that $\mu_F \ll m$ .
$\qedsymbol$

Remark: As an immediate corollary, we get the following.

Theorem 16   If $f \in L^1(m)$ , then $F(x) := \int_{-\infty}^x f(t) dt$ is in NBV and $F'(x) = f(x)$ for almost all $x$ . Conversely, if $F \in NBV$ and is absolutely continuous, then $F' \in L^1(m)$ and

$$F(x) = \int_{-\infty}^x F'(t) dt.$$

Lemma 3.4   If $F \in AC$ on $[a,b]$ , then $F \in BV[a,b]$ .

Proof. For you. $\qedsymbol$

Monday, 2-14-2005

Theorem 17   If $-\infty < a < b < \infty$ and $F:[a,b] \rightarrow \mathbb{C}$ , then the following are equivalent:

(a)

$F$ is absolutely continuous on $[a,b]$ .

(b)

$$F(x) - F(a) = \int_a^x f(t) dt$$ for some $f \in L^1([a,b],m)$

(c)

$F$ is differentiable for $m$ -almost every $x \in [a,b]$ , $F' \in L'([a,b],m)$ , and $$F(x) - F(a) = \int_a^x F'(t) dt$$ .

Proof. $(c) \implies (b)$ is trivial
$(b) \implies (a)$ Take $f = 0$ outside of $[a,b]$ . Then $f \in L^1(\mathbb{R})$ , and apply previous results.
$(a) \implies (c)$ . Put $G(x) = F(x) - F(a)$ . Then, $G(a) = 0$ and we extend $G$ to all of $\mathbb{R}$ by $G(t) = 0$ for $t < a$ , $G(t) = G(b)$ if $x > b$ . Then, $G \in NBV$ and is absolutely continuous so $G' \in L^1(m)$ and

$$G(x) = \int_{-\infty}^x G'(t) dm(t)$$

i.e.,

$$F(x) - F(a) = \int_a^x F'(t) dt.$$

$\qedsymbol$

Final Remarks:
Let $\mu$ be a Borel measure on $\mathbb{R}^k$ . $\mu$ is called discrete if there exists a countable set $\{ x_j \}_{j=1}^\infty \subset \mathbb{R}^k$ and $c_j \in \mathbb{C}$ such that

1. $$\sum_{j=1}^\infty \vert c_j\vert < \infty$$ and
2. $$\mu = \sum_{j=1}^\infty c_j \delta_{x_j}$$

We say that $\mu$ is continuous if $\mu(\{ x\}) = 0$ for all $x \in \mathbb{R}^k$ .

Observe:
If $\mu$ is a complex Borel measure, then $\mu = \mu_c + \mu_d$ where $\mu_d$ is discrete and $\mu_c$ is continuous. Write

$$E = \{ x : \mu(\{ x \}) \neq 0 \}.$$

If $F \subset E$ is countable, the series

$$\sum_{ x \in F} \mu(\{ x\})$$

converge absolutely. Therefore,

$$\{ x \in E : \vert\mu(\{x\})\vert > \frac1k \}$$    is finite $$$$

and as

$$E = \bigcup\{ x : \vert\mu(\{ x \} ) \vert . \frac1k \},$$

$E$ is countable. Now put $\mu_d(A) = \mu(A \cap E)$ and $\mu_c(A) = \mu(A \cap E^c)$ . Notice that $\mu_d \perp m$ . We can further decompose to get $\mu_c = \mu_{ac} + \mu_{sc}$ where $\mu_{sc} \perp m$ and $\mu_{ac} \ll m$ . Thus,

$$\mu = \mu_d + \mu_{sc} + \mu_{ac}$$

This decomposition is important in function theory (branch of complex analysis). For an example of a singular (with respect to $m$ ) and continuous function, look at the measure generated by the Cantor function.

If $F \in NBV$ , write $\mu_F$ for the associated complex measure. Then, we often write $\int g d\mu_F$ as $\int g dF$ .