Measure theory

Borel algebra

If A and B are any two sets, let $A - B$ be the set of all elements in A that are not in B (This does not imply $B \subset A$ ).

A family $R$ of sets is called a ring if $A, B \in R$ implies

$A \cup B \in R$ ,

. $A - B \in R$ .

Notice that the conditions imply $A \cap B \in R$ if its a ring.

A ring is called a $σ$ ring if

$\cup_{n = 1}^{\infty} A_{n} \in R$ ,

for any countable sequence of sets $A_{n} \in R$ .

If this condition is fulfilled, then

$\cap_{n = 1}^{\infty} A_{n} \in R$

also holds.

A set function $ϕ$ defined on $R$ assigns to every element of a number $ϕ (A)$ of the extended number system. $ϕ$ is additive if $A \cap B = 0$ implies

$ϕ (A \cup B) = ϕ (A) + ϕ (B)$ ,

and it is countably additive if $A_{i} \cap A_{j} = 0$ ( $i \neq j$ ) implies

$ϕ (\cap_{n = 1}^{\infty} A_{n}) = \sum_{n = 1}^{\infty} ϕ (A_{n}) .$

Notice that nonnegative additive set functions are monotonic by inclusion. In addition, for additive set functions,

$ϕ (A - B) = ϕ (A) - ϕ (B)$

if $B \subseteq A$ and $| ϕ B | < + \infty$ .

Suppose $ϕ$ is countably additive on a ring $R$ . If $A_{n} \in R$ and $A_{n} \subseteq A_{n + 1}$ , $A \in R$ and

$A = \cup_{n = 1}^{\infty} A_{n},$

Then as $n \to \infty$ ,

$ϕ (A_{n}) \to ϕ (A)$

Lebesgue measure

An elementary set is a union of a finite element of intervals. The family of all elementary subsets of $R^{p}$ is denoted $E$ .

The following properties may be verified,

$E$ is a ring but not a $σ$ -ring.
$m$ defined by $m ([a, b]) = Π_{i = 1}^{p} (b_{i} - a_{i})$ is well defined and additive on $E$ .

A nonnegative additive set function is regular if for every $A \in E$ and to every $ε > 0$ there exist sets $F, G \in E$ such that $F$ is closed, $G$ is open, $F \subset A \subset G$ , and

$ϕ (G) - ε \leq ϕ (A) \leq ϕ (F) + ε$

$m$ is regular, $μ (a, b) = α (b) - α (a)$ where $α$ is monotonically increasing is regular on the real line.

Every regular set function on $E$ can be extended to a countably additive set function on a $σ$ -ring which contains $E$ .

Let $μ$ be additive, regular, nonnegative and finite on $E$ . Consider countable coverings of any set $E \subseteq R^{p}$ by open elementary sets $A_{n}$ :

$E \subseteq \cup_{n = 1}^{\infty} A_{n}$

Define $μ^{*} (E) = inf \sum_{n = 1}^{\infty} μ (A_{n}),$ the inf being taken over all countable coverings of $E$ by open elementary sets. $μ^{*} (E)$ is called the outer measure of $E$ , corresponding to $μ$ .

Properties of an outer measure.

The outer measure is nonnegative for all $E$ .
It is monotonic by inclusion.
Its restriction to elementary set is equivalent to the original measure.
Subadditivity: If $E = \cup_{1}^{\infty} E_{n}$ , then

$μ^{*} (E) \leq \sum_{n = 1}^{\infty} μ^{*} (E_{n})$

Distances between sets

For any $A, B \subseteq R^{p}$ , we define the symmetric difference and the distance functions

$S (A, B) = (A - B) \cup (B - A)$ , $d (A, B) = μ^{*} (S (A, B)) .$

We write $A_{n} \to A$ if $lim_{n \to \infty} d (A, A_{n}) = 0$ .

If there is a sequence of elementary sets converging to $A$ , we say its finitely $μ$ measurable

and write $A \in M_{F} (μ)$ .

If $A$ is the union of a countable collection of finitely $μ$ measurable sets, then $A$ is $μ$ measureable and write $A \in M (μ)$ .

Properties of the symmetric difference and the distance function.

$S (A, B) \subseteq S (A, C) \cup S (C, B)$ .
$S (A_{1} \cup A_{2}, B_{1} \cup B_{2}), S (A_{1} \cap A_{2}, B_{1} \cap B_{2}), S (A_{1} - A_{2}, B_{1} - B_{2}) \subseteq S (A_{1}, B_{1}) \cup S (A_{2}, B_{2})$

The distance function follow analagous relations with the relevant set operations becoming the appropriate arithmetic operations.

There is also the property

$| μ^{*} (A) - μ^{*} (B) | \leq d (A, B)$

if at least one of $μ^{*} (A), μ^{*} (B)$ is finite.

The distance function is almost a true metric but different sets can have zero distance. Thus we may instead define equivalence in terms of vanishing distance. This equivalence relation makes $M_{F} (μ)$ be the closure of $E$ .j

$M (μ)$ is a $σ$ -ring and $μ^{*}$ is countably additive on $M (μ)$ .

The extended set function is called a measure, and the Lebesgue measure on $R^{p}$ is thus the special case $μ = m$ .

Borel sets

A borel set is a set that can be obtained by a countable number of unions,intersectoins or complements starting from open sets. The collection of all Borel sets is a $σ$ -ring, in fact the smallest one which contains all open sets. An element of a borel set is $μ$ measureable. Every $μ$ measurable set is the union of a Borel set and a set of measure zero.

Measure spaces

A set $X$ is a measure space if there exists a $σ$ ring $M$ of subsets of $X$ (the measurable sets) and a non-negative countably additive measure defined on $M$ . If, in adddition, $X \in M$ , then $X$ is said to be a measurable space.

Measurable functions

A function is measurable if the set ${x | f (x) > a}$ is measurable for every real $a$ . Equality can be included.

Making more measurable functions.

If $f$ is measurable, then $| f |$ is measurable.
Let ${f_{n}}$ be a sequence of measurable functions. For $x \in X$ , put

$g (x) = sup f_{n} (x)$

$h (x) = lim_{n \to \infty} sup f_{n} (x)$ , Then $g$ and $h$ are measurable.

This results in the following corollaries,

If $f$ and $g$ are measurable, then $max (f, g)$ and $min (f, g)$ are measurable
The limit of a convergent sequence of measurable functions is measurable.

Common binary operations are also preserve measurability.

Let $f$ and $g$ be measurable real-valued functions defiend on $X$ , let $F$ be real and continuous on $R^{2}$ . Then $F (f (x), g (x))$ is measurable.

Thus common operations of analysis can be applied to measurable functions to give measurable functions. An example where measurability doesn’t carry over is $h (x) = f (g (x))$ , where $f$ is measurable and $g$ is continuous, but $h$ is not necessarily measurable. Interestingly, we were able to define measurable functions without referencing any particular measure. Thus, the class of measurable functions on $X$ depends on the $σ$ ring $M$ .

Simple functions

Let $s$ be a real-valued function defined on $X$ . If the range of $s$ is finite, then $s$ is a simple function. Let $E \subseteq X$ and put define $K_{E} (x) = 1$ when $x \in E$ and zero otherwise as the characteristic function of $E$ . Denoting $E_{i} = s^{- 1} ({c_{i}})$ where $c_{i}$ make up the range of $s$ , we can decompose any simple function into a linear combinations of characteristic functions, $s = \sum_{i = 1}^{n} c_{i} K_{E_{i}} .$

$s$ is measurable iff $E_{i}$ are measurable.

This allows us to approximate every function by simple functions.

Let $f$ be a real function on $X$ . There exists a sequence ${s_{n}}$ of simple funcitons such that $s_{n} (x) \to f (x)$ as $n \to \infty$ , for every $x \in X$ . If $f$ is measurable, ${s_{n}}$ may be chosen to be a sequence of measurable functions. If $f \geq 0$ , ${s_{n}}$ may be chosen to be a monotonically increasing sequence.

$s_{n}$ can be constructed as

$s_{n} = \sum_{i = 1}^{n 2^{n}} \frac{i - 1}{2^{n}} K_{E_{n i}} + n K_{F_{n}}$

for non-negative $f$ . For general functions, the function can be split into the difference between its positive and negative components. The sequence converges uniformly to $f$ if $f$ is bounded.

Lebesgue Integration

Suppose $(x \in X, c_{i} > 0)$ $s (x) = \sum_{i = 1}^{n} c_{i} K_{E_{i}} (x)$ is measurable and $E \in M$ . We define

$I_{E} (s) = \sum_{i = 1}^{n} c_{i} μ (E \cap E_{i}) .$ If $f$ is measurable and non-negative, we define

$\int_{E} f d μ = sup I_{E} (s),$

where the sup is taken over all measurable simple functions $s$ such that $0 \leq s \leq f$ .

The integral may have the value $+ \infty$ . If the integral of each of the positive and negative components of a function is finite, then $f$ is said to be integrable, or $f \in L (μ)$ . Integrability carries over to subsets.

Suppose $f$ is measurable and non-negative on $X$ . For $A \in M$ , define

$ϕ (A) = \int_{A} f d μ .$

Then $ϕ$ is countably additive on $M$ . The same conclusion holds for any $f \in L (μ)$ .

Suppose $E \in M$ . Let ${f_{n}}$ be a sequence of measurable functions such that $0 \leq f_{1} (x) \leq f_{2} (x) \leq \dots$ . Let $f$ be defined by $f_{n} (x) \to f (x)$ as $n \to \infty$ . Then

$\int_{E} f_{n} d μ \to \int_{E} f d μ$

A corollary of the result is that the lebesgue integral and summation of a sequence of measurable functions commute.

Suppose $E \in M$ . If ${f_{n}}$ is a sequence of non-negative measurable functions and

$f (x) = lim_{n \to \infty} inf f_{n} (x),$

then

$\int_{E} f d μ \leq lim_{n \to \infty} inf \int_{E} f_{n} d μ$

Suppose $E \in M$ . Let ${f_{n}}$ be a sequence of measurable functions such that $f_{n} (x) \to f (x)$ as $n \to \infty$ . If there exists a function $g \in L (μ)$ such that

$| f_{n} (x) | \leq g (x)$

for all $n$ , then,

$lim_{n \to \infty} \int_{E} f_{n} d μ = \int_{E} f d μ$

In summary, the limits of measurable functions are always measurable, whereas the limits of Riemann-integrable functions may fail to be Riemann-integrable.

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