Measure and Measure Spaces

The idea behind (Lebesgue-) measures is to extend the notion of lengths for a larger class of subsets of $\mathbb{R}$ than just finite intervals. This is particularly useful for integration as it forms the basis for Lebesgue Integration which allows one to integrate many more functions than Riemann integration as all Riemann-integrable functions are Lebesgue-integrable (as long as the integral is proper). Also, much of probability is based on measure theory. For further help in starting a wiki page, check out Wiki Guidelines and Wiki Formatting or come chat with us.

1. $\digamma$ is called a $\sigma$-algebra on a set $\Omega$ (generally referred to as the sample space) if it satisfies the following 3 properties:
   $\hspace{5mm}$ (i) $\varnothing, \Omega \in \digamma.$
   $\hspace{5mm}$ (ii) $A,B \in \digamma \implies \Omega \setminus A : = A^{c} \in \digamma \ \text{ and }\ A \cup B \in \digamma.$
   $\hspace{5mm}$ (iii) $A_{1},A_{2},... \in \digamma \implies \cup_{n=1}^{\infty} A_{n} \in \digamma.$
   The first two properties define an algebra on $\digamma$, whilst property (iii) adds that sets in this algebra must also be closed under countable union. This leads naturally to the definition of a measure space:

2. A pair $(\Omega, \digamma),$ where $\digamma$ is a $\sigma$-algebra on $\Omega,$ is called a measurable space and any set in $\digamma$ is called a measurable subset of $\Omega$ with respect to $\digamma .$

3. A function $\mu$: $\digamma \rightarrow [0, \infty]$ is called a measure on $(\Omega, \digamma)$ if
   $\hspace{5mm}$ (a) $\mu ( \varnothing ) = 0$ and $\mu (A) \geq 0 \ \ \forall A \in \digamma$
   $\hspace{5mm}$ (b) $A_{n} \in \digamma$ with $A_{n}$ disjoint $\implies \mu( \cup_{n=1}^{\infty} A_{n} ) = \displaystyle\sum_{n=1}^{\infty} \mu (A_{n}),$
   where condition (b) says that a measure must be countably additive (over disjoint sets). Also, note that the codomain is the extended real number line [0, $\infty$ ] rather than [0, $\infty$ ) such that $\mu (A) = \infty$ is permitted.
   

Probability Space

Consider the triple $( \Omega , \digamma , \mathbb{P})$. If $(\Omega, \digamma)$ is a measurable space and $\mathbb{P}$ is a measure with $\mathbb{P} ( \Omega ) = 1,$ then we have a probability space where $\Omega$ is the sample space and $\digamma$ is a set of subsets of $\Omega$ containing events.

Let $L(I)$ be the length of an interval $I.$ For example, $L\big((a,b]\big) = b - a.$

The (Lebesgue) outer measure of any set $A \subseteq \mathbb{R}$ is given by $$m^{*}(A) = \inf Z_{A},$$ where $$Z_{A} := \left\{ \displaystyle\sum_{n=1}^{\infty} L(I_n) : A \subseteq \cup_{n=1}^{\infty} I_{n} \right\}$$ for intervals $I_{n}$ which (we say) form a covering of $A.$

$\ m^{*} (A) = 0$ if and only if $A$ is a null set. One example of a null set is a singleton ${a}$ for $\ a \in \mathbb{R}$. Now suppose we have a uniform distribution on $[0,1].$ Then as a single element forms a null set, we have that $\mathbb{P} ({a}) = 0 \ \ \forall a \in [0,1],$ which comes as a bit of a shock as the event of picking $a$ is not impossible, yet the probability of it occurring is zero. More surprisingly $\mathbb{P} \big( \mathbb{Q}_{[0,1]} \big) = 0$ despite the fact that there are infinitely many rationals in $[0,1].$ This is because there are countably many rationals and infinitely many reals (on any non-empty interval) so the set of rationals is 'negligible' compared to the set of reals. Thus, if $\mathbb{P} (A) = 0,$ we say that "$A$ almost surely won't occur."

Show that if $A \subset B,$ then $\ m^{*} (A) \leq m^{*} (B).$

ANSWER

We will show that $Z_{B} \subset Z_{A},$ and then use the definition of the infimum of a set. Let $I_{n}$ be a sequence of covering intervals for $B,$ then $A \subset B \subseteq \cup_{n=1}^{\infty} I_{n},$ so $I_{n}$ is also a covering sequence of intervals for $A,$ i.e. $x \in Z_{B} \implies x \in Z_{A},$ so $Z_{B} \subset Z_{A}$. Taking the same $x,$ we have that $m^{*} (A) \leq x$ as the infimum is a lower bound for a set, so as $x$ was taken from $Z_{B},$ $m^{*} (A)$ is a lower for $Z_{B}$. Hence, as the infimum is the greatest lower bound (of a set), $m^{*} (A) \leq m^{*} (B)$. $_\square$