## Probability

### The Analysis of Data, volume 1

Continuity of Measure

## E.2. The Measure Function

Definition E.2.1. Let $\mathcal{F}$ be an algebra. A measure is a function $\mu:\mathcal{F}\to [0,\infty) \cup \{+\infty\}$ that satisfies:
• Empty set: $\mu(\emptyset)=0$
• Countable additivity: If $E_n, n\in\mathbb{N}$ is a sequence of disjoint events ($E_i\cap E_j=\emptyset$ whenever $i\neq j$) $\mu\left(\bigcup_{i=1}^{\infty} E_i\right) = \sum_{i=1}^{\infty} \mu(E_i).$

If $\mathcal{F}$ is a $\sigma$-algebra, the triplet $(\Omega,\mathcal{F},\mu)$ is called a measure space.

Contrasting this with Definition 1.2.1, we see that a probability is a measure function that satisfies $\mu(\Omega)=1$.

Proposition E.2.1. (The Continuity of Measure). Any measure with $\mu(\Omega) < \infty$ satisfies the following properties
• Finite Additivity: For a finite sequence of disjoint sets $E_1,\ldots,E_k$, $\mu\left(\sum_{i=1}^k E_k\right)=\sum_{i=1}^k \mu(E_i).$
• Continuity from Below: If $E_n\nearrow E$, $E\in\mathcal{F}$, and $E_n\in\mathcal{F}$ for all $n\in\mathbb{N}$, then $\mu(E_n)\nearrow \mu(E)$.
• Continuity from above: If $E_n\searrow E$, $E\in\mathcal{F}$, $E_n\in\mathcal{F}$ for all $n\in\mathbb{N}$, and $\mu(E_1) < \infty$, then $\mu(E_n)\searrow \mu(E)$.
• (4 Countable sub-additivity) If $E_n\in\mathcal{F}$ and $\cup_n E_n\in\mathcal{F}$ then $\mu\left(\bigcup_{i=1}^{\infty} E_i\right) \leq \sum_{i=1}^{\infty} \mu(E_i).$

Note that the notations $\nearrow, \searrow$ mean convergence of monotonic sequences (Definition B.2.3) and convergence of monotonic sets (Definition A.4.3), depending on whether they are associated with sequences of numbers or sets.

Proof. The first property follows from the countable additivity of the measure function and setting $E_l=\emptyset$ for $l > k$.

We prove next the second property. We define the following sequence of disjoint sets $B_1=E_1$ and $B_k=E_k\setminus E_{k-1}$ where $E=\cup_{n\in\mathbb{N}} B_n$ and $E_n=\cup_{k=1}^n B_k$. By the countable and finite additivity of $\mu$ $\mu(E)=\sum_{n\in\mathbb{N}} \mu(B_n) =\lim_{n\to\infty}\, \sum_{k=1}^n \mu(B_k) = \lim_{n\to\infty} \mu(E_n)$ proving (2). To prove property (3) note that $E_1\setminus E_n\nearrow E_1 \setminus E$, which together with property (2) implies $\mu(E_1) -\mu(E_n) = \mu (E_1\setminus E_n)\nearrow \mu(E_1\setminus E) =\mu(E_1 )-\mu(E).$ To prove (4), we denote $B=E_1$ and $B_n=E_n\cap E_1^c\cap\cdots\cap E_{n-1}^c$ for $n\in\mathbb{N}$. Note that $B_k$ is a disjoint sequence of sets and $\cup_{n=1}^k B_n=\cup_{n=1}^k E_n$ for all $k\in\mathbb{N}$. Since $B_k\subset E_k$ we have $\mu\left(\bigcup_{n=1}^k E_k\right) = \mu\left(\bigcup_{n=1}^k B_k\right) = \sum_{n=1}^k \mu(B_k) \leq \sum_{n=1}^k \mu(E_k).$ It remains to let $k\to\infty$ and use property (2) on the left hand side, applied to the sets $E_k'=\cup_{n=1}^k E_n\nearrow \cup_{n\in\mathbb{N}} E_k$.