A probability model begins with a failure of counting. If a dart lands somewhere in the interval $[0,1]$, symmetry says no individual point should be preferred. Giving every point the same positive probability would force finitely many points to have arbitrarily large total probability, while giving every point probability $0$ cannot explain why the whole interval has probability $1$ by countable summation alone. Probability therefore needs more than a set of outcomes: it needs a chosen class of events and a countably additive way to measure them.

[example: A Uniform Point Cannot Be Built from Equal Point Masses] Let $\Omega=[0,1]$, and suppose first that every singleton has one common positive probability: \begin{align*} \mathbb P(\{x\})=c>0\quad\text{for every }x\in[0,1]. \end{align*} Fix $n\in\mathbb N$ and choose distinct points $x_1,\ldots,x_n\in[0,1]$. Because the points are distinct, if $i\ne j$ then $\{x_i\}\cap\{x_j\}=\varnothing$, so the singleton events are pairwise disjoint. To apply countable additivity to this finite family, define \begin{align*} A_k=\{x_k\}\quad\text{for }1\le k\le n \end{align*} and \begin{align*} A_k=\varnothing\quad\text{for }k>n. \end{align*} The sequence $(A_k)_{k\in\mathbb N}$ is pairwise disjoint, and its union is \begin{align*} \bigcup_{k=1}^{\infty}A_k=\{x_1\}\cup\cdots\cup\{x_n\}=\{x_1,\ldots,x_n\}. \end{align*} Countable additivity gives \begin{align*} \mathbb P(\{x_1,\ldots,x_n\})=\sum_{k=1}^{\infty}\mathbb P(A_k). \end{align*} For $k>n$, $A_k=\varnothing$. Also, since $\varnothing$ is disjoint from itself, countable additivity applied to $\varnothing=\bigcup_{k=1}^{\infty}\varnothing$ gives \begin{align*} \mathbb P(\varnothing)=\sum_{k=1}^{\infty}\mathbb P(\varnothing). \end{align*} The left side is finite because $\mathbb P$ takes values in $[0,1]$, so this equality forces $\mathbb P(\varnothing)=0$. Hence the tail terms vanish, and \begin{align*} \sum_{k=1}^{\infty}\mathbb P(A_k)=\mathbb P(\{x_1\})+\cdots+\mathbb P(\{x_n\}). \end{align*} Each singleton has probability $c$, so \begin{align*} \mathbb P(\{x_1,\ldots,x_n\})=c+\cdots+c. \end{align*} There are $n$ summands, each equal to $c$, therefore \begin{align*} c+\cdots+c=nc. \end{align*} Thus \begin{align*} \mathbb P(\{x_1,\ldots,x_n\})=nc. \end{align*} Since $c>0$, choose $n\in\mathbb N$ with $n>1/c$. Multiplying the inequality $n>1/c$ by the positive number $c$ gives \begin{align*} nc>1. \end{align*} For this choice of $n$, \begin{align*} \mathbb P(\{x_1,\ldots,x_n\})=nc>1. \end{align*} This contradicts the requirement that $\mathbb P$ take values in $[0,1]$. Thus an equal-mass model on $[0,1]$ cannot assign a positive common probability to all singletons. A uniform probability model on $[0,1]$ must instead have $\mathbb P(\{x\})=0$ for each point $x$, while still having $\mathbb P([0,1])=1$; probabilities on continuous spaces therefore cannot be reconstructed by summing equal point masses. [/example]

The example raises the central question for this page: what structure must be present before probability statements have mathematical meaning? The next definitions separate the raw outcomes, the observable events, and the probability measure.

## Definition

The opening failure shows why probability cannot be just a function on individual outcomes. A model must say what the possible outcomes are, which subsets are legitimate events, and how probabilities are assigned to those events. The central object packages these three choices at once.

[definition: Probability Space] A probability space is a triple $(\Omega,\mathcal F,\mathbb P)$ where $\Omega$ is a set, $\mathcal F$ is a $\sigma$-algebra on $\Omega$, and $\mathbb P:\mathcal F\to[0,1]$ is a probability measure on $(\Omega,\mathcal F)$. [/definition]

The definition is compact, but each component has a distinct job. The first component answers the question of what could happen. It carries no probabilities and no information structure by itself, so the definition below is intentionally minimal.

[definition: Sample Space] A sample space is a set $\Omega$ whose elements are called outcomes. [/definition]

A sample space is too raw for probability because most probabilistic statements concern subsets of outcomes. If one can ask whether an event happens, one should also be able to ask whether it does not happen and whether at least one event in a countable list happens. The event collection must therefore be closed under these logical operations.

[definition: $\sigma$-Algebra] Let $\Omega$ be a set. A $\sigma$-algebra on $\Omega$ is a collection $\mathcal F\subset\mathcal P(\Omega)$ such that: 1. $\Omega\in\mathcal F$. 2. If $A\in\mathcal F$, then $A^c\in\mathcal F$, where the complement is taken in $\Omega$. 3. If $A_1,A_2,\ldots\in\mathcal F$, then $\bigcup_{n=1}^{\infty}A_n\in\mathcal F$. [/definition]

The sets in $\mathcal F$ are the events whose probabilities the model is allowed to discuss. To assign numerical probabilities to those events, we need the following normalized version of a measure.

[definition: Probability Measure] Let $(\Omega,\mathcal F)$ be a measurable space. A probability measure on $(\Omega,\mathcal F)$ is a function $\mathbb P:\mathcal F\to[0,1]$ such that: 1. $\mathbb P(\Omega)=1$. 2. If $A_1,A_2,\ldots\in\mathcal F$ are pairwise disjoint, then \begin{align*} \mathbb P\left(\bigcup_{n=1}^{\infty}A_n\right)=\sum_{n=1}^{\infty}\mathbb P(A_n). \end{align*} [/definition]

This definition is the foundation for the rest of probability. The next example shows that even the smallest familiar experiment already uses all three parts of the triple.

[example: A Biased Coin] Let $\Omega=\{H,T\}$, let $\mathcal F=\mathcal P(\Omega)$, and fix $p\in[0,1]$. Since $\mathcal F$ is the full power set, it contains $\Omega$, is closed under complements in $\Omega$, and is closed under countable unions of subsets of $\Omega$; hence $\mathcal F$ is a $\sigma$-algebra. Define \begin{align*} \mathbb P(\varnothing)=0,\quad \mathbb P(\{H\})=p,\quad \mathbb P(\{T\})=1-p,\quad \mathbb P(\Omega)=1. \end{align*} Equivalently, for every $A\subset\Omega$, \begin{align*} \mathbb P(A)=p\,\mathbf 1_{\{H\in A\}}+(1-p)\,\mathbf 1_{\{T\in A\}}. \end{align*} Indeed, for $A=\varnothing$ both indicators are $0$, so the formula gives $0$. For $A=\{H\}$ it gives $p\cdot1+(1-p)\cdot0=p$. For $A=\{T\}$ it gives $p\cdot0+(1-p)\cdot1=1-p$. For $A=\Omega$ it gives \begin{align*} p\cdot1+(1-p)\cdot1=p+1-p=1. \end{align*} Since $p\in[0,1]$, we have $0\le p\le1$. Subtracting $p\le1$ from $1$ gives $1-p\ge0$, and subtracting $0\le p$ from $1$ gives $1-p\le1$, so $1-p\in[0,1]$. Thus every assigned value lies in $[0,1]$, and \begin{align*} \mathbb P(\Omega)=1. \end{align*} It remains to verify countable additivity. Let $(A_n)_{n\in\mathbb N}$ be pairwise disjoint subsets of $\Omega$, and put \begin{align*} U=\bigcup_{n=1}^{\infty}A_n. \end{align*} For the outcome $H$, either $H\notin U$, in which case every term $\mathbf 1_{\{H\in A_n\}}$ is $0$, or $H\in U$, in which case $H\in A_m$ for exactly one index $m$ because the sets $A_n$ are pairwise disjoint. Therefore \begin{align*} \mathbf 1_{\{H\in U\}}=\sum_{n=1}^{\infty}\mathbf 1_{\{H\in A_n\}}. \end{align*} The same argument for $T$ gives \begin{align*} \mathbf 1_{\{T\in U\}}=\sum_{n=1}^{\infty}\mathbf 1_{\{T\in A_n\}}. \end{align*} Using the indicator formula for $\mathbb P$ on $U$, \begin{align*} \mathbb P(U)=p\,\mathbf 1_{\{H\in U\}}+(1-p)\,\mathbf 1_{\{T\in U\}}. \end{align*} Substituting the two indicator identities gives \begin{align*} \mathbb P(U)=p\sum_{n=1}^{\infty}\mathbf 1_{\{H\in A_n\}}+(1-p)\sum_{n=1}^{\infty}\mathbf 1_{\{T\in A_n\}}. \end{align*} Each of the two indicator sums has at most one nonzero term, so multiplying by the constants and adding term by term gives \begin{align*} \mathbb P(U)=\sum_{n=1}^{\infty}\left(p\,\mathbf 1_{\{H\in A_n\}}+(1-p)\,\mathbf 1_{\{T\in A_n\}}\right). \end{align*} For each $n$, the expression inside the parentheses is exactly $\mathbb P(A_n)$ by the indicator formula. Hence \begin{align*} \mathbb P\left(\bigcup_{n=1}^{\infty}A_n\right)=\sum_{n=1}^{\infty}\mathbb P(A_n). \end{align*} Thus $(\Omega,\mathcal F,\mathbb P)$ is a probability space, and the parameter $p$ is precisely the probability assigned to heads. [/example]

The coin example hides a simplification: every subset is measurable. The next section explains why the choice of $\mathcal F$ becomes a real part of the model once the outcome space is infinite.

## Event Structures

The event collection controls what the model can observe. A very small $\sigma$-algebra represents limited information, while a very large one allows many distinctions among outcomes. The first extreme is useful because it shows that outcomes need not be individually observable.

[definition: Indiscrete $\sigma$-Algebra] Let $\Omega$ be a set. The indiscrete $\sigma$-algebra on $\Omega$ is $\{\varnothing,\Omega\}$. [/definition]

With the indiscrete $\sigma$-algebra, the model distinguishes only impossibility from certainty. To model experiments where every subset is observable, we need the opposite extreme below.

[definition: Discrete $\sigma$-Algebra] Let $\Omega$ be a set. The discrete $\sigma$-algebra on $\Omega$ is the power set $\mathcal P(\Omega)$. [/definition]