Why do we require the event collection in a probability space to be a sigma-algebra?

Finite unions are not enough because many natural events in probability involve infinitely many outcomes. A sigma-algebra must be closed under countable unions to ensure that events formed by collecting infinitely many individual outcomes remain measurable.

Conditions: The sample space may be countably infinite.; Events of interest may require combining infinitely many outcomes.

A sigma-algebra is closed under countable intersections because any countable intersection can be rewritten in terms of complements and countable unions using De Morgan's law. Since a sigma-algebra is already closed under complements and countable unions, it must also be closed under countable intersections.

Conditions: Each $A_k \in \mathcal{F}$.; $\mathcal{F}$ is closed under complements and countable unions.

The Borel sigma-algebra on an interval $I \subseteq \mathbb{R}$ is defined as the smallest sigma-algebra that contains all open intervals $(a,b)$ contained in $I$. It is denoted as $\mathcal{B}(I)$.

Conditions: $I$ is an interval on the real line $\mathbb{R}$.; The generating sets are open intervals $(a,b) \subseteq I$.

When the sample space $\Omega$ is finite or countable, a suitable sigma-algebra should typically contain all singletons, meaning sets of the form $\{x\}$ for every $x \in \Omega$. This allows probabilities to be assigned to all individual outcomes.

Conditions: $\Omega$ is finite or countable.; The goal is to assign probabilities to individual outcomes.

The Borel sigma-algebra is the standard choice in continuous probability because there exists a unique probability measure on $\mathcal{B}([0,1])$ that assigns to each half-open interval $(a,b]$ its exact length $b-a$. This measure, known as Lebesgue measure, corresponds to the uniform distribution and avoids the non-measurability issues of the power set.

Conditions: The context is the continuous setting.; The measure is defined on $\mathcal{B}([0,1])$.; The endpoint condition is $0 \leq a \leq b \leq 1$.