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Since the $B_i$ form a partition of a set of the sample space, the events $A\cap B_i$ are pairwise disjoint and their set union is $A$. By countable additivity of probability, followed by the definition of conditional probability,Every ratio is legitimate because each $\mathbb{P}(B_i)>0$. This proves the countable law of total probability.

The probability axioms first give $\mathbb{P}(\varnothing )=0$: apply countable additivity to $\Omega ,\varnothing ,\varnothing ,\dots$ and use normalization and nonnegativity. If $A\subseteq B$, then $B=A\dot{\cup }(B\setminus A)$ is a disjoint union of events. Finite additivity, as a special case of countable additivity, givesThus a probability measure is monotone with respect to set inclusion. The difference event is measurable because the events belong to a sigma-algebra.

The probability measure $\mathbb{P}$ maps the sigma-algebra $\mathcal{F}$ of events into $[0,1]$. Its probability axioms are nonnegativity, normalization $\mathbb{P}(\Omega )=1$, and countable additivity: for every pairwise disjoint sequence of events $A_1,A_2,\dots$,The sample space is $\Omega$, and $\mathcal{F}$ specifies which subsets are events. Requiring countable additivity, rather than only its finite version, is part of the probability space definition.

The disjoint decomposition into sets $A=(A\cap B)\cup (A\setminus B)$ uses only events in $\mathcal{F}$. Finite additivity and nonnegativity giveThus $\mathbb{P}(A\cap B)\le \mathbb{P}(A)$ already follows from the two stated requirements. Equality is attained, for example, by choosing $B=A$, since then $A\cap B=A$. More generally it holds whenever $\mathbb{P}(A\setminus B)=0$. Neither normalization nor countable additivity is needed for this particular inequality.

The missing axiom is normalization: $\mathbb{P}(\Omega )=1$. The zero set function satisfies both displayed requirements but is not a probability measure, showing why this condition is essential.

The incomplete axiom needs countable additivity: for every sequence of pairwise disjoint events $A_1,A_2,\dots \in \mathcal{F}$,The stated two-event version implies finite additivity, but does not supply this requirement for a countably infinite collection. Together with nonnegativity and normalization, this gives the usual probability measure axioms on the sigma-algebra $\mathcal{F}$.

Let $A_i$ be the event that candidate $i$ succeeds. The countable additivity axiom adds probabilities only for mutually disjoint events. Successes of different candidates are not mutually exclusive, so replacing the probability of their union by ${\sum }_i\mathbb{P}(A_i)=1$ misuses that axiom. Without disjointness one obtains only the union bound, not certainty. The same sum is the expected value of the number of successes, which also need not imply at least one.

If the ten outcomes are mutually independent events, with each success probability $0.1$, all fail with probability $0.9^{10}$. HenceThe independence assumption is essential. The marginal probabilities alone allow any union probability from $0.1$ up to $1$; perfectly overlapping success events give the lower endpoint and disjoint successes give the upper endpoint.

For each integer $x$, the events $\{X=x,Y=y\}$, indexed by the integers $y$, are disjoint and their union is $\{X=x\}$. Therefore countable additivity gives the marginal probability mass functionThe nonnegative terms make this sum well-defined, even when the support is infinite.

Let $(B_j)$ be a finite or countable partition of a set of the sample space into disjoint events, with $\mathbb{P}(B_j)>0$. For any event $A$, the law of total probability isTo prove it, the intersections $A\cap B_j$ are disjoint and have union $A$. Countable additivity and the definition of conditional probability therefore giveIt is enough that the union of the partition events has probability one. Parts of probability zero contribute zero in the intersection formula and can be omitted before writing conditional probabilities; conditioning by division on a zero-probability part would be undefined.

The events $\{X=k\}$, for $k\in \mathbb{Z}$, are pairwise disjoint. Their union is $\{X\in \mathbb{Z}\}$, and the integers form a countable set. By countable additivity,Countability is the necessary justification: one cannot apply this argument to an arbitrary uncountable union of zero-probability singletons. This is a special case of countable sets have zero probability under a density.

A probability measure is a nonnegative, countably additive measure with total mass one. Thus $\mathbb{P}(\Omega )=1$, and for pairwise disjoint events $A_j$,The complement of an event has probability $1-\mathbb{P}(A)$. The probability of a union of overlapping events needs a correction: $\mathbb{P}(A\cup B)=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A\cap B)$. Counting the overlap once is the simplest instance of the inclusion-exclusion principle.

A zero-probability event need not be impossible as a member of the sample space. Under the continuous uniform distribution on $[0,1]$, every individual point has probability zero, while the whole interval has probability one. Countable additivity applies to countable unions, not arbitrary uncountable ones; there is no contradiction.

For a measure space $(\Omega ,\mathcal{F},\mu )$ and $E\in \mathcal{F}$, the restriction of a measure to a measurable subset is the measure on $(E,\mathcal{F}{|}_E)$ given by the displayed formula. The trace sigma-algebra consists of the measurable sets contained in $E$. The empty-set axiom and countable additivity are inherited directly from $\mu$. An associated measure on the original sigma-algebra is $\nu (A)=\mu (A\cap E)$ for $A\in \mathcal{F}$; the two constructions have different underlying measurable spaces.

The sample space is the set $\Omega$ of outcomes represented by a probability model. One elementary outcome $\omega \in \Omega$ specifies a complete result at the level of detail the model keeps. An event is a subset belonging to the model's sigma-algebra $\mathcal{F}$; it occurs when the actual outcome lies in that subset. Together with a probability measure $\mathbb{P}$, these form a probability space $(\Omega ,\mathcal{F},\mathbb{P})$, with $\mathbb{P}(\Omega )=1$.^[ref]

Choosing a sample space means deciding which distinctions matter. For two tosses, recording the ordered head-or-tail sequence distinguishes $HT$ from $TH$; recording only the number of heads merges them. Neither representation is automatically wrong, but the second cannot answer a question about which toss was heads. Keeping a detailed sample space and defining a random variable for the desired observation lets several questions share one model.

A sample space alone assigns no probabilities. Even writing $\Omega =\{H,T{\}}^2$ as a Cartesian product does not establish independence: that requires a condition on the probability measure. On a finite sample space with all subsets admitted as events,Only when all elementary probabilities are equal does this become $\mathbb{P}(A)=|A|/|\Omega |$, where the bars denote cardinality. This is the uniform distribution on a finite set, an additional modelling choice.

Sample spaces can also be infinite. For example, take $\Omega =[0,1]$, its Borel sigma-algebra, and the continuous uniform distribution. Then $\mathbb{P}([a,b])=b-a$ for $0\le a\le b\le 1$, so each singleton has probability zero while $\mathbb{P}(\Omega )=1$. Thus an elementary outcome need not carry positive probability, and a continuous model cannot be understood by counting equally likely points. The distinction is developed in zero-probability event and countable additivity.

A sigma-algebra on a set $\Omega$ is a collection $\mathcal{F}$ of subsets such that $\Omega \in \mathcal{F}$, $A\in \mathcal{F}$ implies $\Omega \setminus A\in \mathcal{F}$, and $A_n\in \mathcal{F}$ for all $n$ implies ${\bigcup }_n{A}_n\in \mathcal{F}$. Thus it contains the empty set and is closed under countable intersections as well. This is precisely the closure needed to apply countable additivity without leaving the domain of a measure.

For $\Omega =\{a,b,c\}$,is a sigma-algebra. It describes observations that distinguish $a$ from the pair $\{b,c\}$, but cannot distinguish $b$ from $c$. The power set provides a finer sigma-algebra with all eight subsets. Both are valid; choosing a sigma-algebra records which questions the model can answer.

On the real numbers, the Borel sigma-algebra contains all open sets and everything generated from them by these operations. The Lebesgue measurable sets form a larger sigma-algebra, including all subsets of Borel null sets. Closure under countable unions accommodates constructions involving an entire sequence of observations or approximations.