Probability Spaces

A probability space lists every possible outcome, declares which collections of outcomes you may ask about, and assigns each one a size between 0 and 1.

Two coin flips give four outcomes and sixteen legal events, with the three-outcome event of at least one head carrying probability 3/4

Why Does This Exist?

Every probability calculation quietly assumes three things: the list of outcomes is complete, the question you asked is answerable, and the numbers add up. Skip any one and you get nonsense like probabilities above 1 or events with no assigned chance.

A probability space makes the three assumptions explicit as the triple $(\\Omega, \\mathcal{F}, P)$ . $\\Omega$ is the sample space of all outcomes. $\\mathcal{F}$ is the $\\sigma$ -algebra of legal events. $P$ is the probability measure. This page builds each piece with a two-coin model you can check by hand, grounding the probability foundations you already use.

The one idea you need first

An event is a set of outcomes. With one die roll, $\\{2, 4, 6\\}$ is the event "the roll is even". Probabilities attach to events, never to bare outcomes in the continuous case. This page defines exactly which sets qualify.

Think of It Like This

A library with a catalog rule

$\\Omega$ is every book in the building. $\\mathcal{F}$ is the catalog: the rule says if a shelf list is in the catalog, its complement is too, and any countable stack of lists can be merged into one catalog entry. You may only borrow through the catalog.

$P$ is the librarian's size stamp on each catalog entry: nonnegative, the whole building stamps 1, and disjoint entries add. A book that never made it into the catalog cannot be borrowed, no matter how loudly you ask. That is a non-measurable set: it exists, but the theory refuses to price it.

How It Actually Works

$\\mathcal{F}$ must satisfy three rules: it contains $\\Omega$ , it is closed under complements (if $A$ is legal, so is $A^c$ ), and it is closed under countable unions. $P$ must satisfy three more: $P(A) \\geq 0$ , $P(\\Omega) = 1$ , and countable additivity, so $P(\\cup_i A_i) = \\sum_i P(A_i)$ for disjoint $A_i$ .

Why closure matters

Finite unions feel like enough until you take a limit. The event "heads eventually appear" is a countable union of finite-horizon events. Without countable closure it would fall outside $\\mathcal{F}$ and have no probability. Closure is what lets limits stay inside the theory.

Worked example

Two fair flips: $\\Omega = \\{HH, HT, TH, TT\\}$ . The power set has $2^4 = 16$ events, so $\\mathcal{F}$ can be all of them. Uniform $P$ gives each outcome $1/4$ . The event $A = \\{HH, HT, TH\\}$ ("at least one head") has $P(A) = 3/4 = 0.75$ . Complement check: $A^c = \\{TT\\}$ with $P = 1/4$ , and $3/4 + 1/4 = 1$ . Union check: $\\{HH\\} \\cup \\{TT\\} = \\{HH, TT\\}$ with $P = 1/2$ , the sum of the parts.

Code

1import itertools2
3omega = [''.join(r) for r in itertools.product('HT', repeat=2)]4prob = {w: 1 / 4 for w in omega}  # uniform over 4 outcomes5event = [w for w in omega if 'H' in w]  # at least one head6print(len(omega), len(event))  # -> (4, 3)7print(sum(prob[w] for w in event))  # -> 0.75

Four outcomes, three in the event, total mass 0.75. The complement carries the remaining 0.25.

Watch Out For

Assuming every subset is an event

On finite $\\Omega$ the power set works, so people assume it always does. On uncountable $\\Omega$ like $[0, 1]$ it does not: some subsets are non-measurable and admit no consistent probability. Assigning them one breaks additivity.

Fix: work with the Borel $\\sigma$ -algebra generated by intervals. If your event can be built from intervals with complements and countable unions, it is legal. Otherwise rephrase the question.

Adding probabilities of overlapping events

$P(A \\cup B) = P(A) + P(B)$ holds only when $A$ and $B$ are disjoint. With overlap you double-count the intersection. The general rule subtracts it: $P(A \\cup B) = P(A) + P(B) - P(A \\cap B)$ .

Symptom: union probabilities above 1 or above either part. Always ask whether the pieces overlap before adding.

The Quick Version

A probability space is the triple $(\\Omega, \\mathcal{F}, P)$ : outcomes, legal events, and a size function.
The $\\sigma$ -algebra is closed under complements and countable unions, which keeps limits answerable.
The measure is nonnegative, totals 1, and adds over disjoint pieces, even infinitely many.
Two fair flips give 4 outcomes, 16 events, and $P(\\text{at least one head}) = 0.75$ .
Not every subset of an uncountable space is an event; build events from intervals.
Add probabilities only for disjoint events, else subtract the overlap.