Probability space — sample space Ω, sigma-algebra F, measure P — Kolmogorov axioms: Question bank

Before we start, symbols the questions lean on — earn them once so no line uses them cold:

$Ω$ (capital omega) is the sample space: the bag of every outcome that could happen.
$F$ (fancy F) is the sigma-algebra: the list of subsets of $Ω$ we are allowed to assign a probability to. Each such subset is an event.
$P$ is the probability measure: the rule $P : F \to [0, 1]$ obeying Kolmogorov's three axioms (A1 non-negativity, A2 normalization $P (Ω) = 1$ , A3 countable additivity for disjoint events).
$A^{c}$ means "not $A$ " (the complement), $A \cup B$ means " $A$ or $B$ ", $A \cap B$ means " $A$ and $B$ ", $\emptyset$ is the empty set (the event that never happens).
$A ⊔ B$ (union with a bar under it) is a disjoint union: exactly $A \cup B$ but with the extra promise that $A \cap B = \emptyset$ , i.e. the pieces do not overlap. We write $⊔$ instead of $\cup$ precisely to signal "safe to just add the probabilities".

True or false — justify

True or false: On a finite sample space you may always take

F = 2^{Ω}

(every subset an event).

True — with finitely many outcomes there are no pathological (non-measurable) sets, so the power set is a valid sigma-algebra.

True or false: On

Ω = [0, 1]

you may always take

F = 2^{[0, 1]}

False — a Vitali set

V

is built by picking one representative from each group of points differing by a rational; its rational shifts

V \oplus q

are disjoint, all have equal "length", and countably many of them tile

[0, 1]

— so A3 would force

P ([0, 1])

to equal

\sum

of infinitely many equal numbers, which is either

0

\infty

, never

1

. So

V

cannot be measured, and we shrink

F

to the Borel sigma-algebra.

True or false: If

P (A) = 0

then

A

can never occur.

False — in continuous spaces a single point like

{0.5}

has probability

0

yet is a perfectly possible outcome; "probability

0

" means "almost never", not "impossible".

True or false:

P (A \cup B) = P (A) + P (B)

holds for every pair of events.

False — it holds only when

A, B

are disjoint; if they overlap you double-count

A \cap B

and must subtract it (inclusion–exclusion).

True or false: The range

[0, 1]

for

P

is an extra axiom Kolmogorov imposed.

False — only

P \geq 0

(A1) and

P (Ω) = 1

(A2) are axioms;

P (A) \leq 1

is derived from the complement rule, so

[0, 1]

is a theorem.

True or false: A sigma-algebra must be closed under countable intersections.

True — but it is not a separate axiom; De Morgan turns

⋂_{i} A_{i} = (⋃_{i} A_{i}^{c})^{c}

into complements and a union, both already guaranteed.

True or false:

\emptyset

must belong to

F

True —

F

contains

Ω

and is closed under complement, so

\emptyset = Ω^{c} \in F

automatically.

True or false: Finite additivity is enough to build all of probability theory.

False — countable additivity (A3) is strictly stronger; it gives continuity of

P

and underpins limits, Random variables, and expectations.

True or false: If

A \subseteq B

then

P (A) \leq P (B)

True — write

B = A ⊔ (B ∖ A)

(disjoint since

B ∖ A

shares no point with

A

), so

P (B) = P (A) + P (B ∖ A)

and the second term is

\geq 0

by A1 (monotonicity).

Spot the error

Error hunt: "

P (\emptyset) = 0

is an axiom, so we just assume it."

Wrong — it is derived: take the disjoint sequence

Ω, \emptyset, \emptyset, \dots

, so A3 gives

P (Ω) = P (Ω) + \sum_{i \geq 2} P (\emptyset)

, hence

\sum_{i \geq 2} P (\emptyset) = 0

; since every term equals the same value

P (\emptyset) \geq 0

(A1), an infinite sum of one repeated non-negative number can only total

0

if that number is

0

Error hunt: "

F

is the set of all outcomes."

Wrong — outcomes

ω

live in

Ω

;

F

is a collection of subsets of

Ω

(events), one level up.

Error hunt: "Since

P

maps into

[0, 1]

, negative values are impossible by definition, so A1 is redundant."

Wrong — the

[0, 1]

codomain is the conclusion we want to justify; without A1 you couldn't prove non-negativity, so A1 is doing real work.

Error hunt: "

A

and

A^{c}

overlap at the boundary, so

P (A) + P (A^{c})

over-counts."

Wrong —

A

and

A^{c}

are disjoint by definition of complement (nothing is both in and out), so their probabilities add cleanly to

P (Ω) = 1

Error hunt: "For inclusion–exclusion we split

A \cup B

into

A

and

B

, giving

P (A) + P (B)

Wrong —

A

and

B

are not disjoint; the correct disjoint split is

(A ∖ B) ⊔ (A \cap B) ⊔ (B ∖ A)

, which yields the

- P (A \cap B)

correction.

Error hunt: "

{1, 2, 3}

is a sigma-algebra on

Ω = {1, 2, 3}

Wrong — that is a subset of

Ω

, not a family of subsets; the smallest sigma-algebra here is

{\emptyset, Ω}

Error hunt: "Under the uniform (Lebesgue) measure on

[0, 1]

P ([0.5, 0.5 + \frac{1}{n}]) = \frac{1}{n}

never reaches

0

, so

P ({0.5}) > 0

Wrong — with

P ([a, b]) = b - a

, we have

{0.5} = ⋂_{n} [0.5, 0.5 + \frac{1}{n}]

and continuity from above gives

P ({0.5}) = lim \frac{1}{n} = 0

Why questions

Why must

F

be closed under complement?

Because if you can ask "did

A

happen?" you must also be able to ask "did

A

not happen?"; a measurable question with an unmeasurable negation would be logically incomplete.

Why is A3 stated for countable (not just finite) unions?

Because continuous distributions, limits, and integrals require adding infinitely many disjoint pieces; finite additivity alone cannot express

P

of a limit of shrinking sets. See Measure theory & Lebesgue integration.

Why do we need

F

at all instead of measuring every subset?

On uncountable

Ω

, insisting every subset be measurable contradicts countable additivity (Vitali's construction), so we restrict to a well-behaved family.

Why does A2 fix

P (Ω) = 1

specifically, not some other number?

It is a normalization choice: "something happens" is certain, and calling certainty

1

makes probabilities behave as fractions of the whole. Any total could be rescaled, but

1

is the convention that makes ratios read as chances.

Why can two different events both have probability

0

yet one be

\emptyset

and the other not?

Probability measures size, not possibility;

\emptyset

contains no outcomes (truly impossible) while

{0.5}

contains one outcome of zero length (possible but negligible).

Why does independence not appear among Kolmogorov's axioms?

Independence is an extra relationship between events defined via

P

, not a rule every

P

must satisfy; it lives one layer up in Conditional probability & independence.

Edge cases

Edge case: What is the smallest possible sigma-algebra on any

Ω

The trivial one

{\emptyset, Ω}

— it satisfies all three closure rules and lets you ask only "did anything happen?".

Edge case: On

Ω = {a}

(a single outcome) what must

P

be?

P ({a}) = P (Ω) = 1

by A2 and

P (\emptyset) = 0

; the space is forced, with no freedom.

Edge case: Can an event

A

satisfy

P (A) = 1

while

A \neq = Ω

Yes — in

[0, 1]

the set

[0, 1] ∖ {0.5}

has probability

1

but is not the whole space; such

A

is called "almost sure".

Edge case: If

A_{1} \supseteq A_{2} \supseteq \dots

shrink toward

\emptyset

, what happens to

P (A_{n})

By continuity from above (a consequence of A3),

P (A_{n}) \to P (⋂_{n} A_{n}) = P (\emptyset) = 0

Edge case: If

A_{1} \subseteq A_{2} \subseteq \dots

grow upward, what is

P (⋃_{n} A_{n})

By continuity from below (the twin property, also from A3),

P (⋃_{n} A_{n}) = lim_{n} P (A_{n})

— the probability of the growing limit is the limit of the probabilities; e.g.

P ([0, 1 - \frac{1}{n}]) = 1 - \frac{1}{n} \to 1 = P ([0, 1))

Edge case: Is

P (A \cup B) \leq P (A) + P (B)

ever violated?

Never — inclusion–exclusion gives

P (A \cup B) = P (A) + P (B) - P (A \cap B)

and

P (A \cap B) \geq 0

, so the union bound always holds; equality iff

A, B

are disjoint (up to a probability-

0

overlap). This generalizes via the Inclusion–exclusion principle.

Edge case: Does the union bound survive for countably many events?

Yes —

P (⋃_{i = 1}^{\infty} A_{i}) \leq \sum_{i = 1}^{\infty} P (A_{i})

for any countable family (overlapping or not); split the union into disjoint shells

A_{i} ∖ (A_{1} \cup \dots \cup A_{i - 1})

, apply A3, and each shell has probability

\leq P (A_{i})

by monotonicity. This is "countable subadditivity" (Boole's inequality).

Recall One-line summary to lock in

Axioms constrain the measure $P$ ; closure rules constrain the event list $F$ ; probability- $0$ is about size, not possibility; and "always additive" is the single most common trap.