Exercises — Probability space — sample space Ω, sigma-algebra F, measure P — Kolmogorov axioms

Check the three rules each time: contains $\Omega$; closed under complement; closed under (countable) union.

• (i) Contains $\Omega$ ✓. Complements: ${\Omega }^c=\varnothing$ ✓, ${\varnothing }^c=\Omega$ ✓. Unions of these stay inside ✓. Yes, it is a sigma-algebra (the smallest possible, the "trivial" one).
• (ii) Contains $\{a\}$ but $\{a{\}}^c=\{b,c\}$ is missing. Complement rule fails. Not a sigma-algebra.
• (iii) Contains $\Omega$ ✓. Complements: $\{a\}\leftrightarrow \{b,c\}$ both present ✓, $\varnothing \leftrightarrow \Omega$ ✓. Unions: $\{a\}\cup \{b,c\}=\Omega$ ✓. Yes, it is a sigma-algebra.

$\{1\}$ and $\{2\}$ are disjoint and $\{1\}\cup \{2\}=\Omega$. Finite additivity (from A3) demands $P(\Omega )=P(\{1\})+P(\{2\})=0.7+0.7=1.4\ne 1.$ But A2 says $P(\Omega )=1$. Contradiction. Not valid — it violates additivity / normalization consistency (the numbers overshoot 1). ✗

• (a) Complement rule $P(A^c)=1-P(A)$ with $A=\text{rain}$: $P(\text{no rain})=1-0.4=\mathrm{0.6.}$
• (b) Note $\text{rain}\cap \text{wind}\subseteq \text{rain}$ (every rainy-and-windy day is in particular a rainy day). We first prove monotonicity from the axioms (it is a theorem, not an axiom). Write the bigger set as a disjoint split at the boundary: $\text{rain}=(\text{rain}\cap \text{wind})\bigsqcup (\text{rain}\setminus \text{wind}).$ Finite additivity (from A3) gives $P(\text{rain})=P(\text{rain}\cap \text{wind})+P(\text{rain}\setminus \text{wind})$. The last term is $\ge 0$ by A1, so $P(\text{rain}\cap \text{wind})\le P(\text{rain}),\text{i.e. }0.25\le \mathrm{0.40.}✓$ In general the same argument proves monotonicity: $A\subseteq B\Rightarrow P(A)\le P(B)$. A joint event can never be more likely than one of its parts.

$A=\{2,4,6\}$, $B=\{4,5,6\}$, $A\cap B=\{4,6\}$.

• Counting: $A\cup B=\{2,4,5,6\}$, so $P(A\cup B)=\frac{4}{6}=\frac{2}{3}$.
• Inclusion–exclusion $P(A\cup B)=P(A)+P(B)-P(A\cap B)$: $\frac{3}{6}+\frac{3}{6}-\frac{2}{6}=\frac{4}{6}=\frac{2}{3}.✓$ The subtraction of $P(A\cap B)=2/6$ removes the double-counted overlap $\{4,6\}$.

Proof. Inclusion–exclusion gives $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. By A1 (non-negativity), $P(A\cap B)\ge 0$, so subtracting it can only decrease the right side: $P(A\cup B)=P(A)+P(B)-\underset{\ge 0}{\underbrace{P(A\cap B)}}\le P(A)+P(B).■$ Application. With $A=\text{overheat}$, $B=\text{power cut}$: $P(A\cup B)\le 0.03+0.02=\mathrm{0.05.}$ So the failure-from-either probability is at most $0.05$. (Equality holds only if $A,B$ are disjoint; if they overlap it is strictly less.)

Nest intervals that squeeze onto the point: $C_n=\left[0.5,0.5+\frac{1}{n}\right],C_1\supseteq C_2\supseteq \cdots ,\{0.5\}={\bigcap }_{n=1}^{\infty }{C}_n.$ Each interval has length $P(C_n)=\frac{1}{n}$. As $n\to \infty$ (the arrow $\to$ means "gets arbitrarily close to"), $\frac{1}{n}\to 0$.

Why we may pass to the limit — continuity from above, proved here. We must justify $P\left({\bigcap }_n{C}_n\right)={\lim }_{n}P(C_n)$; it is a theorem, not an axiom. Set $D_n=C_1\setminus C_n$. Since the $C_n$ shrink, the $D_n$ grow ($D_1\subseteq D_2\subseteq \cdots$) with union $C_1\setminus {\bigcap }_n{C}_n$. By continuity from below (proved from A3 in L5.1 below), $P(C_1\setminus {\bigcap }_n{C}_n)={\lim }_{n\to \infty }P(D_n).$ Finite additivity applied to the disjoint split $C_1=C_n\bigsqcup D_n$ (valid since $C_n\subseteq C_1$) gives $P(D_n)=P(C_1)-P(C_n)$, and likewise $P(C_1\setminus {\bigcap }_n{C}_n)=P(C_1)-P\left({\bigcap }_n{C}_n\right)$. Substituting and cancelling $P(C_1)$: $P\left({\bigcap }_n{C}_n\right)={\lim }_{n\to \infty }P(C_n).$ That is continuity from above. Applying it here, $P(\{0.5\})={\lim }_{n\to \infty }\frac{1}{n}=0.$ Why not impossible: the point $0.5$ is a genuine element of $\Omega$, so it can be the outcome; "probability 0" here means "almost never", not "never" — only $\varnothing$ is truly impossible.

Derivation. Treat $B\cup C$ as one event and apply the two-event rule to $A$ and $(B\cup C)$: $P(A\cup (B\cup C))=P(A)+P(B\cup C)-P(A\cap (B\cup C)).$ Expand $P(B\cup C)=P(B)+P(C)-P(B\cap C)$. For the last term use the distributive law $A\cap (B\cup C)=(A\cap B)\cup (A\cap C)$, then two-event again: $P((A\cap B)\cup (A\cap C))=P(A\cap B)+P(A\cap C)-P\underset{(A\cap B)\cap (A\cap C)}{\underbrace{(A\cap B\cap C)}}.$ Substitute all pieces back: $P(A\cup B\cup C)=P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C).■$ This matches the Inclusion–exclusion principle: add singles, subtract pairs, add the triple. Numbers. $0.6+0.5+0.4-0.3-0.2-0.2+0.1=\mathrm{0.9.}$ So $P(\text{at least one})=0.9$.

Start with $A=\{1,2\}$ and force the closure rules:

• Must contain $\Omega =\{1,2,3,4\}$ (rule 1).
• Must contain $A^c=\{3,4\}$ (complement rule).
• Must contain $\varnothing$ ($={\Omega }^c$). Now check unions/intersections: $A\cup A^c=\Omega$ (present), $A\cap A^c=\varnothing$ (present). Nothing new is generated. So $\sigma (A)=\{\varnothing ,\{1,2\},\{3,4\},\Omega \}.$ Why nothing smaller: drop any of these four and a closure rule breaks — e.g. removing $\{3,4\}$ loses the complement of $A$. It has exactly $4=2^2$ elements, matching the two "atoms" $\{1,2\}$ and $\{3,4\}$.

Idea: turn the nested sets into disjoint rings so A3 (countable additivity) applies. Define the "new material added at step $n$": $B_1=A_1,B_n=A_n\setminus A_{n-1}(n\ge 2).$ These $B_n$ are pairwise disjoint (each captures only what wasn't there before), and ${\bigcup }_{n=1}^{\infty }{B}_n=A,{\bigcup }_{k=1}^n{B}_k=A_n.$ Apply A3 to the disjoint $B_n$: $P(A)={\sum }_{n=1}^{\infty }P(B_n)={\lim }_{N\to \infty }{\sum }_{n=1}^{N}P(B_n).$ But by finite additivity the partial sum is exactly $P(A_N)$: ${\sum }_{n=1}^{N}P(B_n)=P\left({\bigcup }_{k=1}^N{B}_k\right)=P(A_N).$ Therefore $P(A)={\lim }_{N\to \infty }P(A_N).■$ This is the engine behind L3.2's shrinking-interval trick (its "from above" twin, which we derived from this result there) and behind limits used by Random variables and Expectation and Lebesgue integral.

Set-up (what "translation invariant" means). Work on the circle: identify $[0,1)$ with points on a loop, so adding a number $t$ means rotating every point by $t$ and wrapping any part past $1$ back to the start (this is addition modulo 1, written $x\oplus t$). "Translation invariant" means a set and its rotated copy get the same probability: $P(S\oplus t)=P(S)$ — the natural demand for a uniform law, since no spot on the loop is special. Construction (this step needs the axiom of choice). Call two points equivalent if they differ by a rational. This chops $[0,1)$ into disjoint equivalence classes. Form a set $V$ by picking exactly one representative from each class — selecting one point out of each of uncountably many classes at once is precisely the axiom of choice (without it no such $V$ need exist; this is why $V$ is "non-constructive" and slippery). Contradiction. Rotating $V$ by every rational $q\in [0,1)$ yields countably many disjoint copies $V_q=V\oplus q$ whose union is all of $[0,1)$. By translation invariance each copy has the same probability $p=P(V)$. Countable additivity (A3) then forces $1=P([0,1))={\sum }_{q\in \mathbb{Q}\cap [0,1)}P(V_q)={\sum }_{q}p.$ If $p=0$ the sum is $0\ne 1$; if $p>0$ the sum is $+\infty \ne 1$. No value of $p$ works. So $V$ cannot be assigned a consistent probability — it must be excluded. Hence $\mathcal{F}$ must be smaller than $2^{\Omega }$; we use the Borel sets. (This is the rigorous backbone under Measure theory & Lebesgue integration.)

• L1: check all three closure rules / all three axioms — presence of $\varnothing ,\Omega$ alone is not enough.
• L2: complement $=1-P$; union $=$ sum $-$ overlap; monotonicity comes from a disjoint split $+$ A1.
• L3: $P(A\cup B)\le P(A)+P(B)$; a point can have $P=0$ yet occur (continuity from above).
• L4: three-event I–E = singles $-$ pairs $+$ triple; smallest $\sigma (A)=\{\varnothing ,A,A^c,\Omega \}$.
• L5: continuity from below via disjoint rings $B_n$; Vitali set (axiom of choice) forces $\mathcal{F}⊊2^{\Omega }$.