[example: Why Events Need a Sigma Algebra]
Let $\Omega=[0,1]$. Start with the open subintervals of $[0,1]$, and close them under complements in $\Omega$ and countable unions; the resulting collection is the Borel sigma algebra $\mathcal{B}([0,1])$. Because a sigma algebra is also closed under countable intersections, intervals such as $[a,b]$ are Borel: for $0\le a\le b\le 1$,
\begin{align*}
[a,b]=(a,1]\cap[0,b)
\end{align*}
where $(a,1]=\Omega\setminus[0,a]$ and $[0,b]=\Omega\setminus(b,1]$, and these sets are obtained from open intervals by complements and countable unions.
On $\mathcal{B}([0,1])$, the uniform probability measure is Lebesgue length restricted to $[0,1]$, so
\begin{align*}
P([a,b])=b-a
\end{align*}
whenever $0\le a\le b\le 1$; in particular,
\begin{align*}
P([0,1])=1-0=1.
\end{align*}
The restriction to a sigma algebra is not cosmetic. If one tried to assign a probability to every subset while keeping countable additivity and [translation invariance](/theorems/4911) modulo $1$, a Vitali-type obstruction appears. Choose one representative from each equivalence class of $[0,1)$ under $x\sim y$ when $x-y\in\mathbb{Q}$, and call the resulting set $V$. For rational $r\in[0,1)\cap\mathbb{Q}$, let
\begin{align*}
V+r=\{(v+r)\bmod 1:v\in V\}.
\end{align*}
If $(V+r)\cap(V+s)$ contained a point, then for some $v,w\in V$ we would have $(v+r)\bmod 1=(w+s)\bmod 1$, hence $v-w\in\mathbb{Q}$, so $v=w$ by the choice of one representative from each equivalence class, and then $r=s$. Thus the sets $V+r$ are pairwise disjoint. They also cover $[0,1)$, because every $x\in[0,1)$ is rationally equivalent to exactly one representative $v\in V$, so $x=(v+r)\bmod 1$ for some rational $r\in[0,1)$.
If such a probability measure measured all subsets and were invariant under these translations, then every set $V+r$ would have the same value $P(V)$. Countable additivity would give
\begin{align*}
1=P([0,1))=\sum_{r\in[0,1)\cap\mathbb{Q}}P(V+r)=\sum_{r\in[0,1)\cap\mathbb{Q}}P(V).
\end{align*}
If $P(V)=0$, the right side is $0$; if $P(V)>0$, the right side diverges to infinity. Both contradict $P([0,1))=1$. The sigma algebra therefore records exactly which events the model is prepared to measure while preserving the probability rules.
[/example]