Not all sets are measurable. Within $[0,1]$ there exists unmeasurable sets. We’ll construct one here. First define an equivalence class $x \sim y \iff x - y \in \mathbb{Q}$. The set $[0,1]$ contains many such equivalence classes(which by definition are disjoint from each other). By axiom of choice, we can choose a representative from each equivalence class. Let $A$ be a set containing the representatives. We construct a disjoint cover:
\[[0, 1] \subset \bigcup_{r \in \mathbb{Q}_{[-1,1]}}\{a+r: a \in A\} := U\]Where \(\mathbb{Q}_{[-1,1]} := \{x: x \in \mathbb{Q}, -1 \leq x \leq 1\}\). Denote each \(X_r := \{a+r:a \in A\}\). $U$ covers $[0,1]$ because \(\forall x \in [0,1], x \sim a\) for some $a \in A$ by definition, and so \(x - a \in \mathbb{Q}_{[-1,1]}\). For $r \neq s$, $X_r \cap X_s = \emptyset$ because suppose they’re not disjoint, then:
\[\begin{align} x \in X_r \cap X_s \\ \implies \exists a_r, a_s \in A \text{ s.t. } a_r+r = a_s + s = x \\ \implies a_r - a_s = s-r \in \mathbb{Q} \\ \implies a_r \sim a_s \\ \implies a_r = a_s \\ \implies r = s \end{align}\]Which is a contradiction. Also, $U \subset [-1, 2]$ because
\[r \geq -1, a \geq 0 \implies a+r \geq -1 \forall a \in A, r \in \mathbb{Q}_{[-1,1]}\]and
\[r \leq 1, a \leq 1 \implies a+r \leq 2 \forall a \in A, r \in \mathbb{Q}_{[-1,1]}\]Because lebesgue measures are translation invariant:
\(\begin{align} \mu(X_r) = \mu(A) \quad \forall r \in \mathbb{Q}_{[-1,1]} \\ \implies \mu([0,1]) \leq \mu(\bigcup_{r \in \mathbb{Q}_{[-1,1]}} X_r) \leq \mu([-1,2]) \\ 1 \leq \sum_{r \in \mathbb{Q}_{[-1,1]}} \mu(A) \leq 3 \end{align}\) which is impossible.
Knowing that some sets aren’t measurable is important - it means we can’t assign a probability to specific sets of outcomes. We need a more rigorous definition for probability theory, and that’s where measure theory comes in.