The Lebesgue measure $\lambda$ on the real line is a countably additive measure that assigns to each interval $[a,b]$ with $a \leq b$ its length $b-a$. Why construct the Lebesgue measure? It’s so that we can get around a blemish of the Riemann integral: namely, that the Riemann integration theory does not know how to integrate the characteristic function of the rational numbers, since the rationals and irrationals are dense in every interval.
However, the Lebesgue measure on the real line only is defined for certain subsets on the real line. The Lebesgue measurable subsets of the real line form a $\sigma$-algebra that contains the Borel $\sigma$-algebra generated by all the open intervals. That is certainly quite a lot of sets. But today, we are going to see a subset of $\R$ that is not Lebesgue-measurable.
We define an equivalence relation $\sim$ on $\R$ by setting $x\sim y$ if and only if $x-y\in \Q$. Let $A$ be a set contained in $(0,1)$ that contains exactly one element from each equivalence class. Such a set $A$ exists because every equivalence class intersects $(0,1)$, and we can select one element from each equivalence class by the axiom of choice, the greatest of all the set-theoretic axioms.
Suppose $A$ is Lebesgue measurable. Then each $A_i$, being a translation of $A$, is also Lebesgue measurable. Since the $A_i$ are pairwise disjoint,
$$\lambda(\cup A_i) = \sum \lambda(A).$$
Let $x\in (0,1)$. Then by definition of $A$ there exists a $b\in A$ such that $x-b\in \Q$ and moreover $x-b\in (-1,1)$ since both $x$ and $b$ are in $(0,1)$. Therefore, $x-b=r_k$ for some $k$ and hence $x\in \cup A_i$. That means $(0,1)\subseteq \cup A_i$ since $x$ was arbitrary, so the set $\cup A_i$ cannot have measure zero by monotonicity of the Lebesgue measure. So $\lambda(A)\not= 0$.
On the other hand, if $\lambda(A)\not=0$ then $\lambda(\cup A_i) = \infty.$ But each $A_i$ is contained in $(-1,2)$ since each $A_i$ is a translation of the set $A$ contained in $(0,1)$ by an element contained in $(-1,1)$, Therefore $\cup A_i$ must have finite Lebesgue measure, a contradiction.
Hence, $A$ actually is not Lebesgue measurable at all!
We should be aware that the construction of a set that is not Lebesgue measurable depended on the axiom of choice, which states that for each set of nonempty sets, there exists a set that contains exactly one element from each set. In fact, if we assume only the ZF axioms of set theory, it is impossible to construct a set that is not Lebesgue measurable. In fact, there is an alternate axiom that you can tack onto ZF, called the Axiom of Determinateness (AD), which still gives us some power for reasoning about sets and under which every subset of $\R$ is Lebesgue-measurable.
In fact, reading a lot about alternative axioms gives me the impression that these axiom systems were constructed because these results feel ugly to some. In fact, Herrlich in his book Axiom of Choice puts the result about every set being Lebesgue measurable under ZF+AD in the chapter entitled “Beauty without choice”. I may be in the minority, but I actually like this result and I think the Banach-Tarski “paradox” is also cool.
Moreover, the axiom of choice seems intuitive to me so I would not want to do mathematics without it.