Thomas J. Jech, on the second page of his The Axiom of Choice (1973) proves the existence of a nonmeasurable set of real numbers thus:
Let $\mu(X)$ denote the Lebesgue measure of a set $X$ of real numbers. We know that $\mu$ is countably additive and translation invariant, and that $\mu([a, b]) = b - a$ for every closed real interval $[a, b]$.
For $x, y \in [0, 1]$, define the relation $x \sim y$ if and only if $x - y$ is rational.
$\sim$ is a equivalence relation, so denote $[[x]]$ as the equivalence class with respect to $\sim$ for each $x \in [0, 1]$.
By the Axiom of Choice we can choose one element out of each equivalence class.
Thus we have a set $M$ of real numbers $M \subseteq [0, 1]$ which has the property that for each real $x$ there exists a unique $y \in M$ and a unique rational $r$ such that $x = y + r$.
Let $M_r := \lbrace{y + r: y \in M}\rbrace$ for each rational number $r$.
Then we have a partition of $\mathbb R$ into a countable number of disjoint sets: $$(1.1) \qquad \mathbb R = \bigcup \lbrace{M_r: r \in \mathbb Q}\rbrace$$ where $\mathbb Q$ denotes the set of rationals.
Suppose $M$ is measurable.
First we note that $\mu(M) = 0$ is impossible as that would mean $\mu(\mathbb R) = 0$ using $(1.1)$.
But then $\mu(M) > 0$ means: $$\mu([0, 1]) \ge \mu \left({\bigcup \lbrace{M_r: \text{$r \in \mathbb Q$ and $0 \le r \le 1$} }\rbrace }\right)$$
$$= \sum_{r \in \mathbb Q: 0 \le r \le 1} \mu(M_r) = \infty$$
because each $M_r$ would have to have the same measure as $M$. $\blacksquare$
That's the gist of it.
The question has been raised as to the validity of the line:
$$\mu([0, 1]) \ge \mu \left({\bigcup \lbrace{M_r: \text{$r \in \mathbb Q$ and $0 \le r \le 1$} }\rbrace }\right)$$
A respondent claims that the above is wrong and a mistake in Jech, and that it should be:
$$\mu([0, 2]) \ge \mu \left({\bigcup \lbrace{M_r: \text{$r \in \mathbb Q$ and $0 \le r \le 1$} }\rbrace }\right)$$
While that latter statement is indeed obvious, it has been pointed out that it is in no way clear that: $$\mu([0, 1]) \ge \mu \left({\bigcup \lbrace{M_r: \text{$r \in \mathbb Q$ and $0 \le r \le 1$} }\rbrace }\right)$$ because while:
$$[0, 2] \supseteq \bigcup \lbrace{M_r: \text{$r \in \mathbb Q$ and $0 \le r \le 1$} }\rbrace$$
it is most definitely not the case that:
$$[0, 1] \supseteq \bigcup \lbrace{M_r: \text{$r \in \mathbb Q$ and $0 \le r \le 1$} }\rbrace$$
because elements of $M_r$ are of the form $y + r$ where $0 \le y \le 1$ and $0 \le r \le 1$.
So, did Jech make a mistake when he wrote:
$$\mu([0, 1]) \ge \mu \left({\bigcup \lbrace{M_r: \text{$r \in \mathbb Q$ and $0 \le r \le 1$} }\rbrace }\right)$$
... or is it just that he used something subtle in the above?
I can see where my respondent is coming from, but before I am able to declare that Jech is actually wrong I would welcome a second opinion and perhaps a proof as to exactly why he is wrong, by demonstrating that:
$$\mu([0, 1]) < \mu \left({\bigcup \lbrace{M_r: \text{$r \in \mathbb Q$ and $0 \le r \le 1$} }\rbrace }\right)$$
