Probability with Measure

1.6 An example of a non-measurable set \((\star )\)

Note that this section has a \((\star )\), meaning that it is off-syllabus. It is included for interest.

We might wonder, why go to all the trouble of defining the Borel \(\sigma \)-field? In other words, why can’t we measure (the ‘size’ of) every possible subset of \(\R \)? We will answer these questions by constructing a strange looking set \(\mathscr {V}\sw \R \); we will then show that it is not possible to define the Lebesgue measure of \(\mathscr {V}\).

As usual, let \(\Q \) denote the rational numbers. For any \(x\in \R \) we define

\begin{equation} \label {eq:Qx_def} \Q _x=\{x+q\-q\in \Q \}. \end{equation}

Note that different \(x\) values may give the same \(\Q _x\). For example, an exercise for you is to prove that \(\Q _{\sqrt {2}}=\Q _{1+\sqrt {2}}\). You can think of \(\Q _x\) as the set \(\Q \) translated by \(x\).

It is easily seen that \(\Q _x\cap [0,1]\) is non-empty; just pick some rational \(q\) that is slightly less than \(x\) and note that \(x+(-q)\in \Q _x\cap [0,1]\). Now, for each set \(\Q _x\), we pick precisely one element \(r\in \Q _x\cap [0,1]\) (it does not matter which element we pick). We write this number \(r\) as \(r(\Q _x)\). Define

\[\mathscr {V}=\{r(\Q _x)\-x\in \mathbb {R}\},\]

which is a subset of \([0,1]\). For each \(q\in \Q \) define

\[\mathscr {V}_q=\{q+m\-m\in \mathscr {V}\}.\]

Clearly \(\mathscr {V}=\mathscr {V}_0\), and \(\mathscr {V}_q\) is precisely the set \(\mathscr {V}\) translated by \(q\). Now, let us record some facts about \(\mathscr {V}_q\).

Before we prove this lemma, let us use it to show that \(\mathscr {V}\) cannot have a Lebesgue measure. We will do this by contradiction: assume that \(\lambda (\mathscr {V})\) is defined.

Since \(\mathscr {V}\) and \(\mathscr {V}_q\) are translations of each other, they must have the same Lebesgue measure. We write \(c=\lambda (\mathscr {V})=\lambda (\mathscr {V}_q)\), which does not depend on \(q\). Let us write \(\Q \cap [-1,1]=\{q_1,q_2,\ldots ,\}\), which we may do because \(\Q \) is countable. By parts (1) and (3) of Lemma 1.6.1 and property (M2) we have

\[\lambda \l (\bigcup _{q\in \Q \cap [-1,1]} \mathscr {V}_q\r )=\sum \limits _{i=1}^\infty \lambda (\mathscr {V}_{q_i})=\sum \limits _{i=1}^\infty c.\]

Using the monotonicity property of measures (see Section 1.7) and part (3) of Lemma 1.6.1 we thus have

\[1\leq \sum \limits _{i=1}^\infty c\leq 3.\]

However, there is no value of \(c\) which can satisfy this equation! So it is not possible to define of the Lebesgue measure of \(\mathscr {V}\). Since we know that we can define the Lebesgue measure on all Borel sets, the set \(\mathscr {V}\) is not a Borel set.

The set \(\mathscr {V}\) is known as a Vitali set. In higher dimensions even stranger things can happen with non-measurable sets; you might like to investigate the Banach-Tarski paradox.

Proof:[Of Lemma 1.6.1.] We prove the three claims in turn.

(1) Let \(q_1,q_2\in \Q \) be unequal. Suppose that some \(x\in \mathscr {V}_{q_1}\cap \mathscr {V}_{q_2}\) exists – and we now look for a contradiction. By definition of \(\mathscr {V}_q\) we have

\begin{equation} \label {eq:xqr} x=q_1+r(\Q _{x_1})=q_2+r(\Q _{x_2}). \end{equation}

By definition of \(\Q _x\) we may write \(r(\Q _{x_1})=x_1+q'_1\) for some \(q'_1\in \Q \), and similarly for \(x_2\), so we obtain \(x=q_1+x_1+q'_1=q_2+x_2+q'_2\) where \(q,q'\in \Q \). Hence, setting \(q=q_2-q_1+q'_2-q'_1\in \Q \), we have \(x_1+q=x_2\), which by (1.8) means that \(\Q _{x_1}=\Q _{x_2}\). Thus \(r(\Q _{x_1})=r(\Q _{x_2})\), so going back to (1.9) we obtain that \(q_1=q_2\). But this contradicts our assumption that \(q_1\neq q_2\). Hence \(x\) does not exist and \(\mathscr {V}_{q_1}\cap \mathscr {V}_{q_2}=\emptyset \).

(2) We will show \(\supseteq \) and \(\subseteq \). The first is easy: since \(\mathscr {V}_q\sw \R \) it is immediate that \(\R \supseteq \bigcup _{q\in \Q } \mathscr {V}_q\).

Now take some \(x\in \R \). Since we may take \(q=0\) in (1.8) we have \(x\in \Q _x\). By definition of \(r(\Q _x)\) we have \(r(\Q _x)=x+q'\) for some \(q'\in \Q \). By definition of \(\mathscr {V}\) we have \(r(\Q _x)\in \mathscr {V}\) and since \(x=r(\Q _x)-q'\) we have \(x\in \mathscr {V}_{-q'}\). Hence \(x\in \bigcup _{q\in \Q } \mathscr {V}_q\).

(3) Since \(\mathscr {V}\sw [0,1]\), we have \(\mathscr {V}_q\cap [0,1]=\emptyset \) whenever \(q\notin [-1,1]\). Hence, from part (2) and set algebra we have

\[\R \cap [0,1]\;=\;\l (\bigcup _{q\in \Q }\mathscr {V}_q\r )\cap [0,1]\;=\;\bigcup _{q\in \Q }\mathscr {V}_q\cap [0,1] \;=\; \bigcup _{q\in \Q \cap [-1,1]}\mathscr {V}_q\cap [0,1]\;\sw \; \bigcup _{q\in \Q \cap [-1,1]}\mathscr {V}_q.\]

This proves the first \(\subseteq \) of (3). For the second simply note that \(\mathscr {V}\sw [0,1]\) so \(\mathscr {V}_q\sw [-1,2]\) whenever \(q\in [-1,1]\).   ∎