Ngô Quốc Anh

August 15, 2009

Several questions involving the Vitali set

We denote by V the Vitali set which is defined as follows:

We say that x, y \in  [0, 1) are equivalent, and write x  \sim y, if and only if x - y is a rational number. This equivalence relation partitions [0, 1) into an uncountable family of disjoint equivalence classes. By the axiom of choice there is a set V which contains exactly one element from each equivalence class.

Now let \{ \tau_n\} be a sequence of all rationals in [0, 1) with \tau_0 =0 and define V_n = V + \tau_n (mod 1).

Now we show that the V_n are pairwise disjoint and

\bigcup\limits_n {V_n }  = \left[ {0,1} \right).

Indeed, if x \in V_i \cap V_j, then x = v_i+\tau_i (mod 1) and x = v_j+\tau_j (mod 1), with v_i and v_j belonging to V = V_0. Consequently, v_i - v_j \in \mathbb Q, which means that v_i \sim v_j and therefore i = j. This shows that V_i \cap V_j = 0 if i \ne j. Since each x  \in [0, 1) is in some equivalence class, x differs modulo 1 from an element in V by a rational number, say r, in [0, 1). Thus x \in V_k, which proves that

[0, 1) \subset \bigcup\limits_n {V_n }.

The opposite inclusion is obvious.

Question 1. Show that there exist sets E_1, E_2, ...,E_k,... such that E_k  \searrow E, and |E_k|_e < \infty and

\lim_{k \to \infty} |E_k|_e > |E|_e

with strict inequality.

Solution. We put E_n  = \bigcup\limits_{k \geq n} {V_k }. Clearly, \{E_n\} is a decreasing sequence. Since the V_k are pairwise disjoint, we see that E: = \bigcap\limits_n {E_n }  = \emptyset and E_n  \searrow E. Moreover,

\left| {E_n } \right|_e  \geq  \left| {V_n } \right|_e  = \left| V \right|_e  > 0

(the last inequality comes from the fact that V is not measurable). It is now enough to show that

\mathop {\lim }\limits_{n \to \infty } \left| {E_n } \right|_e  \geq  \left| V \right|_e  > 0 = \left| E \right|_e

and the proof is complete.

Question 2. Show that there exist disjoint E_1, E_2, ...,E_k,... such that

\left| { \bigcup_k E_k } \right|_e  < \sum_k {\left| {E_k } \right|_e }

with strict inequality.

Solution. We put E_n = V_n then E_n are pairwise disjoint and obviously

\left| {\bigcup_n {E_n } } \right|_e  = \left| {\left[ {0,1} \right)} \right|_e  = 1.

Moreover, all the E_n are of the same outer measure. Thus \sum_n {\left| {E_n } \right|_e }  =  + \infty which completes the proof.

Question 3. Show that each of the sets

{E_n} = \bigcup\limits_{k = 0}^n {{V_k}}

is non-measurable.

Question 4. Show that if E is a measurable subset of the Vitali set V, then |E|=0.

Question 5. Show that there exist sets A and B such that

{\left| {A \cup B} \right|_i} = {\left| A \right|_i} + {\left| B \right|_i}


{\left| {A \cup B} \right|_e} < {\left| A \right|_e} + {\left| B \right|_e}.

Question 6. Show that any set of positive outer measure contains a non-measurable subset.

Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Blog at

%d bloggers like this: