If $V$ is a vector space and $m$ is the counting measure, then $$\dim(V) = \inf \{m(U) : U \subset V, \text{ span}(U) = V\}.$$

Given a measure space $(V, \mathcal M, \mu)$ such that $V$ is a vector space, have people considered the following definition? $$\dim_\mu(V) := \inf \{\mu(U) : U \in \mathcal M, \text{ span}(U) = V\}$$

In particular, considering $\mathbb R$ as a $\mathbb Q$-vector space, is there a measure $\mu$ such that $0 < \dim_\mu(\mathbb R/\mathbb Q) < \infty$ ?

  • $\begingroup$ You probably want to rule out trivial examples, for example when $\mu$ is concentrated on a finite set. $\endgroup$ Commented Mar 29, 2013 at 6:49
  • $\begingroup$ @Evan, You can throw a Vitali set (Hamel basis for $\mathbb R / \mathbb Q$) into any arbitarily small interval, so you can avoid any finite set. $\endgroup$ Commented Mar 29, 2013 at 13:56
  • $\begingroup$ @Ewan, sorry Ewan, not Evan. $\endgroup$ Commented Mar 29, 2013 at 14:17

2 Answers 2


Consider the $\mathbb Q$-vector space $\mathbb R$. Of course $1 \in \mathbb R$. Let $V \subseteq \mathbb R$ be a $\mathbb Q$-subspace complementary to $\mathbb Q\cdot 1$. (A subspace of codimension $1$. In fact, any subspace of codimension $1$ will work for this. Of course some form of AC is needed to get $V$. For example, take a Hamel base and then use the span of all but one element. The existence of $V$ may be considerably weaker than the existence of a Hamel base, though.)

Now let $\mu$ be counting measuse on the complement $\mathbb R \setminus V$. Note that every subset of $\mathbb R$ is $\mu$-measurable. Two observations: (a) $V \cup \{1\}$ is a spanning set, and has measure $1$. (b) No subset of $V$ is a spanning set, so for every spanning set $E$ we have $\mu(E) \ge 1$.

Therefore $\dim_\mu(\mathbb R / \mathbb Q) = 1$.

  • $\begingroup$ If you insist on specifying that we use the Borel sets of $\mathbb R$, this will not do the same thing. Then $V$ would have infinite outer measure. $\endgroup$
    – GEdgar
    Commented Mar 29, 2013 at 16:16
  • $\begingroup$ Thanks for your answer. Do you know what will happen if we allow countable linear combinations? I guess you can do the same trick with a Schauder basis instead of a Hamel base. But does $\mathbb R$ have a Schauder basis? $\endgroup$ Commented Mar 29, 2013 at 18:23
  • $\begingroup$ Over $\mathbb Q$, of course. $\endgroup$ Commented Mar 29, 2013 at 18:31
  • $\begingroup$ These questions are probably obvious to anyone who knows functional analysis. Maybe I'll revisit this once I've taken a functional analysis class! $\endgroup$ Commented Mar 29, 2013 at 18:35

EDIT: as GEdgar pointed out, this does not answer the question.

This question is actually not very interesting; maybe it can be modified. Just let $\mu$ have weight $1$ on each of the rationals. Any Vitali set contains exactly one rational, so will have measure 1.

  • 1
    $\begingroup$ But there are spanning sets that contain no rationals. So we get $\dim_\mu(\mathbb R/\mathbb Q) = 0$. $\endgroup$
    – GEdgar
    Commented Mar 29, 2013 at 15:48
  • $\begingroup$ Oops, I was mistakenly assuming linear independence. $\endgroup$ Commented Mar 29, 2013 at 18:10

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