I'm teaching a basic measure theory course using Royden's book, and told my students that there does not exist a set function $M : \mathcal{P}(\mathbb{R}) \rightarrow [0, \infty]$ such that

$1) M(I) = \ell(I)$ where $I$ is an interval and $\ell(I)$ is the length of $I$

$2) M\left(\bigcup_{k = 1}^\infty E_k\right) = \sum_{k = 1}^\infty M(E_k)$ for pairwise disjoint $E_k$.

$3) M(E + x) = M(E)$ for all $E \subseteq \mathbb{R}$ and $x \in \mathbb{R}$.

I just proved the existence of a non Lebesgue measurable set and wanted to go back and show above, but hit a roadblock when trying to prove this and I can't seem to find a good reference.

In particular, the strategy is to prove that the conditions above imply that $ m^* (E) = M(E)$ for all $E \subseteq \mathbb{R}$ where $m^*$ is Lebesgue outer measure. Then clearly these conditions (namely $2)$ and $3)$) would imply that every set is Lebesgue measurable, which is false.

This doesn't seem to hard to prove by Dynkin's Lemma for all Borel sets $E$, and in general it's easy to prove that $m^* (E) \geq M(E)$ but I'm having a hard time proving $m^* (E) \leq M(E)$ for all $E \subset \mathbb{R}$. Is this even true? Is there a better way to do this?



2 Answers 2


Let $\sim$ be the equivalence relation on $[0, 1]$ given by $x \sim y$ iff $x - y \in \mathbb{Q} \cap [-1, 1]$. Use the Axiom of Choice to choose a set $E \subset \mathbb{R}$ which contains exactly one element from each equivalence class. If such an $M$ exists, then $M(E)$ is well-defined and $M(E + q) = M(E)$ for all $q \in \mathbb{Q} \cap [-1, 1]$. But by definition of $E$, $(E + q_1) \cap (E + q_2) = \varnothing$ whenever $q_1 \neq q_2 \in \mathbb{Q} \cap [-1, 1]$. Thus,

$$M(\cup_{q \in \mathbb{Q} \cap [-1, 1]} (E + q)) = \sum_{q \in \mathbb{Q} \cap [-1, 1]} M(E + q) = \sum_{q \in \mathbb{Q} \cap [-1, 1]} M(E) = \infty \cdot M(E)$$

We observe that $\cup_{q \in \mathbb{Q} \cap [-1, 1]} (E + q) \subset [-1, 2]$, so $M(\cup_{q \in \mathbb{Q} \cap [-1, 1]} (E + q)) < \infty$ and therefore $M(E) = 0$, whence $M(\cup_{q \in \mathbb{Q} \cap [-1, 1]} (E + q)) = 0$. On the otherhand, $\cup_{q \in \mathbb{Q} \cap [-1, 1]} (E + q) \supset [0, 1]$, so we must have $1 = M([0, 1]) \leq M(\cup_{q \in \mathbb{Q} \cap [-1, 1]} (E + q)) = 0$, a contradiction.

  • $\begingroup$ This is, of course, just the Vitali set. $\endgroup$
    – David Gao
    Commented Feb 22 at 23:34
  • $\begingroup$ Thanks! Seems like this was a dumb question after all and had the obvious answer of just applying the Vitali set construction to $M$ $\endgroup$ Commented Feb 23 at 0:01

If you're defining the outer measure in the standard way, i.e. $$m^*(A):=\inf \Bigl\{\sum_{n \in \mathbb{N}} M(A_n): A_n \in \mathcal{P(\mathbb{R})}, \: A \subset \bigcup_{n \in \mathbb{N}}A_n \Bigr\}$$ the inequality follows trivially since given any set $E \subset \mathbb{R}$, then $E \in \mathcal{P}(\mathbb{R})$ thus the set is a cover itself which leads to $m^*(E) \leq M(E)$.
I think the confusion is that the usual formulation is to start with a pre-measure defined on some subset $\mathcal{A}$ of $\mathcal{P}(\mathbb{R})$ which means the outer measure is $$m^*(A):=\inf \Bigl\{\sum_{n \in \mathbb{N}} M(A_n): A_n \in \mathcal{A}, \: A \subset \bigcup_{n \in \mathbb{N}}A_n \Bigr\}$$ But in your case, you're assuming $\mathcal{A}=\mathcal{P}(\mathbb{R})$ and then deriving a contradiction.


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