# Existence of countably additive, translation invariant measure on real line

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?

Thanks!

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.
• Thanks! Seems like this was a dumb question after all and had the obvious answer of just applying the Vitali set construction to $M$ 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.