# A set that is not Lebesgue measurable

The Lebesgue measure $\lambda$ on the real line is a countably additive measure that assigns to each interval $[a,b]$ with $a \leq b$ its length $b-a$. Why construct the Lebesgue measure? It’s so that we can get around a blemish of the Riemann integral: namely, that the Riemann integration theory does not know how to integrate the characteristic function of the rational numbers, since the rationals and irrationals are dense in every interval.

However, the Lebesgue measure on the real line only is defined for certain subsets on the real line. The Lebesgue measurable subsets of the real line form a $\sigma$-algebra that contains the Borel $\sigma$-algebra generated by all the open intervals. That is certainly quite a lot of sets. But today, we are going to see a subset of $\R$ that is not Lebesgue-measurable.

## The set

We define an equivalence relation $\sim$ on $\R$ by setting $x\sim y$ if and only if $x-y\in \Q$. Let $A$ be a set contained in $(0,1)$ that contains exactly one element from each equivalence class. Such a set $A$ exists because every equivalence class intersects $(0,1)$, and we can select one element from each equivalence class by the axiom of choice, the greatest of all the set-theoretic axioms.

Theorem. The set $A$ is not Lebesgue measurable.

Proof.Let $r_1,r_2,\dots$ be an enumeration of every rational number in $\Q\cap (-1,1)$ and set $A_i = A + r_i$. It is easy to see that $A_i\cap A_j = \varnothing$ whenever $i\not= j$.

Suppose $A$ is Lebesgue measurable. Then each $A_i$, being a translation of $A$, is also Lebesgue measurable. Since the $A_i$ are pairwise disjoint,
$$\lambda(\cup A_i) = \sum \lambda(A).$$

Let $x\in (0,1)$. Then by definition of $A$ there exists a $b\in A$ such that $x-b\in \Q$ and moreover $x-b\in (-1,1)$ since both $x$ and $b$ are in $(0,1)$. Therefore, $x-b=r_k$ for some $k$ and hence $x\in \cup A_i$. That means $(0,1)\subseteq \cup A_i$ since $x$ was arbitrary, so the set $\cup A_i$ cannot have measure zero by monotonicity of the Lebesgue measure. So $\lambda(A)\not= 0$.

On the other hand, if $\lambda(A)\not=0$ then $\lambda(\cup A_i) = \infty.$ But each $A_i$ is contained in $(-1,2)$ since each $A_i$ is a translation of the set $A$ contained in $(0,1)$ by an element contained in $(-1,1)$, Therefore $\cup A_i$ must have finite Lebesgue measure, a contradiction.

Hence, $A$ actually is not Lebesgue measurable at all!

We should be aware that the construction of a set that is not Lebesgue measurable depended on the axiom of choice, which states that for each set of nonempty sets, there exists a set that contains exactly one element from each set. In fact, if we assume only the ZF axioms of set theory, it is impossible to construct a set that is not Lebesgue measurable. In fact, there is an alternate axiom that you can tack onto ZF, called the Axiom of Determinateness (AD), which still gives us some power for reasoning about sets and under which every subset of $\R$ is Lebesgue-measurable.

In fact, reading a lot about alternative axioms gives me the impression that these axiom systems were constructed because these results feel ugly to some. In fact, Herrlich in his book Axiom of Choice puts the result about every set being Lebesgue measurable under ZF+AD in the chapter entitled “Beauty without choice”. I may be in the minority, but I actually like this result and I think the Banach-Tarski “paradox” is also cool.

Moreover, the axiom of choice seems intuitive to me so I would not want to do mathematics without it.

### 2 Comments

• One nice way around nonmeasurable sets is to look at the reals as a locale rather than a mere set. Alex Simpson’s ‘Measure, randomness and sublocales’ (https://www.sciencedirect.com/science/article/pii/S0168007211001874?via%3Dihub) shows that you can extend the Lebesgue measure to every sublocale of the reals in a countably additive and translation invariant way. The usual constructions of nonmeasurable sets fail because the sublocales corresponding to the nonmeasurable sets are not disjoint. All of this is true regardless of whether or not we take the Axiom of Choice to be true.

I’ve been wondering if we can do the same trick without the extra machinery of locales. Instead of pointless topology just use pointless sets. If we have a topos with a natural numbers object but we don’t assume well pointedness, is it possible to construct nonmeasurable sets even with the Axiom of Choice?

• Jason Polak says:

Thanks for the link to Simpson’s paper. I’ll have to take a closer look as it is a very interesting idea. Based on an initial reading, it seems like the locale approach is like the usual construction of the Lebesgue measure through Lebesgue outer measure, yet the restriction to Lebesgue measurable sets in the locale approach is *sort of* like just having the Lebesgue measurable sets as locales, whereas the non-measurable sets aren’t locales in the sense that sets like $\Q$ and $\R-\Q$ (as mentioned in the paper) are not disjoint *as locales*. So it seems like the moral of the paper is that plain old sets are not really the way we should be thinking of Euclidean space when it comes to measure theory and analysis at all.

I think this is a nicer way of approaching how ZFC causes some oddities in analysis, compared to just throwing away choice altogether and replacing it with some more ‘intuitive’ notion of choice like dependent choice or some other axiom altogether (none of which really seems very intuitive to me).

As for your question, I think that is also quite interesting. I’d have to learn more about toposes as I am not too familiar with this type of mathematics, but from what little I know I would certainly like to learn if such a construction is possible.