# Pursuing the ideal measure

In the last post, I have listed 4 properties I want an ideal measure on the real line to fulfill. The properties were:

1. For every $E\subset \mathbb{R}$, $m(E)$ is defined and satisfies $0\leq m(E)\leq \infty$
2. For every interval $E\subset \mathbb{R}$, $m(E)=|E|$.
3. If we ‘shift’ / ‘slide’ a set, we don’t want the output of the measure to change.
4. If we consider the disjoint union $E=\bigcup_{n=1}^{\infty}E_{n}$ we want that: $m(E)=\sum_{n=1}^{\infty}m(E_{n})$.

However, at the end of the post, I said that there is no such a ‘perfect’ function. Here I want to explain why is that using a proof of the mathematician Giuseppe Vitali. He found a problematic set(s), and by that, showed why such a function doesn’t exist.

Before I am going into the details of the proof I want to give some notes of it

1. The construction of the set is not really intuitive, and actually uses the axiom of choice. You will see in the proof that it probably took some time to come up with such an idea.
2. The proof aims for contradiction, We are going to get bounds for an infinte series that just won’t make sense. Even though the proof is nice, elegant and most important – it works, I still don’t ‘feel’ why it works. It seems like the set was made for nothing but to be a counter-example. However, this is not the first time I am seeing such a proof with similar purpose.

## The proof

Aiming for contradiction, suppose that there is such a function $m$ that fulfills all of the 4 properties in the above. I am going now to define an equivalence relation on $\mathbb{R}$:

x\sim y \iff x-y\in\mathbb{Q}

There are uncountably many equivalence classes, and it is not hard to see that each class intersects with the interval $(0,1)$.

Using the axiom of choice, we can create a set $E\sub (0,1)$ which is made of only one elements from each class. This set is going to cause us problems soon… Now I want to prove 2 things:

1. if $x\in (0,1)$, there exists $r\in \mathbb{Q}\cap (-1,1)$ such that $x\in E + r$.
2. if $r \neq s$ are two rationals, then $(E+r)\cap(E+s)=\emptyset$.

Let’s prove those statements:

1. if $x\in (0,1)$, then there is a unique $y\in E$ such that $x\sim y$, then $x-y=r\in\mathbb{Q}$, thus, $x=y+r\in E+r$ and since $y\in(0,1)$ we get $r = x- y \in (-1,1)$ and that completes the proof.
2. Suppose that $x\in (E+r)\cap(E+s)$. Then there are $y,z\in E$ such that $x=y+r=z+s\Rightarrow y-z=s-r\in\mathbb{Q}$ thus $y\sim z$ which is a contradiction to the construction of $E$ (only one element from each class).

Great, we are now equipped with two statements that will assist us soon. I am going to write all the rationals in $(-1,1)$ as a sequence $\{r_n\}_{n\in\mathbb{N}}$. Using the first statement we get that $(0,1) \sub \bigcup_n E+r_n$.

On the other hand, We also know that $E\sub (0,1), r_n\in (-1,1)$ for any natural $n$, thus $E+r_n\sub (-1,2)$. We can now conclude that $\bigcup_n E+r_n\sub (-1,2)$. So far we have found that :

(0,1)\sub\bigcup_nE+r_n\sub (-1,2)

The natural thing to do now is to apply the measure function, and use the property: $A\sub B \Rightarrow m(A) \leq m(B)$ to get:

m((0,1))\leq m(\bigcup_nE+r_n)\leq m((-1,2))

Now I’ll use the second property to get $m((0,1)) = |(0,1)| = 1, m((-1,2)) = |(-1,2)| = 3$. Thus:

1\leq m(\bigcup_nE+r_n)\leq 3

Because of the second lemma we now that the union $\bigcup_nE+r_n$ is disjoint, so we can use property 4 to get:

1\leq \sum_nm(E+r_n)\leq 3

We now use the third property to get $m(E+r_n) = m(E)$, which gives us:

1\leq \sum_nm(E)\leq 3

Because of the first property we know that $0\leq m(E) \leq \infty$. If $m(E) = 0$, the sum equals 0 which is a contradiction to it being at least one. Otherwise, the sum is infinity, which is a contradiction to it being bounded by 3. And This completes the proof.

## proof review

After you have seen the complete proof, what do you have to say about it? is it intuitive? do you thinks it explains well why such a function can not exist? Think about it.

Im my opinion, after a few rewinds and reviews, I think this proof is kind of not intuitive only because the use of the axiom of choice. However, once we are done using the axiom of choice, I feel like the proof is pretty straightforward and I can understand how someone thought on these two specific short statements in the proof.

So we have proved this theorem that killed our dream of finding an “ideal” measure function, and that also explains why I haven’t fully proved that property number 4 hold for the outer measure, it turns out that it is just not true, and there are counter-examples for that!

## Now what?

So as it turns out, unfortunately we can’t find such an ideal measure, our only choice now is to give up on one of the properties, and that’s what we are going to do to in the next post, we are going to make property 1 weaker. How much weaker? will it be enough? all of that will be in the next one!