# Measure Theory/There is No Total Measure of Real Numbers
## There Is no Total Measure of Real Numbers

In this lesson we are going to start by assuming that there is a "total measure of real numbers", and then derive a contradiction.
(Note that I am making up the phrase "total measure of real numbers", as this is not common language. Especially the word "total" is unusual here, but I am use it to emphasize that the measure we are trying to discuss here, has domain equal to ${\mathcal {P}}({\mathbb {R}})$. That is to say, its domain is EVERY set of real numbers -- the "total powerset", so to speak. It will become apparent later on why I am being emphatic about this.)
The proof will start out assuming that the measure function, $\lambda$, which we discussed in the previous lesson, has certain natural properties. Because I want this proof to appear natural, we will go through the proof first, and only later will we inspect which properties were used.
Then we will have a discussion about which properties we may be willing to compromise, in order to avoid this impossibility proof.

### Vitali Looms

The following proof is due to Vitali, more or less.
Consider the interval $(0,1)$. We are going to try to "feather" this set out into a new set -- exactly what that means will come soon.
First define a relation, $\sim$, on the set $(0,1)$ defined by
    $x\sim y{\text{ if and only if }}y-x\in {\mathbb {Q}}$
* Exercise 1. A Few Rational Differences
* 1. Show that any two rational numbers are related to each other by $\sim$.
* 2. Prove that $\sim$ is an equivalence relation.
* 3. From part (2.) we now know that $\sim$ is an equivalence relation and therefore determines a partition of $(0,1)$.
* Define F to be any set of representatives for every cell of the partition. That is to say, define F as any arbitrary set with the property
* $c\in F{\text{ if and only if }}c\in (0,1){\text{ and for all }}c\sim d,{\text{ if }}c\neq d{\text{ then }}d\not \in F$
* Argue that $x\in (0,1)$ if and only if there is some $c\in F$ such that $x\in [c]$ where $[c]$ denotes the equivalence class for the element $c\in (0,1)$.
* 4. For any $x\in {\mathbb {R}}$, define $F+x=\{c+x|c\in F\}$. Prove that if $c\not \sim d$ (i.e. if c and d are not similar) for $c,d\in (0,1)$, then
* $(F+c)\cap (F+d)=\emptyset$
* 5. Define ${\mathcal {R}}=(-1,1)\cap {\mathbb {Q}}$ and then $A=\bigsqcup _{r\in {\mathcal {R}}}(F+r)$. Argue that this is a "countable union", which is to say that the indexing set ${\mathcal {R}}$ is a countable set.
* Also show that no element of A can be less than -1, nor can it be greater than 2. (Hint: consider the bounds on the values of F and ${\mathcal {R}}$.)
* Also show that $(0,1)\subseteq A$. Hint: For a given $x\in (0,1)$ there is some $c\in F$ such that $x\sim c$ as shown in (3.) Let $r=c-x$. Now show that $r\in {\mathcal {R}}$ and infer that $c-x\in F_{r}$. You're almost home.
Now that you've done all that work, here is how it assembles to a contradiction: We will show that $\lambda (A)$ both
* must be between 1 and 3, and
* must be 0 or $\infty$.
In (5.) of the exercise above, you showed that $(0,1)\subseteq A\subseteq (-1,2)$.
Since $\lambda ((0,1))=1$ then $1\leq \lambda (A)$.
Since $\lambda ((-1,2))=3$ then $\lambda (A)\leq 3$.
This establishes the first bullet point above. For the second, recall that in part (5.) you showed that the sets in the union $\bigsqcup _{r\in {\mathcal {R}}}(F+r)$ are disjoint (hence justifying the use of the square union symbol). Therefore the length of the whole set should be just the sum of its parts.
    $\lambda (A)=\sum _{r\in {\mathcal {R}}}\lambda (F+r)$
Now since F is just a translation of the set $F+r$, these two should have the same measure. Therefore, we may replace $\lambda (F+r)$ with the same shared constant, which I'll call $\alpha =\lambda (F)$.
This shows that $\lambda (A)=\sum _{r\in {\mathcal {R}}}\alpha$.
Now if $\alpha =0$ then $\lambda (A)=\sum _{r\in {\mathcal {R}}}0=0$.
The only other option is that $\alpha >0$ in which case $\lambda (A)=\sum _{r\in {\mathcal {R}}}\alpha =\infty$.
This now shows the second bullet point. The two bullet points together form a contradiction, and so the proof is complete.
$\Box$

### Compromise Your Values

Well, that does it! We can't have what we want.
There is no total measure on all real numbers.
But in Vitali's proof, we made several implicit assumptions about the properties of $\lambda$.
* Length-measure: A foundational assumption was that $\lambda$ assigns to an interval its length, which we used to compute that $\lambda ((0,1))=1$, and a few other measures.
* Monotonicity: When we reasoned that since $(0,1)\subseteq A$ therefore $\lambda ((0,1))\leq \lambda (A)$, this was an instance of a property that we call "monotonicity". It is the idea that, if you start from a set and add points, the larger set could only possibly have larger measure. We used it again when we reasoned that since $A\subseteq (-1,2)$ therefore $\lambda (A)\leq \lambda ((-1,2))$.
* Translation invariance: When we reasoned that since F is a translation of $F+r$, therefore $\lambda (F)=\lambda (F+r)$, this was an instance of "translation invariance". It is the idea that a set's measure should not change simply by placing it in a different location along the real line.
* Countable additivity: When we reasoned that, since $A=\bigsqcup _{r\in {\mathcal {R}}}(F+r)$, therefore $\lambda (A)=\sum _{r\in {\mathcal {R}}}\lambda (F+r)$ -- this was an instance of "countable additivity". It is the idea that a set's measure should be the sum of its disjoint parts.
If we would like to block the proof given above, can we compromise on any of these values? That is to say, can we seek out a function $\lambda$ which does not have all of these properties but perhaps has slightly weaker but good enough properties?
Well of course we cannot compromise length-measure. That is the very motivation and backbone of the entire project!
We actually cannot compromise monotonicity on its own. This is because (as we shall argue later on) monotonicity follows from countable additivity. Therefore we should really debate countable additivity before questioning monotonicity.
Translation invariance just seems too intuitive for what we mean by length-measure. This seems entirely non-negotiable.
Finally there is countable additivity. This one truly is debatable, and some mathematicians have chosen to study measures which only satisfy the weaker condition, finite additivity.
However, the much more popular way to approach this problem is to refuse to abandon countable additivity.
So are we at a loss? If we are unwilling to give up anything, has the project finally and utterly failed?
To be continued ...