Intuitive Explanation of Why the Power Set of $\mathbb{R}$ is "too big" for the Lebesgue Measure?

I've been working with the construction of measures for a little bit, and I understand that in order for the Lebesgue measure to be an official measure on $\mathbb{R}$, we need to restrict it to a certain $\sigma$-algebra, namely the one generated by $\tau \cup \mathcal{N}$, where $\tau$ is our topology and $\mathcal{N}$ is the collection of all null sets.

I have been looking at a proof as to why the Lebesgue measure "fails" when we consider it as a mapping from $2^{\mathbb{R}}$, and it seems to be more algebraic in nature even though it is an Analysis book. Condensed, it basically defines a relation $x\sim y$ if $x - y \in \mathbb{Q}$ for $x,y \in [0,1]$. We then consider the set of equivalence classes (basically quotient $[0,1]$ by this equivalence relation), and the rest of the proof is over my head, in the sense that I have no idea where the rest of the steps are coming from. In the end, we get a contradiction, so our measure doesn't work, basically.

My question is about any kind of intuition behind why we need to restrict our domain? The Lebesgue $\sigma$-algebra is still an uncountable set, but it seems as though if we allow all possible subsets, then there is too much "overlap" for our intervals, but I do not really know how to formulate this rigorously.

Thanks!

Solution 1:

Let me try to explain the Vitali construction by breaking everything into small pieces. (Usually I find this presented quite rapidly in books, so that everything seems to come all at once and it's not clear what all the pieces even were.)

You begin by quotienting $[0,1]$ by this equivalence relation that you stated. You get the full collection of equivalence classes. Now you use the axiom of choice to get a set which contains exactly one element of each equivalence class. This set is called a Vitali set; let's denote one such set by $V$.

Let's define the translation of a set of real numbers $A$ by a real number $b$: $A+b=\{ a+b : a \in A \}$. From basic properties of equivalence relations, we find that a Vitali set has the property that $V+q$ is disjoint from $V+q'$ for any rational numbers $q \neq q'$.

This translation property by itself is not a problem. For example, singletons also have this property, and yet they are of course measurable. But a countable union of singletons is a null set under Lebesgue measure. By contrast, when we take the union of $V+q$ when $q$ ranges over $\mathbb{Q} \cap [-1,1]$, we do not get a null set. In fact, this set must contain all of $[0,1]$. Why? Because 1. every difference between two points in $[0,1]$ is in $[-1,1]$ and 2. every point in $[0,1]$ is a rational translate of a point in $V$. On the other hand, the union is contained in $[0,1]+[-1,1]=[-1,2]$. So if $V$ is measurable then the measure of this union is some number in the interval $[1,3]$.

These two things together still do not constitute a problem. What we have now is a countable collection of disjoint sets whose union is contained between two sets, each of which have finite, positive measure. Where the problem comes in is when we postulate that the Lebesgue measure should be translation-invariant: $m(A+b)=m(A)$. If this is the case, then countable additivity implies that either the measure of this union is zero (if all the things in the union have measure zero) or the measure is infinite (if all the things have some fixed, positive measure). But we just said that neither of these can be the case, so we have a contradiction.

A summary of what we did: we got this set $V$. We built a set, say $U$, comprised of a countable union of translates of $V$. $U$ contains $[0,1]$, so its measure must be at least $1$. $U$ is contained in $[-1,2]$, so its measure must be at most $3$. On the other hand, $U$ is a countable union of disjoint translates of the same set, so since the Lebesgue measure is supposed to be translation-invariant, its measure must be either $0$ or $\infty$. Since neither $0$ nor $\infty$ is between $1$ and $3$, $V$ must not be Lebesgue measurable. So in particular the problem is not so much that the power set is too big as that we insist on Lebesgue measure being translation-invariant.

Solution 2:

I know this question was asked a long time ago but I just came across this question and I think it is a good one that deserves an attempt at an answer. I'm kind of new to measure theory but I think I can give some degree of insight on this question. When it comes to the Lebesgue measure, we want the smallest $\sigma$-algebra, say $\mathcal{B}$ generated by a set we know a priori is Borel measurable. Take $(a,b)$ for any $a,b \in \mathbb{R}, a \neq b$ and then we have that $\sigma(\{x:x \in (a,b)\})$ is given by $$ \bigcap^{\infty}_{n=1} (a,b+1/n)=(a,b] $$ which implies that $(a,b] \in \mathcal{B}$, also we have $$ \bigcap^{\infty}_{n=1} (a-1/n,b)=[a,b) $$ so $[a,b) \in \mathcal{B}$. Now suppose that $a=b$, then we have that $$ \bigcap^{\infty}_{n=1}(a-1/n,a+1/n)=\{a\} $$ so every singleton $\{a:a \in \mathbb{R}\} \in\mathcal{B}$ and the same argument can be made for $a \neq b$ to show that every closed set $[a,b] \in \mathcal{B}$. Now there are still non-measurable subsets of every open set $(a,b) \in \mathcal{B}$ (any open interval always has non-measurable sets) so this set is not quite "big enough", as you put it, for the Lebesgue measure. So to make it so, we have to extend this measure to include all the subsets of sets whose Borel measure is zero (all non-measurable sets on an interval are subsets of sets of measure zero), which we can call $\mathcal{N}=\{N \subset \mathbb{R}: \exists M \in \mathcal{B}:\mu(M)=0,N \subseteq M\}$. Now we need a $\sigma$-algebra $\mathcal{B}_{0}=\{B \cup N:B \in \mathcal{B},N \in \mathcal{N}\}$ and from there we know that there is an extension of $\mu$ to $\mathcal{B}_{0}$ given by the outer measure $$\mu^{*}(E)=\inf\{\mu(A):E \subseteq A \in \mathcal{B}\}$$ which gives us $$ \mu^{*}(B \cup N)=\mu(B) $$ from which we get the familiar Lebesgue measure $\mu$ since the outer measure equals the Lebesgue measure $\mu^{*}(E)=\mu(E)$ as long as for $A \subseteq \mathbb{R}$ $$ \mu(A)=\mu(A \cap E)+\mu(A \cap E^{c}) $$

Now, you have already stated yourself (at least I think) that every power set is a $\sigma$-algebra (though the converse is obviously not true in general) so there are clearly a notion that some $\sigma$-algebras can be bigger than others. The key is to find a $\sigma$-algebra which contains the right sets for which a measure makes sense. It is not that we couldn't take the power set $2^{\mathbb{R}}$ instead of $\mathcal{B}$ since $\mathcal{B} \subset 2^{\mathbb{R}}$ but if we took the power set, then the Lebesgue measure would only be defined for a very small subset of $2^{\mathbb{R}}$ since $\#(2^{\mathbb{R}})>\#(\mathcal{B})$ so it is much more useful to restrict our focus to just the sets which have a meaningful (Lebesgue or otherwise) measure defined on them. Non-measurable sets were very confusing to me at first and it took me awhile to come to terms with it. It doesn't help that you can't really define these sets either so while I'm sure you've already figured it out by now, hopefully this will help you with the intuition part.