## “Measuring” Sets: An Introduction to Measure Theory.

Greetings! I’m excited to author my first post and correct an imbalance in the amount of analysis present in the posts on this blog. I will try to continue the tradition of the quality, depth, and exposition of the posts before me.

This will be the first of what I anticipate to be three posts that will introduce the basics of  measure theory, which will go something like:

1. This post; which will outline the problem of measure.
2. A deeper discussion of (pre-/outer-)measures, an introduction to $\sigma$-algebras, a statement (and maybe proof of) Carathédory’s theorem, and an introduction to Lebesgue measure and a brief discussion of product measures.
3. Constructing the Lebesgue integral.

After which I hope to be regularly posting on topics in functional analysis, ergodic theory, and (as I learn it next semester) harmonic analysis.

Analysis is the art of approximation, and no problem is more fundamental in geometry and analysis than assigning a size or a measure to a set in a meaningful way. When asked, one might be quick to answer that area of a circle with radius one is $\pi$ or the length of the interval $(3,4)$ is $1$, but one would naturally struggle trying to calculate the area of this:

Since each of these sets are merely a collection of points in the plane, one may ask a deeper question: can we measure all subsets of the plane, or in space? This natural question has a surprising answer.

We’ll first consider the problem of measuring sets descriptively. What we’d like to construct is a function $\mu$ that takes a subset of our given set and produce a non-negative number (possibly $\infty$, accounting for our inclination to assign infinite size to the entire real line). Let us consider for a moment what we’d expect from such a measure. Certainly the set containing no elements should receive zero measure. We might demand that if we take two completely disjoint sets $A$ and $B$, that $\mu(A\cup B)=\mu(A)+\mu(B)$. This certainly agrees with our physical intuition. However, in the fashion of the Ancient Greeks, we’d like to measure sets exhaustively: taking a simple set (like the circle) and approximating its area with a sequence of (let’s say triangulated) polygons. Thus, we might require that if $\{A_n: n\in \mathbb{N}\}$ is a countable family of disjoint sets, that their union is measured similarly: $\mu(\cup_{n=1}^{\infty}A_n)=\sum_{n=1}^{\infty}\mu(A_n)$.

In the special case of $\mathbb{R}$ we might further request from our measure that it the unit interval $(0,1)$ receive measure (length) one, and that $\mu$ is invariant under translations $\mu(E+r)=\mu(E)$ for every $r\in \mathbb{R}$. These requests from our measure are seemingly lax, surely we can find some measure that will assign every subset of $\mathbb{R}$ a size. Yet even with this collection of geometrically appealing axioms, we find difficulty. Consider the following construction of Giuseppe Vitali.

Define an equivalence relation on $[0,1)$  by declaring that $x\sim y$ if and only if $x-y$ is a rational number. Let $N$ be a subset of $[0,1)$ that contains exactly one member from each equivalence class (I used the Axiom of Choice here, can you determine how?). Now for each $r\in \mathbb{Q}\cap [0,1)$, let

$N_r=\{x+r: x\in N\cap [0,1-r)\}\cup \{x+r-1: x\in N\cap [1-r,1)\}$

Think of $N_r$ as the following: we take $N$ and shift it to the right by $r$ units, and we shift the part that sticks our past $1$ precisely $1$ unit to the left. Then certainly $N_r\subset [0,1)$. Moreover, one can show that the family $\{N_r: r\in \mathbb{Q}\cap [0,1)\}$ partitions $[0,1)$. Suppose that $N$ can be measured by our magic measure $\mu$. Then since $N_r$ is really just a shift modulo one, its measure is invariant. More precisely

$\mu(N)=\mu(N\cap [0,1-r))+\mu(N\cap [1-r,1))=\mu(N_r)$

for any $r\in \mathbb{Q}\cap [0,1)$. Moreover, since $\mathbb{Q}\cap [0,1)$ is countable, and $[0,1)$ is the disjoint union of the $N_r$‘s, we see then that

$1=\mu([0,1))=\sum\limits_{r\in Q\cap [0,1)}\mu(N_r)=\sum\limits_{r\in Q\cap [0,1)}\mu(N)$

If $\mu(N)> 0$, then we have that the right hand sum is $\infty$ which is not equal to one. However, if $\mu(N)=0$, then the right hand sum is zero, which is not one. So we have a contradiction. The only assumption we made was that $N$ was could be measured by $\mu$, and so it follows that $N$ is not measurable. As an aside; although we haven’t constructed Lebesgue measure, this argument demonstrates that the axiom of choice implies the existence of a Lebesgue non-measurable set. One might ask if we need the axiom of choice to construct such a set, and the answer is yes! There is a model (The Solovay Model) in which every axiom of ZF holds, absent the axiom of choice, in which every subset of the reals is Lebesgue measurable.

This shows the existence of a non-measurable set given our intuitive geometric requests of our measure. In case you wondered if it gets better in larger dimensions, I’ll refer you to check out a similar result known fondly as the Banach-Tarski Paradox.

So now we’ve come face to face with the problem of measure. To cope with this problem, we reject some of our previous hypothesis about what a meaningful measure might be, and we rightfully (for the time being) do, by ignoring the translation-invariance in the definition of our measure (we’ll return to this discussion) and we define a measure as follows.

Let $X$ be a set and let $S$ be a collection of subsets of $X$. A function $\mu:S\to [0,\infty]$ is called a measure provided that $\mu(\emptyset)=0$ and if $\{A_n: n\in \mathbb{N}\}$ is a countable family of disjoint sets in $S$, then $\mu(\cup_{n=1}^{\infty}A_n)=\sum_{n=1}^{\infty}\mu(A_n)$.

To return to our original question – which sets are in fact measurable – will therefore depend on which measure we choose since our collection of measurable sets will be precisely the domain of our measure. For example, if we assign every set measure zero, then this is trivially verified to be a measure, and thus every set is measurable. But this seems rather silly since we’d like our measure (in the case of $\mathbb{R}^n$; well for $n=1,2,3$) to reflect our geometric intuition (non-empty boxes getting positive measure, etc).

Constructing a meaningful measure on $\mathbb{R}^2$ seems like an onerous task (and the first time one does, it is). But we might instead try to determine the minimum (loosely-speaking) amount of information we’d need to know to construct such a measure. Just in the sense that a continuous, real-valued function is determined by its values on a dense set, or that that a holomorphic function defined on a disk is completely determined by its values on the boundary of the disc, we can completely determine the area (if it exists) of a set with a (relatively) minimal amount of information.

Next time, we’ll delve deeper into exactly what information one needs to generate a measure, and exactly what sets are measurable under such a measure.

This Fall I will be a first year mathematics PhD student at UCLA. I enjoy doing analysis - particularly of the functional variety.
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### 5 Responses to “Measuring” Sets: An Introduction to Measure Theory.

1. JCummings says:

YES!!!! Adam is officially on the blog!!! And doing measure theory as well! I’m excited for you to get to ergodic theory, so make sure you post more often than the rate that the rest of us are going at.

Also, I inserted the Koch snowflake into the post (do you know to do this?) so people don’t have to click on the link. Or did you not want it this way? I think images in posts are nice. While I was there I also linked up “holomorphic”.

O, and I also “continuous real-valued function” became “continuous, real-valued function”. I haven’t yet gotten over the way mathematicians try to make every adjective-adjective-…-adjective-noun sequence into a really long noun. Do you have a feeling on this?

2. Ryan says:

An interesting corollary (i.e. exercise in Folland we had for analysis HW this year) of sorts to the existence of the Lebesgue non-measurable set you described here is that any Lebesgue measurable subset $L\subset N$ must have Lebesgue measure zero (the argument is easy enough, if not then every translate $N_r$ of $N$ must contain a translate $L_r\subset N_r$ of the same measure. Since there countably many such translates, they must have zero measure, otherwise you would get a subset of $[0,1]$ with infinite measure). The corollary to this is that in fact EVERY Lebesgue measurable set with positive measure must contain a non-measurable set. I’ll leave this for you guys to prove, but essentially prove the contrapositive and make use of the fact that I just stated above. Kind of cool when you think about it! P.S. Adam I’m assuming when you talk about Caratheodory you will talk about completions of $\sigma-$algebras and the difference between using these and the original $\sigma-$algebra you start with? If not, let me know and perhaps I can write about this.

• JCummings says:

:-) We did both of those, too.