Posted by Jason Polak on 27. August 2011 · Write a comment · Categories: measure-theory · Tags: , ,

Today I shall continue in the spirit of my last post, which was essentially a revised set of notes on material for my qualifying exam. Here, and in the next post, we shall see two ways to prove that if $ A$ and $ B$ are Lebesgue-measurable subsets of the real line with positive measure, then $ A+B$ contains an interval. The notation $ A+B$ means $ \{ a + b : a\in A, b\in B\}$.

Before we start, let us examine counterexamples. Firstly, the converse to the statement is not true. I’ll leave it as an exercise to show that the Cantor set $ C$ is such that $ C+C$ contains an interval, yet $ C$ has measure zero. On the other hand, $ \mathbb{Q}+\mathbb{Q}=\mathbb{Q}$, and $ \mathbb{Q}$ does not contain an interval. In fact, if we prove the above then we shall have another proof that $ \mathbb{Q}$ has measure zero, although this follows more directly by observing that $ \mathbb{Q}$ is countable, and that points have measure zero.

Both these solutions were hinted at in Rudin’s “Real and Complex Analysis”. I’ve broken up these solutions into two posts for convenience. In the sequel, measurable means Lebesgue measurable on the real line, and $ m$ denotes the Lebesgue measure.

The Direct Approach

We start by using the concept of metric density. Let $ E$ be measurable. If $ x\in E$, we define the metric density of $ E$ at $ x$ to be the limit

$ D_E(x) = \lim_{r\to 0} \dfrac{m(E\cap (x-r,x+r))}{2r}$

if it exists. If not then $ D_E(x)$ is just undefined. In a general metric space, we would replace $ (x-r,x+r)$ by the open ball of radius $ r$ around $ x$ and $ 2r$ by the volume of such a ball. Recall that if $ f:\mathbb{R}\to\mathbb{C}$ is measurable, then a Lebesgue point of $ f$ is an $ x\in \mathbb{R}$ such that

$ \lim_{r\to 0} \dfrac{1}{2r}\int_{x-r}^{x+r} |f(t) – f(x)|dt = 0$.

A basic theorem of Lebesgue points is:

Theorem 1. If $ f\in L^1(m)$ then almost every point is a Lebesgue point of $ f$.

A consequence of the definition of a Lebesgue point is that

$ \lim_{r\to 0} \dfrac{1}{2r}\int_{x-r}^{x+r} f(t)dt = f(x)$

for every Lebesgue point $ x$ of $ f$. Using this, together with the fact that almost every point is a Lebesgue point. If $ E$ is Lebesgue measurable, we would like to apply Theorem 1 to $ f=\chi_E$, the characteristic function of $ E$. Of course, $ \chi_E$ is not necessarily in $ L^1(m)$, but if we consider $ \chi_{E\cap I}$ for a countable covering of $ \mathbb{R}$ with intervals $ I$, then we can get the next theorem.

Theorem 2. If $ E$ is measurable, then $ D_E(x) = 1$ at almost every point of $ E$ and $ D_E(x) = 0$ at almost every point in the complement of $ E$.

That $ \mathbb{R}$ is $ m$-finite is crucial, for the Lebesgue measure of each of the non Lebesgue points in each of the intervals $ I$ has measure zero, and a countable union of such sets still has measure zero.

The Proof

Let $ A,B\subseteq \mathbb{R}$ be sets of positive measure. The positive measure of both sets guarantees, by Theorem 2, the existence of points $ a\in A$ and $ b\in B$ such that $ 1=D_A(a) = D_B(b)$. By definition of metric density, we can choose an $ r > 0$ such that

$ 1 \geq \dfrac{m(A\cap (a-r,a+r))}{2r} > 3/4$
 
$ 1 \geq \dfrac{m(B\cap (b-r,b+r))}{2r} > 3/4$.

Thus $ m(A\cap (a-r,a+r)) + m(B\cap (b-r,b+r)) > 3r$. Set $ c = a + b$. We claim that $ (c-r,c+r)\subseteq A+ B$. We shall proceed by contradiction. Thus, assume that

$ (c-r,c+r)\not\subseteq A + B$.

In this case, there is an $ x\in (c-r,c+r)$ with $ x\not\in A+B$. Without loss of generality, assume that $ x > c$, as we shall see that the case $ x < c$ proceeds in the same manner. Of course $ x\not= c$ (why?). Now, $ (b,b+r)\subseteq (b-r,b+r)$ and it follows that since $ x > c$, there is for each $ y\in (b,b+r)$ a $ z\in (a-r,a+r)$ such that $ y + z = x$. Since $ x\not\in A+B$, either $ y\in A$ or $ z\in B$, but not both!

Thus $ (b,b+r)\cap B$ can be translated onto $ A\cap (a-r,a+r)$ without intersecting it. Thus we get that

$ m(A\cap (a-r,a+r)) + m(B\cap (b-r,b+r)) \leq 3r$

which is a contradiction, because we choose $ r$ so that the negation of this would hold.

Concluding Remarks

This proof is closest to elementary. One proof of the theorem on Lebesgue points, which we needed for Theorem 2, uses the Vitali covering lemma and is a bit technical. In order to make this proof elementary, one would need to prove Theorem 2 directly. If anyone knows of a proof I would like to read it!

Stay tuned for Part 2, which should appear shortly.

Leave a Reply

Your email address will not be published. Required fields are marked *