The Sumset of Sets of Positive Measure, Continued

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

In the previous post, we saw how to use a basic theorem on Lebesgue points to prove that if $A$ and $B$ are measurable subsets of the real line with positive measure, then $A+B$ contains an interval. We shall continue now to prove this again using a different, less involved method. This solution is based on a hint in Problem 19 of Chapter 9 in Rudin’s book.

The Proof

Recall that we are proving: if $A$ and $B$ are measurable sets of real numbers such that $A$ and $B$ have positive measure, then $A+B$ contains an interval.

Unlike the previous proof, which was very direct and essentially involved manipulating sets with our bare hands, this proof using convolutions and Fourier transforms will be deft, clean, and indirect.

We begin by assuming that $A$ and $B$ have finite measure. This is no problem, since we can always find subsets of $A$ and $B$ with finite positive measure (how?) and if we can prove that the sum of these contains an interval, then $A+B$ will as well.

We now invoke the following theorem, which is not difficult to prove and I suggest the reader attempt it before continuing:

Theorem 1. If $A$ and $B$ are measurable sets with finite positive measure then $\chi_A*\chi_B$ is continuous.

Let $\widehat{f}$ denote the Fourier transform of an $f\in L^1(m)$. Recall that the Fourier transform is given by the formula

$\widehat{f}(t) = \int_{\mathbb{R}} e^{-itx}f(x)dx$.

We can thus see that by substituting $t = 0$ into the above that

$\widehat{\chi_A*\chi_B}(0) = \widehat{\chi_A}(0) \widehat{\chi_B}(0) = m(A)m(B) > 0$.

Here we just used that the Fourier transform of a convolution is the product of the Fourier transforms of the two functions. Now, this means that $\chi_A*\chi_B$ itself is not identically zero, since the Fourier transform of the zero function is the zero function. Now, by Theorem 1, and using the definition of the convolution, there is an $\epsilon > 0$ and an $x\in\mathbb{R}$ such that for $z\in (x-\epsilon,x+\epsilon)$,

$0 < \chi_A*\chi_B(z) = \int_{\mathbb{R}} \chi_A(z-y)\chi_B(y)dm(y)$. Since the integrand a non-negative function, there must be some $y\in\mathbb{R}$ such that $\chi_A(z-y)\chi_B(y) > 0$, or that $z-y\in A$ and $y\in B$, and so $z-y + y = z\in A +B$, so that $(x-\epsilon,x+\epsilon)\subseteq A+B$.

Final Remarks

Both of these were just exercises, but it might be useful to have these two different solutions recorded in one place. This second proof following Rudin’s hint seemed particular nice. But now that the qualifying exam is over, I shall return to more algebraic things.