# Dense transfer of a set with positive lebesgue measure: is it conull?

I'm facing a problem in measure theory and I need to prove the following conjecture to move on. Attention: I'm not sure the following statement is true.

Let $A \subset \mathbb{R}$ be a measurable set such that $m(A)>0$ and $H$ be a countable, dense subset of $\mathbb{R}$. If $A+H=\{a+h: a \in A, h \in H\}$, prove that $m((A+H)^c)=0$.

I'm totally stuck. It's easy to see that $A+H=\displaystyle{\bigcup_{h \in H} A+h}$, so it's definitely a measurable set, but that's the only progress I've been able to make. Any help would be greatly appreciated!

• Regularity is your friend . . . – Noah Schweber Mar 25 '17 at 21:16
• I've already tried using theorems that I know of that are relative to regularity. Could you please expand on this? – JustDroppedIn Mar 25 '17 at 21:18
• First, can you show that the statement is true if $A$ is open? If so, think about how $A+H$ is a union of countably many sets, and - assuming $A+H$ had positive measure complement - try to approximate each of those sets by open sets in an appropriate way. – Noah Schweber Mar 25 '17 at 21:21
• Thanks, I'll give it a shot and I'll let you know on my progress. Your comment is greatly appreciated – JustDroppedIn Mar 25 '17 at 21:32
• Okay, i can even show that the statement is true if $A$ has an interior point. I can't see where this is going though and this makes me unable to see the appropriate approximation with open sets.. what is it that is going to lead to a contradiction? – JustDroppedIn Mar 25 '17 at 22:09

Use Lebesgue density theorem (LDT) which has an elementary proof.

Towards a contradiction, suppose $B = \mathbb{R} \setminus (A + H)$ has positive measure. Using LDT, choose open intervals $I, J$ of same length such that $B \cap I$ has $\geq 99$ percent measure of $I$ and $A \cap J$ has $\geq 99$ percent measure of $J$. Choose $h \in H$ (using the density of $H$) such that $J + h$ meets $I$ on a set of measure $\geq 99$ percent of $I$ (which has same length as $J$). Do you see a problem now?

• Awesome! If one already knows the Lebesgue differentiation theorem, this is very nice! – PhoemueX Mar 27 '17 at 17:37
• I must admit, this is a great answer. I'm not familiar with this theorem yet, but I read it and its proof and it is indeed elementary. Thank you very much. – JustDroppedIn Mar 27 '17 at 18:05

Below, I provide two proofs: A short and elegant proof using Fourier techniques and a more elementary but longer probabilistic proof.

Fourier proof

Since $A$ has positive measure, so has $A \cap [-n,n]$ for some $n$. Thus, we can assume without loss of generality that $A$ is bounded.

Now assume towards a contradiction that $M := A + H$ is not conull. Then there is an $L^1$ function $f$ with $f = 0$ on $M^c$ and such that $f$ does not vanish almost everywhere (take e.g. $f = 1_{M^c \cap [-k,k]}$ for a suitable $k$). Then we have for all $h \in H$ that $$0 = \int f(x) 1_{A+h}(x) dx = \int f(x) 1_{-A}(h-x) dx = (f \ast 1_{-A}) (h).$$ But since $1_{-A} \in L^\infty$ and since $f \in L^1$, the convolution from above is a continuous function. Since $H$ is dense, this means $f \ast 1_{-A} \equiv 0$.

Now, take the Fourier transform to get $0 \equiv \widehat{f} \cdot \widehat{1_{-A}}$. But since $1_{-A}$ has compact support, the Fourier transform $\widehat{1_{-A}}$ is a (nonvanishing) analytic function and thus has only isolated zeros. In particular, the set where $\widehat{1_{-A}}$ does not vanish is dense. Hence, we get $\widehat{f} \equiv 0$ by continuity and then $f = 0$ almost everywhere, a contradiction.

Probabilistic proof

Since $A$ has positive measure, so has $A\cap\left[-n,n\right]$ for a suitable $n\in\mathbb{N}$, so that we can assume $A\subset\left[-n,n\right]$. Let $\left(Y_{\ell}\right)_{\ell\in\mathbb{N}}$ be an iid sequence of random variables with $Y_{\ell}\sim U\left(\left[-2n,2n\right]\right)$ (the uniform distribution on $\left[-2n,2n\right]$). Fix $x\in\left[-n,n\right]$ and consider the event $$E_{x}:=\bigcap_{\ell\in\mathbb{N}}\left(x\notin A+Y_{\ell}\right).$$ Because of $x-A\subset\left[-2n,2n\right]$ and by translation inavriance of the Lebesgue measure, the probability of this event is \begin{align*} \mathbb{P}\left(E_{x}\right) & =\prod_{\ell=1}^{\infty}\mathbb{P}\left(x\notin A+Y_{\ell}\right)=\prod_{\ell=1}^{\infty}\mathbb{P}\left(Y_{\ell}\notin x-A\right)\\ & =\prod_{\ell=1}^{\infty}\frac{1-\lambda\left(\left[-2n,2n\right]\cap\left(x-A\right)\right)}{4n}\\ & =\prod_{\ell=1}^{\infty}\frac{1-\lambda\left(A\right)}{4n}=0. \end{align*}

Up to now, we have shown for every $x\in\left[-n,n\right]$ that $\mathbb{P}\left(E_{x}\right)=0$. By Fubini's theorem, this shows that for almost every realization $y=\left(y_{\ell}\right)_{\ell\in\mathbb{N}}$ of $\left(Y_{\ell}\right)_{\ell\in\mathbb{N}}$, the set $$N_{y}:=\left\{ x\in\left[-n,n\right]\,:\,\forall\ell\in\mathbb{N}:\, x\notin A+y_{\ell}\right\}$$ is a null-set. Fix one such realization $y$.

Since $A$ is bounded, the map $\mathbb{R}\to L^{1}\left(\mathbb{R}\right),t\mapsto1_{A+t}$ is continuous. By density of $H$, this allows us to choose for arbitrary $\ell,m\in\mathbb{N}$ some $h_{\ell,m}\in H$ with $\lambda\left(\left[A+y_{\ell}\right]\setminus\left[A+h_{\ell,m}\right]\right)\leq\frac{2^{-\ell}}{m}$. Thus, we get \begin{align*} \lambda\left(\left[-n,n\right]\setminus\bigcup_{h\in H}\left(A+h\right)\right) & \leq\lambda\left(\left[-n,n\right]\setminus\bigcup_{\ell,m}\left[A+h_{\ell,m}\right]\right)\\ & \leq\lambda\left(\left[-n,n\right]\setminus\bigcup_{\ell\in\mathbb{N}}\left(A+y_{\ell}\right)\right)+\lambda\left(\bigcup_{\ell\in\mathbb{N}}\left(A+y_{\ell}\right)\setminus\bigcup_{\ell,m}\left(A+h_{\ell,m}\right)\right)\\ \left(\text{with arbitrary }m\in\mathbb{N}\right) & \leq\lambda\left(\bigcup_{\ell\in\mathbb{N}}\left(A+y_{\ell}\right)\setminus\left(A+h_{\ell,m}\right)\right)\leq\sum_{\ell=1}^{\infty}\frac{2^{-\ell}}{m}=\frac{1}{m}. \end{align*} Since $m\in\mathbb{N}$ can be chosen arbitrarily, this implies $\lambda\left(\left[-n,n\right]\setminus\bigcup_{h\in H}\left(A+h\right)\right)=0$.

We are almost done: Above, $H\subset\mathbb{R}$ was an arbitrary dense subset. Now, if $N\in\mathbb{Z}$ is arbitrary, then $H+N\subset\mathbb{R}$ is also dense, so that we get $$0=\lambda\left(\left[-n,n\right]\setminus\bigcup_{h\in H+N}\left(A+h\right)\right)=\lambda\left(\left(\left[-n,n\right]-N\right)\setminus\bigcup_{h\in H}\left(A+h\right)\right)=\lambda\left(\left[-n-N,n-N\right]\setminus\bigcup_{h\in H}\left(A+h\right)\right).$$ Since $\bigcup_{N\in\mathbb{Z}}\left[-n-N,n-N\right]=\mathbb{R}$, this easily implies $\lambda\left(\mathbb{R}\setminus\bigcup_{h\in H}\left(A+h\right)\right)=0$, as desired.

• I'm afraid I have no knowledge of Fourier analysis, so I can't judge your answer. Even so, I'm grateful for your answer. Thing is, the reason I'm asking this question is because I need it in order to solve an exercise in measure theory. So if you have a different proof in mind, please let me know. – JustDroppedIn Mar 26 '17 at 15:57
• @authunchained: I just added another proof :) – PhoemueX Mar 26 '17 at 16:53
• Do you think that the Fourier proof work in any dimension? It should, only that I do not know how to make up for the fact that real analytic functions might have nonisolated zeroes (e.g., $\widehat {1_{-A}}(\xi_1, \xi_2)$ might be equal to $\xi_1^2+\xi_2^2-1$). – Giuseppe Negro Mar 27 '17 at 16:55
• @GiuseppeNegro: Yes, it will work in every dimension. It is still true that the zero-set of an analytic function is a null-set. This can be seen as follows: Fix $x_1,\dots ,x_{n-1}$ and consider $f(x_1, \dots, x_{n-1}, \cdot)$. This is either identically zero or only has isolated zeroes. By Fubini (to calculate the measure of the zero-set), it suffices to show that the "identically zero" case (say, on the set $E \subset \Bbb{R}^{n-1}$) only happens on a set of measure zero. In dimension $2$, this is easy: If not, then the set of $x_1$ with $f(x_1,\cdot) \equiv 0$, has a cluster point – PhoemueX Mar 27 '17 at 17:31
• so that $f(\cdot, x_2) \equiv 0$ for every $x_2$. In general, we proceed by induction: For arbitrary $x_n$ and $g(x_1,\dots,x_{n-1}) := f(x_1,\dots ,x_n)$, we have $g = 0$ on the set $E$ which has positive measure by assumption. Hence, $g \equiv 0$. Since this holds for arbitrary $x_n$, we get $f \equiv 0$.... I think my writeup is a bit confusing, but hopefully one can understand the idea :) – PhoemueX Mar 27 '17 at 17:34