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

Question

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!

Answer 1

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?

Answer 2

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

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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.

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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.

Answer 3

It appears the following elementary approach works without invoking heavy theorems. The main crux is, if we have a dense open set, then its complement, a closed nowhere dense set, might have positive measure. We must show this is not the case here, by using $m(A)>0$.

It suffices to only prove $m\left((A+H)^c \cap [0,1]\right)=0$.

Suppose $h_i\in H$ is so that $m\left((A+h_i) \cap [0,1]\right)>\delta>0$. Then, by the definition of Lebesgue measurable sets, there exist an open set $O_i$ and a closed set $F_i$ such that $F_i \subset (A +h_i) \cap [0,1]\subset O_i$ and $$m\left((A +h_i) \cap [0,1] \setminus F_i\right)<\frac{\epsilon}{2^{i+1}}$$ $$m\left(O_i\setminus (A +h_i) \cap [0,1]\right)<\frac{\epsilon}{2^{i+1}}.$$ Therefore, $O_i \setminus F_i$ is an open set that covers a part of $(A+H)^c \cap [0,1]$ and $m\left(O_i \setminus F_i\right)<\frac{\epsilon}{2^{i}}$.

Since $H$ is a dense set, $A+H$ is dense in $[0,1]$. Therefore, $$\overline{\bigcup_i O_i} \supset \overline{\bigcup_{i}(A+h_i)\cap [0,1]}=[0,1]\supset (A+H)^c \cap [0,1].$$ Also, since $m\left((A+h_i) \cap [0,1]\right)=\delta>0$, we have $m(O_i)\geq \delta>0$. We can therefore choose a subsequence $h_{i_k}\in H$, so that, for every $k\in\mathbb{N}$, $m\left(O_{i_{k+1}} -\bigcup_{j=1}^k O_{i_j}\right)>\frac{\delta}{2}.$ Then, after at most finite $K$ steps, $K\frac{\delta}{2}\geq1$, and so $m\left([0,1]-\bigcup_{j=1}^K O_{i_j}\right)=0$. As a result, the difference set $$(A+H)^c \cap [0,1] - (A+H)^c \cap [0,1] \cap \left(\cup_i O_i\right)\subset [0,1]-\bigcup_{j=1}^K O_{i_j}$$ also has measure zero.

Lastly, since $\cup_i F_i\subset A+H$, we have $$(A+H)^c \cap [0,1]\cap \left(\cup_i O_i\right)\subset \bigcup_i O_i - \bigcup_i F_i \subset \bigcup_i O_i - \bigcap_i F_i=\bigcup_i \left(O_i -F_i\right),$$ and so $$m\left((A+H)^c \cap [0,1]\right) \leq m\left(\bigcup_i \left(O_i -F_i\right)\right)\leq \sum_i \frac{\epsilon}{2^i}<\epsilon.$$