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!
 A: 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?
A: 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.
