Sum of a countable dense set and a set of positive measure Assume $A$ is a countable dense set in $\mathbb{R}$, and set $B$ has positive (Lebesgue) measure. Prove that $A+B=\{a+b:a\in A, b\in B\}=\mathbb{R}\backslash N$, where $N$ is a set of measure zero. 
I haven't come up with a good idea. 
Thanks in advance!
 A: Some more details for the answer by Kena Bel P: $A+B$ is measurable because it's the union of countably many translates of the measurable set $B$.  Let $N$ be its complement and suppose, toward a contradiction, that $N$ had positive measure.  Then by the cited theorem of Steinhaus, there would be an interval included in $\{n-a:a\in A, n\in N\}$.  As $B$ is dense, this interval contains some element $b$ of $B$.  But then we have, for this $b$, that $b=n-a$ for some $a\in A$ and $n\in N$.  That is, $n=a+b$, contradicting the fact that $n\in N=\mathbb R\setminus(A+B)$.
A: This follows from the following Steinhaus theorem: if $A,B$ are subsets of the real line of positive measure then $A-B$ contains an interval.
A: Edit: On further inspection and cleanup, this has pretty much just reduced to a standard proof of the Steinhaus theorem.  The Fourier transform appearing in a previous version was unnecessary.   I'll just leave this in case anyone is interested.
Without loss of generality, we can assume $A$ is bounded.  Suppose $N$ is bounded and disjoint from $A + B$; we will show $N$ has measure zero.  Then the result follows by taking $N = (A+B)^c \cap [-k,k]$, letting $k \to \infty$, and using countable additivity.
Set $f = 1_N$ and $g = 1_{-A}$ (i.e. $g(y) = 1$ if $-y \in A$, and 0 otherwise).  Now if $x \in B$, then $N$ and $x+A$ are disjoint, and so for every $y \in \mathbb{R}$, $f(x-y) g(y) = 0$.  (If $g(y) \ne 0$, then $-y \in A$, so $x-y \notin N$ and $f(x-y) = 0$.)  In particular, for $x \in B$, we have
$$(f * g)(x) := \int f(x-y) g(y)\,dy = 0$$
But now it follows from standard properties of convolution that $f * g$ is continuous.  (If $f$ were continuous and compactly supported, this would follow from dominated convergence.  Now approximate $f$ in $L^1$ norm by a sequence $\{f_n\}$ of continuous, compactly supported functions, and note that $\|f * g - f_n * g\|_{\infty} \le \|f-f_n\|_{1} \|g\|_\infty$)  Since $f*g$ vanishes on the dense set $B$, we have $f*g \equiv 0$.  Now integrating and using Tonelli's theorem,
$$\begin{align*} 0 &= \int (f*g)(x)\,dx \\ &=\int \int f(x-y) g(y)\,dy\,dx \\
&= \int \int f(x-y) \,dx\, g(y)\,dy \\
&= \|f\|_{1} \|g\|_1
\end{align*}$$
Hence either $\|f\|_1 = 0$ or $\|g\|_1=0$, i.e. either $m(N) = 0$ or $m(A) = 0$.  Since by assumption $m(A) \ne 0$, we must have $m(N) = 0$, completing the proof.
A: I wanted to add this as a comment, but I have not required reputation.

This follows from the following Steinhaus theorem: if $A,B$ are
  subsets of the real line of positive measure then $A-B$ contains an
  interval.

The Steinhaus theorem states that $A-A$ contains an open neighborhood of zero and the strong version states that $A+B$ contains an open set.
This desired result result, to my knowledge, was first proved  here.  It is not entirely obvious.
