If $x_n + A = A$ and $x_n \to 0$, then Lebesgue measure of $A$ is $0$ If $A$ is a measurable set, $x_n$ is a sequence converging to $0$ (all $x_n$ different from $0$), and  $x_n +A = A$ for all $n$, then $\lambda(A) = 0$ or $\lambda(A^c) = 0$ where $\lambda$ is the Lebesgue measure.
I have been trying this problem for quite some time but made no progress. Any hints?
 A: Assume $\lambda(A) > 0$, so that we want to prove $\lambda(A^c) = 0$.
First I'll reduce the problem to a [notationally] simpler case:

*

*By replacing $x_n$ with $|x_n|$ (checking $x_n \to 0 \iff |x_n|\to 0$ and $x+A=A \iff |x|+A=A$), we may without loss of generality assume $x_n > 0$ for all $n$.

*By replacing $A$ with $\frac{1}{x_0}A$ and $x_n$ with $\frac{x_n}{x_0}$, we may assume without loss of generality that $x_0 = 1$.  [Checking that $\lambda(\frac{1}{x_0}A) > 0$, that $\frac{x_n}{x_0} + \frac{1}{x_0}A = \frac{1}{x_0}A$, that $\frac{x_n}{x_0} > 0$, that $\frac{x_n}{x_0} \to 0$, and that $\lambda\left(\left(\frac{1}{x_0}A\right)^c\right) = 0 \implies \lambda(A^c) = 0$]

*By Lebesgue's density theorem, there is some $a_0 \in A$ with density $1$.  Since $\lambda$ is translation invariant and $x+A = A \iff x+(A-a_0)=(A-a_0)$, we may replace $A$ with $A-a_0$ to assume, without loss of generality, that $a_0=0$.

Since $1 + A = A$, we have $n+A = A$ and therefore $$[n,n+1)\cap A = [n,n+1)\cap(n+A) = n+[0,1)\cap A$$ for every integer $n$.  Therefore $$\lambda(A^c) = \sum_{n\in\mathbb{Z}}\lambda([n,n+1) \cap A^c) = \sum_{n\in\mathbb{Z}}\left(1 - \lambda([n,n+1)\cap A)\right) = \sum_{n\in\mathbb{Z}}(1 - \lambda([0,1)\cap A)),$$ so it suffices to show $\lambda([0,1) \cap A) = 1$.
To this end, let $N$ be such that $n \geq N \implies x_n < 1/3$.  Then for $n \geq N$,
$$\bigcup_{k=1}^{\lfloor (1-x_n)/(2x_n)\rfloor}\Big(2kx_n + (-x_n, x_n)\cap A\Big) \subseteq [0,1) \cap A \subseteq [0,1)$$
where sets in the union are disjoint, so by applying $\lambda$,
$$\left(\frac{1-3x_n}{2x_n}\right)\lambda((-x_n,x_n)\cap A) \leq \left\lfloor\frac{1-x_n}{2x_n}\right\rfloor\lambda((-x_n,x_n)\cap A) \leq \lambda([0,1)\cap A) \leq 1$$

Now we may finish by appealing to the density of $0 \in A$.
Let $\varepsilon \in (0,1)$.  Then since $0 \in A$ has density $1$ and $x_n > 0$ tends to $0$, there is an $M$ such that $$n \geq M \implies \lambda\Big((-x_n, x_n) \cap A\Big) \geq 2x_n(1-\varepsilon)$$
Therefore, if $n \geq \max(M,N)$,
$$(1-3x_n)(1-\varepsilon) = \left(\frac{1-3x_n}{2x_n}\right)(2x_n(1-\varepsilon)) \leq \lambda([0,1) \cap A) \leq 1$$
Let $n \to \infty$ to show $$1-\varepsilon \leq \lambda([0,1) \cap A) \leq 1$$
Then let $\varepsilon \to 0^+$ to show $\lambda([0,1) \cap A) = 1$ which is what we needed to show.  It follows that $\lambda(A^c) = 0$.
A: Sketch of a solution:
I am assuming the ambient space is $(\mathbb{R},\mathscr{B}(\mathbb{R}),\lambda)$.
Here we prove a little stronger result:
If $f$ is a measurable function and $f(x+x_n)=f(x)$ for all $x\in\mathbb{R}$ and $n\in\mathbb{N}$, then there is $c\in\mathbb{R}$ such that $f(x)=c$ $\lambda$--a.s. By considering bounded functions first, can be extended to any measurable function.
As pointed out in the comments, the set
$$G:=\{a\in\mathbb{R}:f(x+a)=f(x) \,\text{for all}\,x\in\mathbb{R}\}$$
is a additive subgroup of $\mathbb{R}$. By  assumption, $\{x_n:n\in\mathbb{N}\}\subset G$. The assumption  $x_n\neq0$ for all $n$ and $x_n\rightarrow0$, implies  $G$ is not of the form $h\mathbb{Z}$, but  a dense additive subgroup of $\mathbb{R}$.
Consider the function $\phi(x)=\frac{1}{2}\mathbb{1}_{[-1,1]}(x)$, and for any $\varepsilon>0$ let $\phi_\varepsilon(x)=\varepsilon^{-1}\phi(x\varepsilon^{-1})$
Suppose $f\in L_\infty(\mathbb{R})$. Then, $f\in\mathcal{L}^1_{loc}(\mathbb{R})$,
$$f*\phi_\varepsilon(x)=\int \phi_\varepsilon(y)f(x-y)\,dy$$
is a continuous function (in fact uniformly continuous by Young's convolution theorem, but that is not that important for our purposes), and
$f*\phi_{\varepsilon}(x)\xrightarrow{\varepsilon\rightarrow0}f(x)$ at every Lebesgue point of $f$ (and hence a.s.).
For any $a\in G$ and $x\in\mathbb{R}$
$$
f*\phi_\varepsilon(x+a)=\int \phi_\varepsilon(y)f(x+a-y)\,dy=\int \phi_\varepsilon(y)f(x-y)\,dy = f*\phi_\varepsilon(x)
$$
As $G$ is dense in $\mathbb{R}$, this means that $f*\phi_\varepsilon$ is a constant function, say $c_\varepsilon$.   From this, it follows that $f$ is constant a.s. (one can for example substitute $f$ by $f\mathbb{1}_K$ for some compact to get that $(f\mathbb{1}_K)*\phi_\varepsilon\xrightarrow{\varepsilon\rightarrow0}f\mathbb{1}_K$ in $L_1$, and then through a subsequence $\varepsilon_n\rightarrow0$ get convergence a.s.)
To conclude, if $f=\mathbb{1}_A$ with $A+x_n=A$ for all $n\in\mathbb{N}$, then $\mathbb{1}_A$ is constant a.s. Hence, either $\lambda(A)=0$ or $\lambda(A^c)=0$.

Edit: Doing a quick review of the problem again in MSE I found that there are at least three postings 1, 2, and 3 that addressed this questions. The solutions in those postings are different  to the ones presented  here so far, but very interesting too. I also just learnt that the OP is appears as an exercise in Real and Complex Analysis, Rudin, W. chapter 7.
