This question has been reduce to find a compact set $K$ such that for any $x\in K$, there are infinite $n$ such that $x+\frac{1}{n}\not\in K$".

As I have understood David C. Ullrich's answer, it is also required that $m(K)>0$.

I just received a letter from Taras Banakh with the following

**Corollary.** For any decreasing sequence $(a_n)_{n=1}^\infty$ of positive real numbers with $\lim_{n\to\infty}a_n=0$ and $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=1$, there exists a compact subset $K\subset\mathbb R$ of positive Lebesgue measure such that $\bigcap_{n=m}^\infty(K-a_n)=\emptyset$.

Corollary follows from Theorem, see below.

Let $\mathcal K(\mathbb R)$ be the space of non-empty compact subsets of the real line, endowed with the Vietoris topology (which is generated by Hausdorff metric). It is well-known that the space $\mathcal K(\mathbb R)$ is locally compact (more precisely, each closed bounded subset of $\mathcal K(\mathbb R)$ is compact.

For every $c>0$ let $\mathcal K_c(\mathbb R)=\{K\in\mathcal K(\mathbb R):\lambda(K)\ge c\}$ be the subspace of consisting of compact sets $K$ of Lebesgue measure $\lambda(K)\ge c$. The regularity (or countable additivity) of the Lebesgue measure $\lambda$ implies that $\mathcal K_c(\mathbb R)$ is a closed subspace in $\mathcal K(\mathbb R)$. Consequently the space $\mathcal K_c(X)$ is locally compact and Polish.

**Theorem.** Let $(a_n)_{n=1}^\infty$ be a decreasing sequence of positive real numbers such that $\lim_{n\to\infty}a_n=0$ and $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=1$. Then for every $c>0$ the set
$$\{K\in\mathcal K_c(X):\forall m\in\mathbb N\;\bigcap_{n=m}^\infty(K-a_n)=\emptyset\}$$ is dense $G_\delta$ in $\mathcal K_c(X)$.

*Proof*. It suffices to prove that for every $m\in\mathbb N$ the set
$$\mathcal K_m=\{K\in\mathcal K_c(X):\bigcap_{n=m}^\infty(K-a_n)=\emptyset\}$$ is open and dense in $\mathcal K_c(X)$.

To see that $\mathcal K_n$ is open, take any compact set $K\in\mathcal K_m$. By the compactness of $K$, there exists $l>m$ such that $\bigcap_{n=m}^l(K-a_n)=\emptyset$.
Then for every $x\in \mathbb R$ there exists $n_x\in[m,l]$ such that $x\notin K-a_{n_x}$. So, we can find a symmetric neighborhood $U_x\subset[-1,1]$ of zero such that $x+U_x$ is disjoint with $U_x+K-a_{n_x}$. By the compactness of the set $L=[-1,1]+(K-a_m)$, there exists a finite subset $F\subset L$ such that $L\subset\bigcup_{x\in F}(x+U_x)$. We claim that the open neighborhood $\mathcal U=\{K'\in\mathcal K_c(X):K'\subset\bigcap_{x\in F}(K+U_x)\}$ of $K$ in $\mathcal K_c(X)$ is contained in the set $\mathcal K_m$. Assuming that $\mathcal U\not\subset\mathcal K_m$, we can find a compact set $K'\in\mathcal U\setminus \mathcal K_m$. Since $K'\notin\mathcal K_m$, there exists a point $z\in \bigcap_{n=m}^\infty (K'-a_n)$. It follows that $z\in K'-a_m\subset [-1,1]+K-a_m=L$ and hence $z\in x+U_x$ for some $x\in F$. Then $$z\in (x+U_x)\cap\bigcap_{n=m}^\infty K'-a_n\subset (x+U_x)\cap (K'-a_{n_x})\subset (x+U_x)\cap (U_x+K-a_{n_x})=\emptyset,$$ which is a desired contradiction, showing that $\mathcal U\subset\mathcal K_m$ and the set $\mathcal K_m$ is open in $\mathcal K_m$.

Next, we show that $\mathcal K_m$ is dense in $\mathcal K_c(\mathbb R)$. Given any $K\in\mathcal K_c(\mathbb R)$ and $\varepsilon>0$ we need to find a set $K'\in\mathcal K_m$ on the Hausdorff distance $d_H(K',K)<\varepsilon$ for $K$.

Choose $k\ge m$ so large that $\frac1k<\frac12\varepsilon$. Consider the cover $\mathcal I_k=\big\{\big[\frac{n}k,\frac{n+1}k\big]:n\in\mathbb N\big\}$ of $\mathbb R$ by closed intervals of length $\frac1n$. The choice of $k$ ensures that the compact set $\tilde K=\bigcup\{I\in\mathcal I_n:I\cap K\ne\emptyset\}$ has $d_H(\tilde K,K)\le \frac1k<\frac12\varepsilon$. Also it is clear that $\tilde K$ has Lebesgue measure $\lambda(\tilde K)>\lambda(K)\ge c$. Choose $p\ge k$ so large that $(1-\frac1p)\cdot\lambda(\tilde K)\ge c$.

Since $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=1$, there exists $q>p$ such that $\frac{a_{n+1}}{a_{n}}>1-\frac1{4p}$ for all $n\ge q$.
Finally, take $l>q$ such that $\frac2{lk}<a_q$.
Consider the open set $$Z:=\bigcup_{z\in\mathbb Z}{\big]}\tfrac{z}{kl},\tfrac{z}{kl}+\tfrac1{pkl}{\big[}$$ and observe that for every interval $I\in\mathcal I_k$ the set $I\setminus Z$ has Lebesgue measure $\lambda(I\setminus Z)=(1-\frac1p)\lambda(I)$ and Hausdorff distance $d_H(I,I\setminus Z)<\frac12\varepsilon$. Consequently, the compact set $K'=\tilde K\setminus Z$ has Lebesgue measure $\lambda(K')=(1-\frac1p)\cdot\lambda(\tilde K)\ge c$ and $d_H(K',\tilde K)=\frac1{2pkl}<\frac12\varepsilon$. Then $d_H(K',K)\le d_H(K,\tilde K)+d_H(\tilde K,K)<\varepsilon$. It remains to prove that $K'\in\mathcal K_m$. Assuming that $K'\notin\mathcal K_m$, we could find a point $x\in\bigcap_{n=m}^\infty (K'-a_n)$. Then $x+a_n\in K'\subset\mathbb R\setminus Z$ for all $n\ge m$. Let $z\in Z$ be the unique integer number such that $\frac{z-1}{lk}<x\le \frac{z}{lk}$. The choice of $l$ guarantees that $\frac2{lk}<a_q$. Then $\frac{z+1}{lk}=\frac{z-1}{lk}+\frac2{lk}<x+a_q$. Let $i\ge q$ be the smallest number such that $x+a_i\ge \frac{z}{lk}+\frac1{plk}$. Then $x+a_{i+1}<\frac{z}{lk}+\frac1{plk}$ and hence $a_{i+1}<\frac{z}{lk}+\frac1{plk}-x<\frac{z}{lk}+\frac1{plk}-\frac{z-1}{lk}=
\frac1{lk}+\frac1{plk}<\frac2{lk}$.

On the other hand, $$a_i-a_{i+1}=a_{i+1}\tfrac{a_i}{a_{i+1}}
(1-\tfrac{a_{i+1}}{a_i})<\tfrac2{lk}(1-\tfrac1{4p})^{-1}\tfrac1{4p}<\tfrac1{lkp}$$ and hence $x+a_{i+1}>x+a_i-\frac1{lkp}\ge\frac{z}{lk}+\frac1{plk}-\frac1{plk}=\frac{z}{kl}$.
Then $\frac{z}{kl}<x+a_{i+1}<\frac{z}{lk}+\frac1{lkp}$ implies that $x+a_{i+1}\in Z$, which is a desired contradiction, completing the proof of $K'\in\mathcal
K_m$.$\square$

PS. Also Taras Banakh sent to me the reference to the talk “Theorems of H. Steinhaus, S. Picard
and J. Smital” by W. Wilczyński from Ger-Kominek Workshop
in Mathematical Analysis and Real Functions 20-21.11.2015.

At the last page are stated the following results.

**Theorem.** If $A\subset\Bbb R$ is a measurable set, then for each sequence $\{x_n\}_{n\in\Bbb N}$ convergent to $0$ the sequence of characteristic functions
$\{\chi_{A+x_n}\}_{n\in\Bbb N}$ converges in measure to $\chi_A$.

**Remark.** There exists a measurable set $A\subset\Bbb R$ and a sequence
$\{x_n\}_{n\in\Bbb n}$ convergent to $0$ such that $\{\chi_{A+x_n}\}_{n\in\Bbb N}$
does nor converge almost everywhere to $\chi_A$.

**Theorem.** If $A$ has the Baire property, then $\{\chi_{A+x_n}\}_{n\in\Bbb N}$
converges to $\chi_A$ except on a set
of the first category.