Lebesgue null set with meagre complement > There exists a $\lambda$-null comeagre subset of $\mathbb{R}$. I tried to find an example in terms of Cantor sets. Let $C \subseteq [0,1]$ be the ternary Cantor set. Define $$ S = \bigcup_{x \in \mathbb{Q}} C + x $$ Then $S$ is $\lambda$-null, but how do I show that $S^c$ is meagre in $\mathbb{R}$? I only know little about the properties of $[0,1] \setminus C$.

Try another approach:

For every $n$ and rational $q_k$, take a small open interval $I_{n,k}$ such that their union $O_n$ has measure $\le \frac{1}{n}$. Now $\cap_n O_n$ is as required.