A null set can fail to be the discontinuity set of any function

By [L1], fix an enumeration $(q_j{)}_{j\ge 0}$ of $\mathbb{Q}$. For each natural number $k\ge 1$, put $U_k:={\bigcup }_{j\ge 0}(q_j-2^{-j-k-2},q_j+2^{-j-k-2})\mathrm{.}$ Every $U_k$ is open and dense, because it contains every rational. Each interval in the union has length $2^{-j-k-1}$, so [L2], [L3], and [L10] give ${\lambda }_1(U_k)\le {\sum }_{j=0}^{\mathrm{\infty }}{2}^{-j-k-1}=2^{-k-1}{\sum }_{j=0}^{\mathrm{\infty }}{2}^{-j}=2^{-k}\mathrm{.}$ Let $G:={\bigcap }_{k\ge 1}{U}_k\mathrm{.}$ Then [L4] makes $G$ a $G_{\delta }$ set, and [L6] makes it dense. Since $G\subseteq U_k$ for every $k$, [L9] gives ${\lambda }_1(G)\le 2^{-k}$ for all $k$, so ${\lambda }_1(G)=0.$

Suppose, for contradiction, that $G$ is the discontinuity set of some function $f:\mathbb{R}\to \mathbb{R}$. Then [L5] makes $G$ an $F_{\sigma }$ set, so by [L4] write $G={\bigcup }_{n\ge 0}{F}_n$ with each $F_n$ closed. Because $F_n\subseteq G$ and ${\lambda }_1(G)=0$, monotonicity [L9] gives ${\lambda }_1(F_n)=0$ for every $n$. A closed null set cannot contain a nondegenerate interval, by [L8], so each $F_n$ has empty interior and is nowhere dense by [L7]. Therefore $G$ is meager.

Step 1.1 writes $G$ as a dense $G_{\delta }$ set, so $\mathbb{R}\setminus G$ is meager by [L7]. If $G$ were also meager, then $\mathbb{R}=G\cup (\mathbb{R}\setminus G)$ would be a union of two meager sets and therefore meager, contradicting [L6]. So $G$ is a Lebesgue null set that is not the discontinuity set of any real-valued function, and the statement is false.