A positive-measure compact set can miss part of every interval

Enumerate $\mathbb{Q}\cap (0,1)$ as $(q_k{)}_{k\ge 1}$ using [L1]. For each [L1, given, choose, construct] $k$, choose an open interval $I_k$ centred at $q_k$ of length below $2^{-k-2}$, and put $U:={\bigcup }_{k\ge 1}{I}_k\mathrm{.}$ The set $U$ is open and dense in $(0,1)$ because every nonempty open [L1, L2, L3, given, choose, construct, algebra] interval in $(0,1)$ contains a rational point $q_k$ and hence meets $I_k$. Also [L3] and [L2] give $\lambda (U)\le {\sum }_{k=1}^{\mathrm{\infty }}\lambda (I_k)<{\sum }_{k=1}^{\mathrm{\infty }}{2}^{-k-2}=\frac{1}{4}\mathrm{.}$

Define [L4, L5, step 1.1, algebra] $K:=[0,1]\setminus U\mathrm{.}$ Then $K$ is closed and bounded, hence compact by [L4]. By [L5] and step 1.1, $\lambda (K)\ge \lambda ([0,1])-\lambda (U)>1-\frac{1}{4}=\frac{3}{4},$ so $K$ has positive measure. Let $J$ be a nonempty open interval in $\mathbb{R}$. If $J$ is not [L4, L5, step 1.1, algebra] contained in $[0,1]$, then $J$ already contains a point outside $K\subseteq [0,1]$. If $J\subseteq (0,1)$, then step 1.1 gives $J\cap U\mathrm{\ne }\mathrm{\varnothing }$, so $J$ contains a point outside $K$. Thus every nonempty open interval contains a point outside $K$.

The compact set $K$ therefore has positive measure while missing part of [step 2.1] every interval.