A Lebesgue measurable subset of $[0,1]$ that is invariant under changing finitely many binary digits has measure $0$ or $1$

Let $m:=\lambda (A)$ and let $D$ be the set of dyadic rationals in $[0,1]$, that is the numbers $k\mathrm{/}{2}^n$ with $n\in \mathbb{N}$ and $0\le k\le 2^n$. The set $D$ is at most countable, hence null by [L4]. For fixed $n\in \mathbb{N}$ and $0\le k<2^n$, write $I_{n,k}:=[k\mathrm{/}{2}^n,(k+1)\mathrm{/}{2}^n)$, and for $k=2^n-1$ replace the right endpoint by $1$ so that the family $(I_{n,k}{)}_{k<2^n}$ partitions $[0,1]$.

Suppose, for contradiction, that $0<m<1$. Choose a real $\epsilon >0$ with $m-\epsilon >m(m+\epsilon )$. By [L3] choose a compact set $K\subseteq A$ with $\lambda (K)>m-\epsilon$ and an open set $U\supseteq A$ with $\lambda (U)<m+\epsilon$.

Fix $n$ and $k,\mathrm{\ell }<2^n$. Away from the dyadics, translating $I_{n,k}$ onto $I_{n,\mathrm{\ell }}$ changes only the first $n$ binary digits, so the invariance hypothesis and [L1] give $\lambda (A\cap I_{n,k})=\lambda (A\cap I_{n,\mathrm{\ell }})$. Summing over the partition from step 1.1 and using [L2] gives $\lambda (A\cap I_{n,k})=m{2}^{-n}$ for every $k$, and therefore for every union $J$ of generation-$n$ dyadic intervals one has $\lambda (A\cap J)=m\lambda (J)$.

For each $x\in K$, openness of $U$ gives a real $r_x>0$ with $(x-r_x,x+r_x)\subseteq U$. The smaller interval $(x-r_x\mathrm{/}2,x+r_x\mathrm{/}2)$ still contains $x$, so compactness of $K$ gives finitely many points $x_1,\dots ,x_N\in K$ such that the intervals $J_i:=(x_i-r_i\mathrm{/}2,x_i+r_i\mathrm{/}2)$, where $r_i:=r_{x_i}$, cover $K$. Let $\rho :={\min }_{1\le i\le N}(r_i\mathrm{/}2)>0$, choose $n$ with $2^{-n}<\rho$, and let $J$ be the union of the generation-$n$ dyadic intervals meeting $K$. Then $K\subseteq J$. To prove $J\subseteq U$, let $D$ be one of those dyadic intervals and choose $z\in D\cap K$. Pick $i$ with $z\in J_i$. For any $y\in D$ one has $\mathrm{\mid }y-x_i\mathrm{\mid }\le \mathrm{\mid }y-z\mathrm{\mid }+\mathrm{\mid }z-x_i\mathrm{\mid }<2^{-n}+r_i\mathrm{/}2<r_i$, because $D$ has length $2^{-n}$ and $z\in J_i$. Hence $y\in (x_i-r_i,x_i+r_i)\subseteq U$. So every such dyadic interval $D$ lies in $U$, and therefore $J\subseteq U$.

Step 2.1 applied to the set $J$ gives $\lambda (A\cap J)=m\lambda (J)$. Since $K\subseteq A\cap J\subseteq U$, steps 1.2 and 2.2 imply $m-\epsilon <\lambda (K)\le \lambda (A\cap J)=m\lambda (J)\le m\lambda (U)<m(m+\epsilon ),$ contradicting the choice of $\epsilon$. Therefore $m$ cannot lie strictly between $0$ and $1$, and $\lambda (A)\in \{0,1\}$.