A Bernstein set has inner measure $0$, and in every nondegenerate interval its intersection has full outer measure

Every compact subset of $B$ is countable by [L1], hence has Lebesgue measure $0$ by [L2]. Since $0$ is attained by the compact set $\mathrm{\varnothing }$, the supremum in [F1] is exactly $0$. Therefore ${\lambda }_{\ast }(B)=0$.

Let $I$ be a nondegenerate bounded interval, and assume ${\lambda }^{\ast }(B\cap I)<\lambda (I)$. Choose a closed nondegenerate interval $J\subseteq I$ with ${\lambda }^{\ast }(B\cap I)<\lambda (J)$; this is possible by moving any omitted endpoints inward by a sufficiently small amount. By [L3] choose an open set $U\supseteq B\cap I$ with $\lambda (U)<\lambda (J)$. Then $F:=J\setminus U$ is closed in $\mathbb{R}$, satisfies $F\subseteq I\setminus B$, and is measurable. Because $U\cap J$ is measurable and contained in $U$, [L4] gives $\lambda (F)=\lambda (J)-\lambda (U\cap J)\ge \lambda (J)-\lambda (U)>0.$ In particular $F$ is uncountable by [L2]. The closed uncountable set $F$ has a nonempty perfect subset: its set of condensation points is closed, nonempty, and has no isolated points by the argument of Every compact subset of a Bernstein set is countable. That perfect subset lies inside $I\setminus B$, contradicting the Bernstein property. Therefore ${\lambda }^{\ast }(B\cap I)=\lambda (I)$.

Step 1.1 is claim 1, and step 1.2 is claim 2.