If $\{a\}$ is a singleton in $\mathbb{R}$, is the lebesgue outer measure of $A:=\mathbb{R} \times \dots \times\{a\} \times \dots \times \mathbb{R} \subset{\mathbb{R}}^d$ zero?
Let $\delta:=\frac{\epsilon}{2^{n+d+1}n^d}$. Then $(-n,n) \times \dots (a-\delta,a+\delta) \times \dots \times (-n,n)$ is a sequence of open intervals in $\mathbb{R}^d$ covering $A$, each of length $\frac{\epsilon}{2^n}$, so $\lambda^{*}(A)\leq \epsilon$. Since $\epsilon>0$ is arbitrary we conclude $\lambda^{*}(A)=0$.
Is this right?