Prove that $\lambda^{*}(A)=0$, where $\lambda^{*}$ is $n$-dimensional Lebesgue measure on $\mathbb{R}^n$ and $A$ is $k$-dimensional subspace of $\mathbb{R}^n$ and $k<n$.
I've proved this for $n=2$ and $k=1$.
Let $A$ be a linear $1$-dimensional subspace of $\mathbb{R}^2$. Thus, there is an isometry mapping $A$ to $\mathbb{R}$. But we know that $\lambda^{*}$ is invariant under isometries, so $\lambda^{*}(\mathbb{R})=0$ is enough to prove that $\lambda^{*}(A)=0$.
Let $\varepsilon >0$. Consider
$$\mathcal{C}=\{[\frac{1}{n},\frac{1}{n+1}] \times [-\frac{\varepsilon}{n},\frac{\varepsilon}{n}] : 1 \leq n, n \in \mathbb{N} \} \cup \{[-\frac{1}{n},-\frac{1}{n+1}] \times [-\frac{\varepsilon}{n},\frac{\varepsilon}{n}] : 1 \leq n, n \in \mathbb{N} \} \cup [-1,1] \times [-\varepsilon,\varepsilon ]$$
It's an cover of $\mathbb{R}$ (because series $\sum_{n=1}^{\infty}\frac{1}{n}$ is divergent) and:
$$\sum_{B \in \mathcal{C}}\lambda^{*}(B) \to 0$$
when $\varepsilon \to 0$ (because series $\sum_{n=1}^{\infty}\frac{1}{n^2}$ is convergent).
How can I do something similar for other $k,n$?