$E = \mathbb{R}\times e$ is a n-dimensional measure zero if $e \subset \mathbb{R}^{n-1}$ of (n-1) dimensional measure zero

Question

In the book of Mathematical Analysis II by Zurich, at page 116 it is asked that

Show that if a set $E \subset \mathbb{R}^n$ is the direct product of $\mathbb{R}\times e$ of the line $\mathbb{R}$, and a set $e \subset \mathbb{R}^{n-1}$ of (n-1) dimensional measure zero, then $E$ is a set of n-dimensional measure zero.

But, I cannot see how can this be, and do not know how to prove it, so I would appreciate any help for proving this statement.

Edit:

I do not understand the given answer, so please feel free to post another answers.

Answer 1

From the construction of the product measure $\lambda_n$ (Lebesgue measure of $\mathbb{R}^n$),

if $A\in \mathcal{B}(\mathbb{R}^k)$ and $B\in \mathcal{B}(\mathbb{R}^{n-k})$, ($0<k<n$), then $$ \lambda_n(A\times B) =\lambda_k(A)\lambda_{n-k}(B)$$

So here (using also that if $E_m \uparrow E$, $\lim_{m\rightarrow +\infty}\lambda_n(E_m)=\lambda_n(E)$) $$\begin{align*} \lambda_n(E)&= \lambda_n(\mathbb{R}\times e) \\ &=\lim_{m\rightarrow +\infty}\lambda_n((-m,m)\times e)\\ &=\lim_{m\rightarrow +\infty}\lambda_1((-m,m))\times \lambda_{n-1}(e)\\ &=\lim_{m\rightarrow +\infty}\lambda_1((-m,m)) \times 0 \\ &= 0 \end{align*} $$