Definition
Let $A$ be a subset of $\Bbb R^n$. We say $A$ has measure zero in $\Bbb R^n$ if for every $\epsilon>0$ there is a covering $Q_1,Q_2,…$ of $A$ by countably many rectangles such that
$$
\sum_{i=1}^\infty v(Q_i)<\epsilon
$$
If this inequality holds, we often say that the total volume of the rectangles $Q_1,Q_2,…$ is less than $\epsilon$.
Statement
The set $\Bbb R^{n-1}\times\{t\}$ has measure zero in $\Bbb R^{n-1}$ for any $t\in\Bbb R$
Unfortunately I don't be able to prove the statement so I ask to do it. So could someone help me, please?
Best Answer
Consider the rectangles $Q_i = [-i, i]^{n - 1} \times [t - \frac{\epsilon}{2^{n + 1} i^{n - 1} 2^i}, t + \frac{\epsilon}{2^{n + 1}i^{n - 1} 2^i}]$. Then $Q_i$ has volume $\frac{\epsilon}{2^{i + 1}}$, so $\sum v(Q_i) = \frac{\epsilon}{2} < \epsilon$.