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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?

Antonio Maria Di Mauro
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2 Answers2

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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 $\sum v(Q_i) = \frac{\epsilon}{2} < \epsilon$.

Doctor Who
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I'll stick to $t=0$.

If one allows rectangles of zero "height" as rectangles, then one can cover $\Bbb R^{n-1}\times\{0\}$ by rectangles $[-N,N]^{n-1}\times\{0\}$ for $N\in N$.

If one doesn't, use instead $[-N,N]^{n-1}\times[0,\epsilon/((2N)^{n-1}2^N)]$.

Angina Seng
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