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Show that the set $R^n$ x 0 has measure zero in $R^{n+1}$ This question has been asked before, I'm sure all the answers given are great but due to my relative novelty to real analysis I was unable to understand any of the answers. Here is a link to the previous post: $\mathbb{R}^n\times\{0\}$ has measure zero in $\mathbb{R}^{n+1}$ Here is my attempt at the solution : Choose a partition for the given set of the form [k- $\frac{\epsilon}{2^{k+1}}$, k + $\frac{\epsilon}{2^{k+1}}]^n$ x {0} for k $\in \mathbb{R} $ Then the partition covers the entire set, and each interval has length $\frac{\epsilon}{2^n}$ and the total volume $\sum_{i=0}^n \frac{\epsilon}{2^i}$ = $\epsilon$ as required .

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  • Well.. the 'partition' you mention lives in dimension 1 only, and even there, it does not cover the whole one dimensional space (i.e. line). – Berci Dec 26 '15 at 18:38
  • Be careful in your computations; in this question and in previous questions that you've asked, you seem to express subsets of $\mathbb{R}^n$ as intervals, which is not true. – Michael Burr Dec 26 '15 at 18:39
  • The accepted answer of the question that you linked is basically what Noah has suggested below. See if you can connect the dots. In that answer (in the case $n=1$), $\delta$ is the height of the "thin rectangle" covering the line segment $[-k,k] \times {0}$. – angryavian Dec 26 '15 at 18:43
  • @angryavian In the accepted answer of that question I didn't understand their use of ∏∏ and m(Rn×{0})=limk→∞m([−k,k]n×{0})m(Rn×{0})=limk→∞m([−k,k]n×{0}) – ಠ_ಠ Jan 05 '16 at 01:06

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This is almost correct, but $[0-{\epsilon\over 2^{k+1}}, 0+{\epsilon\over 2^{k+1}}]$ is a subset of $\mathbb{R}$, not $\mathbb{R}^{n+1}$.

You've covered $\{0\}$ with small intervals, in $\mathbb{R}$. What you want to do is cover $\mathbb{R}^n\times \{0\}$ with small boxes. Do you see how to fix this? (HINT: If $n=1$, you would use "thin strips" . . .)

EDIT: As user46944 points out below, you seem to be using "$n$" where you mean "$0$" (and I mindlessly copied :P); and you probably want another $n$ to be some other letter, say "$k$". I've edited my answer. Do you see the difference?

Noah Schweber
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Let the Euclidean volume of an $n$-sphere of radius $r$ be $V_n(r).$ Let $\|\cdot \|_n$ denote the Euclidean $n$-dimensional norm. For $j\in N$ let $$S(j)=\{x\in R^n :\|x\|_n<j\}.$$ Let $m$ denote $(n+1)$-dimensional measure. For any $k>0$ and any $ j\in N$ let $$T(j,k)=S(j)\times (-k 2^{-j}/V_n(j),+k 2^{-j}/V_n(j).$$ We have $(R^n\times \{0\})\subset (\cup_{j\in N}T(j,k))$ and $$m(\cup_{j\in N}T(j,k)\leq \sum_{j\in N}m(T(j,k))=2 k.$$ Since $k$ can be arbitrarily small, we have $m(R^n\times \{0\})=0.$

DanielWainfleet
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