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I needed help in showing that the set $R^{n-1} \times 0$ has measure zero in $R^n$.

What I have so far: Let $\epsilon > 0$. If $i_1,\dots,i_{n-1}$ are integers, then define $U_{i_1,\dots,i_{n-1}}=[i_1,i_1+1]\times \cdots \times [i_{n-1},i_{n-1}+1]$. Chose a bijection $f:Z\times Z\times \cdots \times Z\to N$ (the product has $n-1$ factors) where $N$ is the set of positive integers and define $A_{i_1,\dots,i_{n-1}}=U_{i_1,\dots,i_{n-1}} \times [-2^{-f(i_1,\dots,i_{n-1})-1}\epsilon,2^{-f(i_1,\dots,i_{n-1})-1}\epsilon]$. The collection of all such $A_{i_1,\dots,i_{n-1}}$'s covers $R^{n-1}\times 0$ and the sum of the measures of the $A_{i_1,\dots,i_{n-1}}$'s is less than $\epsilon$. (Why?)

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You have asked 10 questions, with 9 of them being answered. But you only accepted 1 answer. Could you improve your accept rate by accepting answers? –  Paul Jan 11 '12 at 2:22
    
Yes, although I am a bit unsure how to accept answers –  Buddy Holly Jan 11 '12 at 2:24
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You can look at here: meta.math.stackexchange.com/questions/3286/… –  Paul Jan 11 '12 at 2:33
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2 Answers

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  • First, let's see that your $U$'s cover $\mathbb{R}^{n-1} \times \{0\}$. Suppose that $(x_1,\dots,x_{n-1},0) \in \mathbb{R}^{n-1} \times \{0\}$. Then let $m_i$ be the largest integers less than or equal to $x_i$ (the floor of $x_i$). Thus $m_i \leq x_i \leq m_i+1$ and so $(x_1,\dots,x_{n-1},0) \in [m_1,m_1+1]\times [m_2,m_2+1]\times \cdots \times [m_{n-1},m_{n-1}+1]\times [-thing,thing]$. Thus $(x_1,\dots,x_{n-1},0) \in U_{m_1,\dots,m_{n-1}}$. So the $U$'s cover our set.

  • Next, the length of each $[i_j,i_j+1]$ is 1. So the volume of $U_{i_1,\dots,i_{n-1}}$ is $1 \times \cdots 1 \times (2^{-f-1}\epsilon-(-2^{-f-1}\epsilon)) = 2\cdot 2^{-f-1} \epsilon = 2^{-f}\epsilon$ where $f=f(i_1,\dots,i_{n-1})$. So the total volume of the $U$'s is $\sum\limits_{i's} 2^{-f(i_1,\dots,i_{n-1})}\epsilon$. But $f$ gives a bijection with the natural numbers so we can rewrite this sum as $\sum\limits_{k=0}^\infty 2^{-k}\epsilon = 2\epsilon$ (geometric series). By the way this series can be rearranged by the function $f$ without worry since its terms are positive and thus it's absolutely convergent.

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An easier way to see this is that we simply need to cover $\mathbb{R}^{n-1}$ with countably many sets that have measure zero. This is because the countable union of sets that have measure zero also has measure zero. First place an integer lattice on $\mathbb{R}^{n-1}$ in the same way that $\mathbb{Z}\times\mathbb{Z}$ is a subset of $\mathbb{R}^{2}$. Then for $2(n-1)$ lattice points that are next to each other, form a cube in $\mathbb{R}^{n-1}$. Each one of these cubes will have measure zero in $\mathbb{R}^{n}$ since they each have a side of length zero. Hence, we have covered $\mathbb{R}^{n-1}$ with a set of measure zero, and we are done.

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