1

Following Lebesgue measure of any line in $\mathbb{R^2}$. I want to show that the lebesgue outer measure of $\mathbb{R}$ in $\mathbb{R}^2$ is zero.

Does the following constitue a good way to show this fact?

Take $R_n = [n - \dfrac{1}{2} , n + \dfrac{1}{2} ] \times [-\epsilon/2^n, \epsilon/2^n], n \in \mathbb{Z}$

This is a cover of $\mathbb{R}$

$\sum_n |R_n| = 2\epsilon$, then $\mu^*(\mathbb{R}) = 0$

Correctomundo?

Sarah Palins Anger
  • 138

1 Answers1

2

In principle, yes.

  1. Usually, an open cover is used.
  2. Remember the nagatives. You need $2^{-|n|}$ if $n$ is negative.

Therefore, I'd replace your expression for $R_n$ by $(n-1,n+1)\times \epsilon (-2^{-|n|},2^{-|n|})$. This has measure $4\epsilon 2^{-|n|}$. Now sum.

Fnacool
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