How to show a set has measure zero?

**Matthollyw00d** · Nov6-09, 01:34 AM

I'm not completely familiar with measures yet, but am trying to be. I'm trying to show a that a few sets in Rⁿ have measure zero and am having difficulty showing this using countable covers with a total volume less than $LaTeX Code: \\epsilon$ . Is there an easier way to show that a set has measure zero? If it's needed, I could give the problems that I'm working with. Any help would be appreciated.

**rochfor1** · Nov6-09, 09:29 AM

The method you mention is really the only general one. If you give the specific sets you're working with I may be able to give a simpler method in that specific situation.

**Matthollyw00d** · Nov6-09, 10:53 AM

Show that R^n-1 x {0} has measure zero in Rⁿ.

This is one of the ones I'm working on.

**wofsy** · Nov6-09, 12:02 PM

Originally Posted by Matthollyw00d

Show that R^n-1 x {0} has measure zero in Rⁿ.

This is one of the ones I'm working on.

This one is easy using epsilon neighborhoods.

**rochfor1** · Nov6-09, 10:01 PM

I'll expand on wofsy's comment a bit. You do use the open cover formulation on this one. Let $LaTeX Code: \\{ q_k \\}_{ k = 1 }^\\infty$ be an enumeration of $LaTeX Code: \\mathbb{Q} \\times \\{ 0 \\}$ . Show that if $LaTeX Code: \\varepsilon > 0$ , $LaTeX Code: \\left \\{ ( q_k - \\frac{ 1 }{ 2 }, q_k + \\frac{ 1 }{ 2 } ) \\times ( -\\varepsilon\\ 2^{ - ( k + 1 ) }, \\varepsilon\\ 2^{ - ( k + 1 ) } ) \\right \\}_{ k = 1 }^\\infty$ is an open cover of $LaTeX Code: \\mathbb{ R } \\times \\{ 0 \\}$ with total measure less than or equal to $LaTeX Code: \\varepsilon$ .

EDIT: This proof works for showing that $LaTeX Code: \\mathbb{ R } \\times \\{ 0 \\} \\subset \\mathbb{ R }^2$ has measure zero. The

case is a straightforward generalization.

**Matthollyw00d** · Nov7-09, 03:45 AM

The way I did it earlier used
$LaTeX Code: \\bigcup$ _n=1 [-n,n]^n-1x{0} which is a countable cover of R^n-1x{0}. Then using the fact that a countable union of sets of measure zero also has measure zero, you can show that each set has measure zero and the rest follows. After reading a bit on it today, I feel like I have a much better understanding of it now.

**rochfor1** · Nov7-09, 10:36 AM

That works too.