How to show a set has measure zero?

[Matthollyw00d]

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 \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]

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]

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

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

 

[wofsy]

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]

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

EDIT: This proof works for showing that \mathbb{ R } \times \{ 0 \} \subset \mathbb{ R }^2 has measure zero. The n > 2 case is a straightforward generalization.

 

[Matthollyw00d]

The way I did it earlier used
\bigcup_n=1^\infty [-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]

That works too.