# How to show a set has measure zero?

1. Nov 6, 2009

### 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.

2. Nov 6, 2009

### 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.

3. Nov 6, 2009

### Matthollyw00d

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

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

4. Nov 6, 2009

### wofsy

This one is easy using epsilon neighborhoods.

5. Nov 6, 2009

### 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.

6. Nov 7, 2009

### Matthollyw00d

The way I did it earlier used
$$\bigcup$$n=1$$\infty$$ [-n,n]n-1x{0} which is a countable cover of Rn-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.

7. Nov 7, 2009

### rochfor1

That works too.