Outer Measure

**Sampras** · July 23rd 2009, 11:51 PM

So one of the axioms of outer measure is the following:

Countable subadditivity axiom: If $A = \bigcup_{n=1}^{\infty} A_n$ then $m^{*}A \leq \sum_{n=1}^{\infty} m^{*}A_n$ .

Well it's not really an axiom since we want/have to prove it. But what is the point of using the " $\varepsilon/2^n$ trick" to prove this? E.g. we have the following:

$\sum_{k=1}^{\infty} |I_{kn}| < m^{*} A_n+ \frac{\varepsilon}{2^n}$ (1)

This is analogous to $\inf S \leq x \Rightarrow x < \inf S + \varepsilon$ for some $\varepsilon >0$ (with $x \in S$ ). But what is special about $\varepsilon/2^n$ ? Why not just use $\varepsilon$ ?

**Moo** · July 24th 2009, 02:24 AM

Hello,

Because $\sum_{n=1}^\infty \frac{\varepsilon}{2^n}=\varepsilon$ , wouldn't it ?

**Sampras** · July 24th 2009, 05:25 AM

Originally Posted by Moo

Hello,

Because $\sum_{n=1}^\infty \frac{\varepsilon}{2^n}=\varepsilon$ , wouldn't it ?

yes, but the sum doesn't have to sum to $\varepsilon$ since it's arbitrary. I guess its convention.

**Moo** · July 24th 2009, 06:44 AM

Originally Posted by Sampras

yes, but the sum doesn't have to sum to $\varepsilon$ since it's arbitrary. I guess its convention.

Not really...

As you're dealing with a limit, you want $\varepsilon$ to appear. This one is arbitrary !

Since you'll sum from 1 to infinity, if you keep $\varepsilon$ , you'll sum something positive an infinite amount of times.
Which is irrelevant for your problem.

$\varepsilon/2^n$ is arbitrary and > 0. It's a good candidate, because of this infinite sum.

And you'll get $\varepsilon$ in the end.

You could also have taken $\varepsilon/4^n$ or whatever you want. The main point is to get a converging series, arbitrary small.

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