Show that E and the closure of E have the same Jordan outer measure

[cragar]

Homework Statement

Let E be a bounded set in $R^n$
Show that E and the closure of E have the same Jordan outer measure

Homework Equations

Jordan outer measure is defined as $m^* J(E)=inf(m(b))$
where $B \supset E$ B is elementary.
3. The Attempt at a Solution [/B]
If E and the closure of E are the same there is nothing to prove.
other case cl(E) contains limit points of E that are not in E.
Let's first look at the measure of E. BY the definition of Jordan outer measure, The closure of E will contain the inf(B) and E is a subset of B and the closure. Let B, be the closure of E, The inf(B) will be the measure of E.
Now this is where I am a little stuck. The Close of E, has all its limit points, so the inf(E) is part of E, its like a closed interval. There is no set B such that inf(m(B)). Unless i pick B to be the set such that
$B=E+ \epsilon$ so the inf of the measure of B is E.

 

[mighty2000]

To prove that E and the closure of E have the same Jordan outer measure, we can use the fact that the Jordan outer measure is monotone, meaning that if A is a subset of B, then m^*(A) <= m^*(B).

Let's assume that E and the closure of E have different Jordan outer measures, meaning that m^*(E) < m^*(cl(E)). Since E is a bounded set, we can find a closed rectangle R such that E is contained in R and m^*(R) < m^*(cl(E)).

Since E is bounded, R is also a bounded set. This means that R is a subset of some elementary set B, which we can write as B = R + ε, where ε is a positive number.

By the monotonicity of the Jordan outer measure, we have m^*(R) <= m^*(B). But since B contains E, we also have m^*(B) <= m^*(cl(E)).

Combining these two inequalities, we get m^*(R) <= m^*(cl(E)). But this contradicts our assumption that m^*(R) < m^*(cl(E)). Therefore, our assumption must be wrong and we can conclude that m^*(E) = m^*(cl(E)).

Hence, E and the closure of E have the same Jordan outer measure.