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This is an exercise in real analysis:

Let $E\subset{\Bbb R}^d$ be Lebesgue measurable. Show that $$ m(E)=\sup\{m(K):K\subset E, K \text{compact}\}. $$

When $E$ is bounded, this can be done by the following proposition:

$E\subset{\Bbb R}^d$ is Lebesgue measurable if and only if for every $\varepsilon>0$, one can find a closed set $F$ contained in $E$ with $m^*(E\setminus F)\leq\varepsilon$.

How can I deal with the case that $E$ is unbounded?

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2 Answers 2

up vote 3 down vote accepted

You've done the case when $E$ is bounded. When $E$ is unbounded, there two cases:

  1. $m(E)=+\infty$
  2. $m(E)<+\infty$

Now we handle with the first case. Consider the closed ball in ${\Bbb R}^d$ $A_m=\{x\in{\Bbb R}^d: |x|\leq m\}$. Then $$ E=\cup_{m=1}^{\infty}E_m $$ with $E_m=E\cap A_m$, and $E_1\subset E_2\subset\cdots\subset{\Bbb R}^d$. By monotone convergence theorem for measurable sets, $$ \lim_{n\to\infty}m(E_n)=m(E)=+\infty. $$ Note that since $E_m$ is bounded, by the case you've done, for every $m$, we have a compact set $K_m\subset E_m\subset E$ such that $m(K_m)+1\geq m(E_m)\to\infty$. Hence $$\sup\{m(K):K\subset E,K\text{compact}\}=+\infty=m(E).$$

Now assume that $m(E)<+\infty$. For any $\varepsilon>0$, we can choose $N$ such that $$ m(E)\leq m(E_N)+\varepsilon/2. $$ We also have a compact set $K\subset E_N\subset E$ with $$ m(E_N)\leq m(K)+\varepsilon/2 $$ since $E_N$ is bounded. It follows that $$ m(E)\leq m(K)+\varepsilon. $$ We are done.

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Same $m$ for index and for measure causes confusion. – Guldam Sep 12 at 11:41

The standard method to deal with the unbounded case is to do something clever involving epsilon and powers of $1/2$. In your case I would suggest letting $A_i=B(0,i)\setminus B(0,i-1)$. Then set $E_n=A_n \cap E$ and find an appropriate $F_n$ such that $m^\ast(E_n \setminus F_n)\leq \varepsilon/2^n$. Then taking unions should give you what you want.

Edit: There's a little more subtlety here than I originally noticed. There's two cases $m(E)$ is finite and $m(E)$ is infinite. If $m(E)$ is finite then notice that

$$m(E)=\sum_{n=1}^\infty m(E_n)$$

so we can find some $N$ such that

$$\sum_{n=N}^\infty m(E_n) < \varepsilon/2.$$ Then we can handle $\bigcup_{i=1}^N E_n$ using the bounded case. In the case that $m(E)$ is finite we pick our $F_n$ appropriately and let $F$ be their union. Now $F$ is not compact, but we do know that

$$m(E \setminus F) < \varepsilon.$$ Furthermore each finite union of the $F_n$ is compact and

$$\lim_{n\rightarrow \infty} \bigcup_{i=1}^n F_i=F$$ in particular $m(\bigcup_{i=1}^n F_i)$ is unbounded so our desired sup is infinte.

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How do you get that the inner approximation is bounded after taking union? – Jack Jan 7 '13 at 0:40
You can take the union and use the subadditivity to get a bound by $\varepsilon$. – Clayton Jan 7 '13 at 0:42
The trick here is that when you use subadditivity you get a geometric series that sums to $1$ times $\varepsilon$, so the bound ends up just being $\varepsilon$. – JSchlather Jan 7 '13 at 0:45
It is the measure that is bounded by $\varepsilon$. I think his question in comment is how you get the $K\subset E$ that is closed and bounded. – Goku Jan 7 '13 at 0:56
Jacob, what do you denote by $B(0,i)$? I don't know this notation. – oeda Jan 7 '13 at 1:25

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