# Thread: How can the measure of a measurable set be bounded?

1. ## How can the measure of a measurable set be bounded?

Helle Everyone,

In one book, the Lebesgue measure is said to possess the following properties (among others):

1) the measure of any measurable set can be approximated from above by open sets; that is, for any measurable M

$\displaystyle \mu (M) = inf \{ \mu(O): M \subset O, O$ is open $\displaystyle \}$

2) the measure of any measurable set can be approximated from below by compact sets; that is, for any measurable M

$\displaystyle \mu(M) = sup \{ \mu( C ): C \subset M, C$ is compact $\displaystyle \}$

Can any one provide a very simple example of this property? Like in $\displaystyle \mathbb{R}^{1}$?

Any one?

4. ## Re: How can the measure of a measurable set be bounded?

Originally Posted by john1844
Helle Everyone,

In one book, the Lebesgue measure is said to possess the following properties (among others):

1) the measure of any measurable set can be approximated from above by open sets; that is, for any measurable M

$\displaystyle \mu (M) = inf \{ \mu(O): M \subset O, O$ is open $\displaystyle \}$
If all you want is an example, take [0, 1]. It can be "approximated from above" by the open sets (-1/n, (n+1)/n). Each of those has measure (n+1)/n+ 1/n= 1+ 2/n which goes to 1, the measure of [0, 1], as n goes to infinity.

2) the measure of any measurable set can be approximated from below by compact sets; that is, for any measurable M

$\displaystyle \mu(M) = sup \{ \mu( C ): C \subset M, C$ is compact $\displaystyle \}$
Again, use [0, 1] which can be "approximated from below" by the compact sets [1/n, (n-1)/n]

Can any one provide a very simple example of this property? Like in $\displaystyle \mathbb{R}^{1}$?