Real analysis: Suppose that $E\subset \mathcal{M}$, show that for each $\epsilon0$, there is an open set, $O$, with $E\subset O$, and $\lambda(O-E)

Question

Real analysis:
Suppose that $E\subset \mathcal{M}$, show that for each $\epsilon>0$, there is an open set, $O$, with $E\subset O$, and $\lambda(O-E)<\epsilon$

zzr0ck3r · Answer

$\mathcal{M}$, is the sigma algebra of Lebegue measurable sets, and $\lambda$ is the Lebegue measure.

zzr0ck3r · Answer

Lebesgue*

fibonaccichick666 · Answer

kind of looks like Lipschitz

zzr0ck3r · Answer

Lippelletz what?

fibonaccichick666 · Answer

What class is this?  (It's a theorem from calculus having to do with a specific type of limit)

fibonaccichick666 · Answer

http://en.wikipedia.org/wiki/Lipschitz_continuity

zzr0ck3r · Answer

called math 511 at my university

fibonaccichick666 · Answer

hmm, interesting.  I'm sorry  can't help, we didn't cover that in my analysis courses

zzr0ck3r · Answer

but tnx for trying:)

abb0t · Answer

@wio

anonymous · Answer

Do you know anything about E or O other than that they are in M?

anonymous · Answer

What about you construct O by taking the bounds of E and then adding $\epsilon/2$ to both sides or something like that?

anonymous · Answer

I mean, adding a length of $\epsilon / 2$

zzr0ck3r · Answer

yeah, it seems super trivial, I just dont know how to show it. We dont know anything, but we do know that since E is measurable, its measure is the same as its outer measure, which is defined to be the infimum of the sums of I_n where I_n is an open cover of E, so right there I have an open set, that covers E. I just dont know how to talk about it.

anonymous · Answer

Unless $E$ spans from infinity to infinity, it will have a max or minimum.  Suppose it has a minimum $a$.  Then let $O =(a-\epsilon,a-\epsilon/2)\cup E$.  Isn't this an open set?

anonymous · Answer

By definition;
$$
\lambda(E)= \inf\{ \sum_{n=1}^\infty Length(I_n)\quad |\quad  E \subset \cup I_n                            \}

$$

anonymous · Answer

Notice that  we can  take each $ I_n$ to be an open interval, because each interval [a,b] and each $ \epsilon >0$ we can find an open interval J such
that 
$$
[a,b]\subset   J=] a-\frac \epsilon 2 , b+ \frac \epsilon 2\
Length[J]= Length[a,b] +\epsilon

$$
Also notice that 
$$
Length[I]= \lambda(I)
$$

anonymous · Answer

By definition of inf, for every $\epsilon >0$, there is a countable union of open intervals $I_n$ that cover E and
$$
\sum\lambda (I_n) \leq \lambda(E) + \epsilon\

$$

anonymous · Answer

But
$$
\lambda(\cup I_n) \leq \sum\lambda (I_n) \leq \lambda(E) + \epsilon\
$$

anonymous · Answer

Let 
$$
O= \cup I_n 
$$
then 
$$
E\subset O\
\lambda(O) \leq \lambda(E) +\epsilon\
\lambda(O\setminus E)\leq \epsilon
$$

zzr0ck3r · Answer

where would the strict inequality come from? I'm thinking the definition of inf, which is what I was having a problem with. I never really knew how to talk about our cover explicitly and now I do. Ty again. can you come teach my class?

anonymous · Answer

You can actually take the inequality to be strict. If a is the inf  of a bunch of numbers, then
for every $ \epsilon > 0$ and if you consider $ a +\epsilon >a  $, then there must be one number of these bunch of numbers that is strictly less than  $ a +\epsilon   $
Thanks for trusting me to teach your class. I love measure theory, I spent all my life teaching it and doing research about it.
Do not hesitate to ask me for help in the future.

anonymous · Answer

Let$$
O= \cup I_n


$$

then
$$

E\subset O\
\lambda(O) <   \lambda(E) +\epsilon\
\lambda(O\setminus E)< \epsilon
$$