# Sandwich measurable set

1. Oct 14, 2013

### Funky1981

Suppose that A is subset of R (real line) with the property for every ε > 0 there are measurable sets B and C s.t. B⊂A⊂C and m(C\B)<ε
Prove A is measurable

By definition A is measurable we need to prove m(E)=m(E∩A)+m(E\A) for all E

the ≤ is trivial enough to show ≥:

Since C is measurable then m(E)= m(E∩C)+m(E\C)
≥ m(E∩A)+m(E\A)-ε (Since A is subset of C)
then move the ε to LHS and since for every ε so, let ε->0 , obtain the result.

is my solution right??? i thought i should use B and m(C\B)<ε somewhere. Could some one help me to check it??

many thanks

2. Oct 15, 2013

### jbunniii

I assume your $m$ denotes outer measure. I'm not sure how you obtained your inequality
$$m(E \cap C) + m(E\setminus C) \geq m(E \cap A) + m(E \setminus A) - \epsilon$$
I agree that $m(E \cap C) \geq m(E \cap A)$, but how did you get the rest?

I was able to solve this pretty straightforwardly by using the fact that $B$ and $C$ are measureable.

Given $\epsilon > 0$, there are $B$ and $C$ such that $B \subset A \subset C$ and $m(C \setminus B) < \epsilon$. Let $E$ be any set. Since $B$ and $C$ are measurable, we have (*)
$$m(E) = m(E \cap C) + m(E \setminus C)$$
and
$$m(E) = m(E \cap B) + m(E \setminus B)$$
We also have the following inequalities due to the nesting of the sets:
$$m(E \cap B) \leq m(E \cap A) \leq m(E \cap C)$$
$$m(E \setminus C) \leq m(E \setminus A) \leq m(E \setminus B)$$
Adding these two inequalities, we get
$$m(E \cap B) + m(E \setminus C) \leq m(E \cap A) + m(E \setminus A) \leq m(E \cap C) + m(E \setminus B)$$
In light of (*) above, it therefore suffices to show that $m(E \setminus B) \leq m(E \setminus C) + \epsilon$. But this follows easily from the fact that $E \setminus B = (E \setminus C) \cup (E \cap (C \setminus B))$.

Last edited: Oct 15, 2013
3. Oct 15, 2013

### jbunniii

One nice consequence of this exercise is that we can immediately see that a measure obtained from an outer measure is complete: in other words, any subset of a set of measure zero is measurable. To see this, simply take $B = \emptyset$ and let $C$ be any set of measure zero.

4. Oct 15, 2013

### Axiomer

Another way to do this:
For any $n\in\mathbb{N}$ $\exists B_n, C_n$ s.t. $B_n \subset A \subset C_n$ and $m(C_n \setminus B_n)<1/n$
Then $m(A \setminus \bigcup _{i=1}^∞B_n)≤m(C_n \setminus B_n)<1/n$ for all n $\Rightarrow$ $m(A \setminus \bigcup _{i=1}^∞B_n)=0$ $\Rightarrow$ $A \setminus \bigcup _{i=1}^∞B_n$ is measurable $\Rightarrow$ $A = (A \setminus \bigcup _{i=1}^∞B_n) \bigcup (\bigcup _{i=1}^∞B_n)$ is measurable, since each $B_n$ is measurable and the measurable sets form a σ-algebra.

5. Oct 15, 2013

### jbunniii

That's a nice proof. It requires knowing that any set with zero outer measure is measurable, but that's easy enough to prove directly from the Caratheodory criterion:

If $E$ is any set and $m(A) = 0$, then $m(E \cap A) \leq m(A) = 0$, so $m(E \cap A) + m(E \setminus A) = m(E \setminus A) \leq m(E)$. The reverse inequality, $m(E) \leq m(E \cap A) + m(E \setminus A)$, is an immediate consequence of subadditivity.