Sandwich measurable set

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Main Question or Discussion Point

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
 

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  • #2
jbunniii
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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))##.
 
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  • #3
jbunniii
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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
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Another way to do this:
For any [itex]n\in\mathbb{N}[/itex] [itex]\exists B_n, C_n[/itex] s.t. [itex]B_n \subset A \subset C_n[/itex] and [itex]m(C_n \setminus B_n)<1/n[/itex]
Then [itex]m(A \setminus \bigcup _{i=1}^∞B_n)≤m(C_n \setminus B_n)<1/n[/itex] for all n [itex]\Rightarrow[/itex] [itex]m(A \setminus \bigcup _{i=1}^∞B_n)=0[/itex] [itex]\Rightarrow[/itex] [itex]A \setminus \bigcup _{i=1}^∞B_n[/itex] is measurable [itex]\Rightarrow[/itex] [itex] A = (A \setminus \bigcup _{i=1}^∞B_n) \bigcup (\bigcup _{i=1}^∞B_n) [/itex] is measurable, since each [itex]B_n[/itex] is measurable and the measurable sets form a σ-algebra.
 
  • #5
jbunniii
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Another way to do this:
For any [itex]n\in\mathbb{N}[/itex] [itex]\exists B_n, C_n[/itex] s.t. [itex]B_n \subset A \subset C_n[/itex] and [itex]m(C_n \setminus B_n)<1/n[/itex]
Then [itex]m(A \setminus \bigcup _{i=1}^∞B_n)≤m(C_n \setminus B_n)<1/n[/itex] for all n [itex]\Rightarrow[/itex] [itex]m(A \setminus \bigcup _{i=1}^∞B_n)=0[/itex] [itex]\Rightarrow[/itex] [itex]A \setminus \bigcup _{i=1}^∞B_n[/itex] is measurable [itex]\Rightarrow[/itex] [itex] A = (A \setminus \bigcup _{i=1}^∞B_n) \bigcup (\bigcup _{i=1}^∞B_n) [/itex] is measurable, since each [itex]B_n[/itex] is measurable and the measurable sets form a σ-algebra.
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.
 
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