sandwich measurable set

• Oct 14th 2013, 02:44 AM
Sonifa
sandwich measurable set
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
• Oct 14th 2013, 05:21 AM
SlipEternal
Re: sandwich measurable set
How do you know that $\displaystyle m(E\cap C) + m(E\setminus C) \ge m(E\cap A) + m(E\setminus A) - \varepsilon$? You just stated what you need to prove, but did not prove it.

You need to show that for all $\displaystyle \varepsilon>0$ and for all measurable sets $\displaystyle E\subseteq \mathbb{R}$, $\displaystyle \left|m(E\cap A) + m(E\setminus A) - m(E) \right| < \varepsilon$. So, you need to come up with a way of rewriting $\displaystyle m(E\cap A)+m(E\setminus A)$ using the terms you know something about: $\displaystyle m(E\cap B), m(E\cap C), m(E\setminus B), m(E\setminus C)$.
• Oct 14th 2013, 07:16 AM
Sonifa
Re: sandwich measurable set
I tried the way u provided, but i cannot get anything relatives to m(C\B)< ε . Could you tell me the proof in details???
• Oct 14th 2013, 09:31 AM
SlipEternal
Re: sandwich measurable set
Since $\displaystyle m(E\cap A) \le \inf\{m^*(E \cap K) \mid A \subseteq K, K\mbox{ is measurable} \}$ and $\displaystyle m(E \setminus A) \le \inf\{m^*(E \setminus K) \mid K \subseteq A, K\mbox{ is measurable} \}$, you can conclude, $\displaystyle m(E \cap A) + m(E \setminus A) \le m(E \cap C) + m(E \setminus B)$. But, $\displaystyle E \cap C = E \cap (B \cup (C \setminus B))$, so $\displaystyle m(E \cap C) = m(E \cap (B \cup (C\setminus B))) = m((E \cap B) \cup (E \cap (C\setminus B)))$. Since $\displaystyle B,C$ are measurable, use the properties of a $\displaystyle \sigma$-algebra to show that $\displaystyle E\cap B, E\cap (C\setminus B)$ are both measurable. Hence $\displaystyle m((E \cap B) \cup (E \cap (C\setminus B))) = m(E \cap B) + m(E \cap (C\setminus B))$.

To recap, we have
\displaystyle \begin{align*}m(E \cap A) + m(E \setminus A) & \le m(E \cap C) + m(E \setminus B) \\ & = m(E \cap B) + m(E \cap (C \setminus B)) + m(E \setminus B) \\ & = (m(E\cap B) + m(E\setminus B)) + m(E \cap (C\setminus B)) \\ & \le (m(E\cap B) + m(E\setminus B)) + m(C\setminus B) \\ & = m(E) + m(C\setminus B) \\ & < m(E) + \varepsilon\end{align*}

This is because $\displaystyle E \cap (C\setminus B) \subseteq C\setminus B$, so $\displaystyle m(E\cap (C\setminus B)) \le m(C \setminus B)$ and since $\displaystyle B$ is measurable, $\displaystyle m(E) = m(E\cap B) + m(E\setminus B)$.

So, finally, we have $\displaystyle m(E\cap A) + m(E \setminus A) - m(E) < \varepsilon$.