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If $E\in L$ and $m(E) > 0$, for any $\alpha < 1$ there is an open interval $I$ such that $m(E\cap I) > \alpha m(I)$.

Attempted proof/brainstorm - Let $E\in L$ with $m(E) > 0$ and suppose there exists an $\alpha\in (0,1)$ such that $m(E\cap I) \leq \alpha m(I)$ for every open interval $I$. With out loss of generality, suppose $m(E) < \infty$. By theorem 1.18, there exists an open $U$ such that $E\subset U$ and $$m(E) \leq m(U) \leq m(E) + \epsilon$$ Let $U = \bigcup_{1}^{\infty}I_j$ where $\{I_j\}_{1}^{\infty}$ are open intervals. Then $E\subset \bigcup_{1}^{\infty}I_j$ and we have $$m(E)\leq m(U)\leq \sum_{1}^{\infty}m(I_j)\leq m(E) + \epsilon$$ Note that $m(U) < \infty$. We can write $$m(I_j) = m(I_j\setminus E) + m(E\cap I_j)\leq m(I_j\setminus E) + \alpha m(I_j)$$ then I believe with what we have above we arrive at a contradiction?

I am not really sure about my approach, any suggestions is greatly appreciated.

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  • $\begingroup$ $m(I_j) = m(I_j\setminus E) + m(E\cap I_j)$ is not correct. $\endgroup$ – Matematleta Jun 17 '16 at 21:48
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Here is a detailed proof following your approach. In particular, showing how to handle the case $\mu(E)=+\infty$.

If $E\in L$ and $m(E) > 0$, for any $\alpha < 1$ there is an open interval $I$ such that $m(E\cap I) > \alpha m(I)$.

Let $E\in L$ with $m(E) > 0$ and suppose there exists an $\alpha\in (0,1)$ such that $m(E\cap I) \leq \alpha m(I)$ for every open interval $I$.

Case 1: Suppose $m(E) <+\infty$.

By theorem 1.18, for all $\epsilon >0$ there exists an open $U$ such that $E\subset U$ and $$m(E) \leq m(U) \leq m(E) + \epsilon$$ Take $\epsilon <(1-\alpha)\mu(E)$.

Let $U = \bigcup_{1}^{\infty}I_j$ where $\{I_j\}_{1}^{\infty}$ are open intervals. Then $E\subset \bigcup_{1}^{\infty}I_j$ and we have $$m(E)\leq m(U)\leq \sum_{1}^{\infty}m(I_j)\leq m(E) + \epsilon <+\infty \tag{1}$$ we also have, since the $I_j$'s may not be disjoint, that $$m(E)\leq \sum_{1}^{\infty}m(E\cap I_j)$$

However, we have that $\mu(E\cap I) \leq \alpha\mu(I_j)$. So we have $$m(E)\leq \alpha \sum_{1}^{\infty}m(I_j)$$ So, from $(1)$, we get $$ m(E)\leq \sum_{1}^{\infty}m(I_j)\leq \alpha \sum_{1}^{\infty}m(I_j) + \epsilon $$ Since $\epsilon <(1-\alpha)\mu(E)$ and $ \alpha \sum_{1}^{\infty}m(I_j) <+\infty$, we have
$$ \sum_{1}^{\infty}m(I_j)\leq \alpha \sum_{1}^{\infty}m(I_j) + \epsilon< \alpha \sum_{1}^{\infty}m(I_j) +(1-\alpha)\mu(E) \leq \\ \leq \alpha \sum_{1}^{\infty}m(I_j) +(1-\alpha)\sum_{1}^{\infty}m(I_j) =\sum_{1}^{\infty}m(I_j)$$ So $$ \sum_{1}^{\infty}m(I_j) <\sum_{1}^{\infty}m(I_j)$$ Contradiction.

Case 2. Suppose $\mu(E)=+\infty$.

Since $E=\bigcup_{n\in \mathbb{Z}}(E\cap(n,n+1])$, there $k \in \mathbb{Z}$ such that $\mu(E\cap (k,k+1])>0$. Let $E_k= E\cap (k,k+1]$. Clearly we have $0<\mu(E_k)\leq 1$.

From case 1, we know that for any $\alpha >0$ there is an open interval I such that $\mu(E_k\cap I)>\alpha \mu(I)$. Let $J=(k,k+1] \cap I$. $J$ is an interval and we have

$$\mu(E\cap J)=\mu(E\cap (k,k+1] \cap I))=\mu(E_k\cap I)>\alpha \mu(I)\geq \alpha \mu(J)$$ So we prove the result for this second case, which completes the proof.

Remark: Note that since we are trying to prove that: for all$E\in L$ and $m(E) > 0$, for any $\alpha < 1$ there is an open interval $I$ such that $m(E\cap I) > \alpha m(I)$, then to work by contradiction we suppose that: there is one $E\in L$ with $m(E) > 0$ such that there exists an $\alpha\in (0,1)$ such that $m(E\cap I) \leq \alpha m(I)$ for every open interval $I$.

Under this hypothesis, we can not simply replace $E$ by another $E$ with finite measure. In this approach, the best thing is to prove by cases (and case 2 is actually an easy consequence of case 1).

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Given $\epsilon > 0$, Proposition 1.20 gives a finite union $A = \cup I_j$ of disjoint open intervals so $m (A - E) +m (E - A) < \epsilon m (E)$ (since $m(E) > 0$). We use this twice. First, $$m (E) = m (E \cap A) + m(E - A) ≤ \sum m(I_j) + \epsilon m(E)$$

so that $(1 −\epsilon )m(E) ≤ \sum m(I_j)$. Second,

$$\sum m(I_j) = \sum m(I_j \cap E) + \sum m(I_j - E) < \sum m (I_j ∩ E) + \epsilon m(E).$$

Using the first estimate on $m$, this gives $$\sum m(I_j) < \sum m(I_j \cap E) + \frac{\epsilon}{1+\epsilon} \sum m(I_j).$$

It follows easily from here if we choose $\epsilon> 0$ small so $\frac{\epsilon}{1+\epsilon} ≥ \alpha$.

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  • $\begingroup$ Ma'am, would you kindly further elaborate how you concluded from $\sum m(I_j) < \sum m(I_j \cap E) + \frac{\epsilon}{1+\epsilon} \sum m(I_j)$ that there is an open interval $I$ such that $m(E\cap I) > \alpha m(I)$? $\endgroup$ – Erfan Oct 17 '17 at 17:17
  • $\begingroup$ @Erfan if for all $j$ we have $m(E\cap I_j)\leq \alpha m(I_j)$ then the expression you mentioned would not hold, but you should assume that $\frac{\epsilon}{1+\epsilon} \leq {1-\alpha}$ $\endgroup$ – user555729 Dec 8 '18 at 20:05
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Fix $0<\alpha<1$ and let $\epsilon>0$.

There is a disjoint union of intervals $\cup I_n\supseteq E$ such that $m(E)+\epsilon>\sum m(I_n)$.

If $m(E\cap I_n) \leq \alpha m(I_n)$, for all $n$, then $E=\bigcup_nE\cap I_n$ so on the one hand

$m(E)=\sum m(E\cap I_n)\leq \alpha \sum m(I_n)$

and on the other,

$m(E)>\sum m(I_n)-\epsilon$,

so if we take $\epsilon$ so small that $\sum m(I_n)-\epsilon>\alpha \sum m(I_n)$

we arrive at a contradiction.

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  • $\begingroup$ Just one little problem I can't figure out, how does $m(E\cap I_n) \leq \alpha m(I_n)$ (for all $n$) imply $E=\bigcup_nE\cap I_n$? $\endgroup$ – Erfan Oct 17 '17 at 17:27

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