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Let $m$ be Lebesgue measure on $\Bbb{R}$ and let $A \subset \Bbb{R}$ be Borel and let. $\delta \in (0,1)$. If for every interval $I$, $m(A \cap I) \leq (1-\delta)m(I)$, then $m(A)=0$.

So first I substituted $m(A \cap I)$ with $m(A)+m(I)-m(A \cup I)$ Then I have

$$m(A)+\delta m(I) \leq m(A \cup I) \leq m(A)+m(I).$$

But this got me nowhere so since my text (Bass) has the covering of $A$ by half open intervals, I suppose I have

$$A \subset \bigcup_n (a_n,b_n].$$

Then I know if I send $\delta$ to $1$ then $m(A \cap I) = 0$ and since this holds for all intervals $I$ can i suppose $I_n=(a_n,b_n]$ and $I$ is the union or?? Any hints greatly appreciated. Or am I allowed to send $\delta$ to $1$?

homosapien
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1 Answers1

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Recall the definition of the Lebesgue measure. We want the infimum of the sum of a sequence of intervals.

Suppose $I_k$ is a sequence of intervals that covers $A$, and the total length is equal to $C$.

But then $m(I_1 \cap A) < (1-\delta)l(I_1)$. So instead of using $I_1$ to cover $I_1 \cap A$, we should use a better covering that has total length at most $(1-\delta/2)l(I_1)$. This covering must exist because $m(I_1 \cap A) < (1-\delta)l(I_1)$.

Replace each $I_k$ with a better covering to arrive at a covering that has total length at most $(1-\delta/2)*C$. A countable number of countable intervals is still countable, so we still have a countable covering.

Since we can repeat this as many times as we want, $m(A) < (1-\delta/2)^n*C$ for any $n$, which implies $m(A)=0$.

Scott Hahn
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