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Let $\lbrace I_1,\ldots I_k \rbrace$ be a collection of bounded intervals. Choose $I_1$ to be of the largest. Denote $T=\lbrace i\in \lbrace 1,\ldots ,k\rbrace \mid (I_1 \cap I_i)\not= \emptyset\rbrace $.

I want to know why is it the case that if $T=\lbrace1,\ldots ,k\rbrace$ then $\mu(I_1)\ge\frac{1}{3}\mu(\cup_{i=1} ^k I_i)$.

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up vote 2 down vote accepted

If $I_1 = [a-r,a+r]$, consider $I_1' = [a-3r,a+3r]$ and note that $\mu(I_1') = 3\mu(I_1)$.

Using the definition of $T$ and the fact that $I_1$ is the largest interval, you can show that $I_i \subset I_1'$ for every $i \in T$.

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What does $I_1'$ have to do with the problem? – cap Dec 6 '12 at 7:55
Ideally, you would like that $I_i \subset I_1$ for every $i \in T$ and you would conclude that $\mu(I_1) \geq \mu(\cup_{i=1}^k I_i)$. Unfortunately, that might be false : for instance, imagine the case where all $I_i$ have the same measure. The idea is then to "enlarge" $I_1$ so that it contains every $I_i$. You should draw a picture to see why one only needs to enlarge it 3 times. – Siméon Dec 6 '12 at 8:03

If all intersections are not empty, the intervals have at least one point in common with $I_1$.

Since $I_1$ is the longest interval you can say that the union is definitely covered by an interval of length $3\mu(I_1)$ and therefore you get the result by translation-invariance, monotonicity, and sub-additivity of the Lebesgue measure.

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Why 3? Is it possible to get sharper bounds? – Learner Dec 6 '12 at 8:42
No. If you choose $k$ to be $3$ and $I_1=[0,1],\ I_2=[1,2],\ I_3=[-1,0]$ then you have equality. So the estimate is sharp. – vanguard2k Dec 6 '12 at 10:26

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