Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\mu(I_1)\ge\frac{1}{3}\mu(\cup_{i=1} ^k I_i)$.

share|improve this question

2 Answers 2

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 $i \in T$.

share|improve this answer
1  
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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.