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Thread: Measurability

  1. #1
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    Measurability

    Let $\displaystyle n,q \in \mathbb{N}$ such that $\displaystyle q \leq n$ and let $\displaystyle E_1,E_2,...,E_n$ be measurable subsets of $\displaystyle [0,1]$. Suppose that for each point $\displaystyle x \in [0,1]$, there are at least $\displaystyle q$ sets in $\displaystyle \{E_1,E_2,...,E_n\}$ that contain $\displaystyle x$. Prove that there exists $\displaystyle 1 \leq i \leq n$ such that $\displaystyle m(E_i) \geq \frac{q}{n}$.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: Measurability

    Quote Originally Posted by Markeur View Post
    Let $\displaystyle n,q \in \mathbb{N}$ such that $\displaystyle q \leq n$ and let $\displaystyle E_1,E_2,...,E_n$ be measurable subsets of $\displaystyle [0,1]$. Suppose that for each point $\displaystyle x \in [0,1]$, there are at least $\displaystyle q$ sets in $\displaystyle \{E_1,E_2,...,E_n\}$ that contain $\displaystyle x$. Prove that there exists $\displaystyle 1 \leq i \leq n$ such that $\displaystyle m(E_i) \geq \frac{q}{n}$.
    Here's the basic idea. Suppose that $\displaystyle m(E_i)<\frac{q}{n}$ then $\displaystyle \sum_{i=1}^{n}m(E_i)<q$, but why is that stupid (hint: use the idea that rougly the $\displaystyle E_i$'s '$\displaystyle q$-fold cover' $\displaystyle [0,1]$).
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  3. #3
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    Re: Measurability

    Okay solved it. Thanks!
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