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Thread: Measure Theory

  1. #1
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    Measure Theory

    Let $\displaystyle E \subset \mathbb {R} $ with $\displaystyle m(E) > 0$. Prove that for each $\displaystyle 0 < c < 1$
    there exists an interval $\displaystyle I$ such that $\displaystyle m(E\cap I) \geq cm(I)$
    ($\displaystyle m$ denotes the exterior measure)
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  2. #2
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    this is how i attempted to solve the problem:
    Choose a decreasing sequence of open covers of $\displaystyle E$ (by open interval) such that
    $\displaystyle m ( E ) = lim (m (I_{n})) $
    Let $\displaystyle I = \cap I_{n} $ so I is an open interval. Then apply continuity etc...
    This seems to work but I'm pretty sure that I make a wrong assumption towards the end
    which means the proof is nonsense. Thanks for your help.
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  3. #3
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    Quote Originally Posted by davidmccormick View Post
    this is how i attempted to solve the problem:
    Choose a decreasing sequence of open covers of $\displaystyle E$ (by open interval) such that
    $\displaystyle m ( E ) = lim (m (I_{n})) $
    Let $\displaystyle I = \cap I_{n} $ so I is an open interval. Then apply continuity etc...
    This seems to work but I'm pretty sure that I make a wrong assumption towards the end
    which means the proof is nonsense. Thanks for your help.
    what did you do after using continuity? i get that $\displaystyle m(E) = m(I) $ so that $\displaystyle m(E \cap I) = m(I) $ and where have you used the value of $\displaystyle c $?
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  4. #4
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    Quote Originally Posted by ramdayal9 View Post
    i get that $\displaystyle m(E) = m(I) $ so that $\displaystyle m(E \cap I) = m(I) $ and where have you used the value of $\displaystyle c $?
    I get the same thing. i.e. $\displaystyle m(E) = m(I) $ but am not sure whether or not that would imply $\displaystyle m(E \cap I) = m(I) $. If it does we're done since c is strictly between 0 and 1 and so the inequality is satisfied.
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  5. #5
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    Try defining $\displaystyle I = (-\infty,a)$, where $\displaystyle a = \sup\{x\in\mathbb{R}:m\bigl((-\infty,x)\cap E\bigr) < cm(E)\}$.
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  6. #6
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    Quote Originally Posted by Opalg View Post
    Try defining $\displaystyle I = (-\infty,a)$, where $\displaystyle a = \sup\{x\in\mathbb{R}:m\bigl((-\infty,x)\cap E\bigr) < cm(E)\}$.
    Can you explain how this works?

    Quote Originally Posted by davidmccormick View Post
    I get the same thing. i.e. $\displaystyle m(E) = m(I) $ but am not sure whether or not that would imply $\displaystyle m(E \cap I) = m(I) $. If it does we're done since c is strictly between 0 and 1 and so the inequality is satisfied.
    why doesn't it work? If I covers E, then $\displaystyle E \cap I = E$
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  7. #7
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    Quote Originally Posted by ramdayal9 View Post
    Quote Originally Posted by Opalg View Post
    Try defining $\displaystyle I = (-\infty,a)$, where $\displaystyle a = \sup\{x\in\mathbb{R}:m\bigl((-\infty,x)\cap E\bigr) < cm(E)\}$.
    Can you explain how this works?
    Think about the problem in an informal, geometric way. You have a set E with positive measure, and you want to find an interval I such that the proportion of E lying inside I is c, in the sense that $\displaystyle m(E\cap I) = cm(E)$.

    One way of doing that is to take a point a that moves along the real line from ∞ to +∞, and to look at the proportion of E that lies to the left of a. Call this proportion f(a), so that $\displaystyle f(a)=m((-\infty,a)\cap E)/m(E)$. When a is very large and negative, practically none of E will lie to the left of a, so f(a) will be very small. As a increases, so does f(a), and as a→+∞, f(a)→1. Also, f(a) is a continuous function of a, so by the intermediate value theorem there will be some point along the line at which f(a)=c. For that value of a, $\displaystyle m((-\infty,a)\cap E) = cm(E)$.

    That was my motivation for suggesting $\displaystyle I = (-\infty,a)$, where $\displaystyle a = \sup\{x\in\mathbb{R}:m\bigl((-\infty,x)\cap E\bigr) < cm(E)\}$. The value of a given by that definition is the sup of all the points for which f(a)<c. At that point, f(a) will be equal to c.
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  8. #8
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    Quote Originally Posted by Opalg View Post
    Think about the problem in an informal, geometric way. You have a set E with positive measure, and you want to find an interval I such that the proportion of E lying inside I is c, in the sense that $\displaystyle m(E\cap I) = cm(E)$.

    One way of doing that is to take a point a that moves along the real line from ∞ to +∞, and to look at the proportion of E that lies to the left of a. Call this proportion f(a), so that $\displaystyle f(a)=m((-\infty,a)\cap E)/m(E)$. When a is very large and negative, practically none of E will lie to the left of a, so f(a) will be very small. As a increases, so does f(a), and as a→+∞, f(a)→1. Also, f(a) is a continuous function of a, so by the intermediate value theorem there will be some point along the line at which f(a)=c. For that value of a, $\displaystyle m((-\infty,a)\cap E) = cm(E)$.

    That was my motivation for suggesting $\displaystyle I = (-\infty,a)$, where $\displaystyle a = \sup\{x\in\mathbb{R}:m\bigl((-\infty,x)\cap E\bigr) < cm(E)\}$. The value of a given by that definition is the sup of all the points for which f(a)<c. At that point, f(a) will be equal to c.
    Thanks for the explanation, but how do we get $\displaystyle m(E \cap I) \geq cm(I) $ from this?
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