Measure Theory

• Nov 5th 2009, 04:54 AM
davidmccormick
Measure Theory
Let $E \subset \mathbb {R}$ with $m(E) > 0$. Prove that for each $0 < c < 1$
there exists an interval $I$ such that $m(E\cap I) \geq cm(I)$
( $m$ denotes the exterior measure)
• Nov 5th 2009, 05:02 AM
davidmccormick
this is how i attempted to solve the problem:
Choose a decreasing sequence of open covers of $E$ (by open interval) such that
$m ( E ) = lim (m (I_{n}))$
Let $I = \cap I_{n}$ 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.
• Nov 5th 2009, 05:59 AM
ramdayal9
Quote:

Originally Posted by davidmccormick
this is how i attempted to solve the problem:
Choose a decreasing sequence of open covers of $E$ (by open interval) such that
$m ( E ) = lim (m (I_{n}))$
Let $I = \cap I_{n}$ 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 $m(E) = m(I)$ so that $m(E \cap I) = m(I)$ and where have you used the value of $c$?
• Nov 5th 2009, 09:10 AM
davidmccormick
Quote:

Originally Posted by ramdayal9
i get that $m(E) = m(I)$ so that $m(E \cap I) = m(I)$ and where have you used the value of $c$?

I get the same thing. i.e. $m(E) = m(I)$ but am not sure whether or not that would imply $m(E \cap I) = m(I)$. If it does we're done since c is strictly between 0 and 1 and so the inequality is satisfied.
• Nov 5th 2009, 12:38 PM
Opalg
Try defining $I = (-\infty,a)$, where $a = \sup\{x\in\mathbb{R}:m\bigl((-\infty,x)\cap E\bigr) < cm(E)\}$.
• Nov 5th 2009, 02:48 PM
ramdayal9
Quote:

Originally Posted by Opalg
Try defining $I = (-\infty,a)$, where $a = \sup\{x\in\mathbb{R}:m\bigl((-\infty,x)\cap E\bigr) < cm(E)\}$.

Can you explain how this works?

Quote:

Originally Posted by davidmccormick
I get the same thing. i.e. $m(E) = m(I)$ but am not sure whether or not that would imply $m(E \cap I) = m(I)$. 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 $E \cap I = E$
• Nov 6th 2009, 12:25 AM
Opalg
Quote:

Originally Posted by ramdayal9
Quote:

Originally Posted by Opalg
Try defining $I = (-\infty,a)$, where $a = \sup\{x\in\mathbb{R}:m\bigl((-\infty,x)\cap E\bigr) < cm(E)\}$.

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 $m(E\cap I) = cm(E)$.

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 $f(a)=m((-\infty,a)\cap E)/m(E)$. 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, $m((-\infty,a)\cap E) = cm(E)$.

That was my motivation for suggesting $I = (-\infty,a)$, where $a = \sup\{x\in\mathbb{R}:m\bigl((-\infty,x)\cap E\bigr) < cm(E)\}$. 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.
• Nov 6th 2009, 01:27 AM
ramdayal9
Quote:

Originally Posted by Opalg
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 $m(E\cap I) = cm(E)$.

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 $f(a)=m((-\infty,a)\cap E)/m(E)$. 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, $m((-\infty,a)\cap E) = cm(E)$.

That was my motivation for suggesting $I = (-\infty,a)$, where $a = \sup\{x\in\mathbb{R}:m\bigl((-\infty,x)\cap E\bigr) < cm(E)\}$. 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 $m(E \cap I) \geq cm(I)$ from this?