Let with . Prove that for each

there exists an interval such that

( denotes the exterior measure)

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- November 5th 2009, 05:54 AMdavidmccormickMeasure Theory
Let with . Prove that for each

there exists an interval such that

( denotes the exterior measure) - November 5th 2009, 06:02 AMdavidmccormick
this is how i attempted to solve the problem:

Choose a decreasing sequence of open covers of (by open interval) such that

Let 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. - November 5th 2009, 06:59 AMramdayal9
- November 5th 2009, 10:10 AMdavidmccormick
- November 5th 2009, 01:38 PMOpalg
Try defining , where .

- November 5th 2009, 03:48 PMramdayal9
- November 6th 2009, 01:25 AMOpalg
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 .

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 . 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*, .

That was my motivation for suggesting , where . 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*. - November 6th 2009, 02:27 AMramdayal9