Let with . Prove that for each
there exists an interval such that
( denotes the exterior measure)
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.
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.