Consider a Cantor-like set $\hat{\mathcal{C}}$ such that $m(\hat{\mathcal{C}}) > 0$. Similar to how the hint prescribes, let $O$ be the union of all of the open intervals removed during the odd stages of the construction of $\hat{\mathcal{C}}$. Next, let $O'$ be the union of all of the open intervals removed during the even stages of the construction of $\hat{\mathcal{C}}$.
Notice $(\dagger)$:
$$ [0,1] \backslash \hat{\mathcal{C}} = O \cup O'$$
Now, given the construction of $\hat{\mathcal{C}}$, $\forall x \in \hat{\mathcal{C}}$, it's easy to see that $\forall r > 0, B_r{(x)} \cap O \neq \emptyset.$
Thus, since every element of $\hat{\mathcal{C}}$ is a limit point of $O$, (given $\dagger$), $\hat{\mathcal{C}} = \overline{O} \backslash O$. Thus, $m(\overline{O} \backslash O) > 0$, as desired.
Thanks for the solution, but it is not complete.
ReplyDeleteIt does not describe why the boundary of the closure has positive measure, it only shows
that the boundary of O has positive measure which is not the same.
For instance, all of your statements would apply correctly to the union of all intervals deleted during the construction
of \hat{\mathcal{C}} - lets it be the new O. Now the closure of O is [0, 1], and the boundary of the closure is {0, 1}, which
measure is 0.
Taking out the open sets deleted at even (or odd) stages assures you that every x \in \hat{\mathcal{C}}
will remain in the boundary of \overline{O}.
Clearly I have failed with the math commands, and there is no edit options.
DeleteHope it is still understandable.
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DeleteI think there may be a typo in here. It's been a while since I've considered this problem, so dissecting it will take me some time.
DeleteThis comment has been removed by the author.
ReplyDeleteThis is my version of the solution:
ReplyDeleteLet O be the union of the intervals deleted at the odd steps, O' the union of the intervals deleted at the even steps and C the cantor-like set. So [0, 1] is partitioned into these three sets.
Now we know that the closure of O is equal to $O \cup C$, since every point in C is a limit point of O and every point in O' is clearly not a limit point of O(O and O' are disjoint).
Also, the interior of $O \cup C$ is just O itself, because C is completely disconnected and no open set entirely inside $O \cup C$ can have even a single element of $C$. So:
$$ boundary(O \cup C) = O \cup C - O = C $$
But C can have a positive measure according to problem 4 part a.