Part a.) Since $E$ is compact, it's clear that $O_n \searrow E$, and $\forall n, m(O_n) < \infty$. Thus, it follows from corollary 3.3 that $m(E) = \lim_{n \to \infty} m(O_n)$.
Part bi.) Let $E = \bigcup_{k=1}^{\infty} [k - \frac{1}{2^{k+1}}, k + \frac{1}{2^{k+1}}].$ Notice that $E^c$ is clearly an open set, so $E$ must be closed. Certainly,
$$m(E) = \sum_{k = 1}^{\infty} \frac{1}{2^k} = 1.$$
However, $\forall n > 5$, (I say 5, but it's probably 2 or 3... who cares, haha.)
$$m(O_n) > \sum_{k = 1}^{\infty} \frac{2}{n} = \infty$$.
Thus, $\lim_{n \to \infty} m(O_n) = \infty \neq m(E)$.
Part bii.) We see in a later exercise (1.9) that there exist bounded open sets $S$ such that $m(\overline{S} \backslash S) > 0.$ Since for any bounded open set $S$, $O_n \searrow \overline{S},$ certainly $\lim_{n \to \infty} m(O_n) \neq m(S).$
Hello, please i want to know why m(O)> \sum_{k=1}^{\infty}\frac{2}{n}=\infty
ReplyDeleteI have the same question
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ReplyDeletePart bi,as n become infinite ,2/n become a infinite small quantity,so infinite infinite small to sum is what?maybe a infinite max or infinite small
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ReplyDeletePart bi,I think the logic is wrong. Because you want to prove when in some state something is wrong.So you must prove don't exist right . But you only prove one state is wrong . You only say that On is wrong,but you must prove all On is wrong.For example,if you want to prove every human don't have super power,you just find a person and say he don't have super power,so every person don't have super power.....
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