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a) $\mu(E)=\lim_{n\rightarrow\infty} \frac{\#[\{2,4,\dots\}\cap \{1,\dots,n\}]}{n}=\lim_{n\rightarrow\infty}\frac{n/2}{n}=1/2$ and $\mu(O)=1/2$ by same argument. $$\mu(S)=\lim_{n\rightarrow\infty}\frac{\#[ \{1,4,9,\dots\}\cap \{1,\dots,n\}]}{n}=\lim_{n\rightarrow\infty}\frac{\sqrt n}{n}=0$$

b) Let $A,B$ be disjoint sets in $\mathcal A$. $$\begin{split}\mu(A\cup B)&=\lim_{n\rightarrow\infty} \frac{\#[(A\cup B)\cap \{1,\dots,n\}]}{n}&&\text{definition}\\ &=\lim_{n\rightarrow\infty} \frac{\#[A\cap \{1,\dots,n\} \cup B \cap \{1,\dots,n\}]}{n}&&\text{DeMorgan's Law}\\ &=\lim_{n\rightarrow\infty} \frac{\#[A\cap \{1,\dots,n\}] +\#[ B \cap \{1,\dots,n\}]}{n}&&\text{$A$ and $B$ are disjoint}\\ &=\mu(A)+\mu(B)\end{split}$$

c) The two requirements for a measure are

  1. $\mu:\mathcal A\rightarrow[0,\infty]$. This is satisfied since $\mu(B)$ is between 0 and 1.

  2. If $B_1,B_2,\dots$ are disjoint elements of $\mathcal A$ then $\mu\left(\bigcup_{i=1}^\infty B_i\right)=\sum_{i=1}^\infty \mu(B_i)$. This follows inductively from part 2.

I recognize this is a pretty simple exercise but have I made any big oversights?

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    $\begingroup$ It does not "follow inductively". What follows is that $\mu$ behaves well with any finite disjoint union, but not necessarily infinite disjoint union. Consider $B_i=\{i\}$. $\endgroup$
    – Kushi
    Jul 12, 2021 at 21:52
  • $\begingroup$ We can prove inductively that $1+1+1+\cdots +1$ is a natural number, but that does not mean an infinite sum $1+1+\cdots$ is a natural number. $\endgroup$ Jul 12, 2021 at 22:15
  • $\begingroup$ @Kushi Ahh thank you. I see from your counter example that $\mu$ is not a measure $\endgroup$
    – Jellyfish
    Jul 12, 2021 at 22:31
  • $\begingroup$ @Thomas thank you for the analogy! $\endgroup$
    – Jellyfish
    Jul 12, 2021 at 22:31

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