Proving that the probability of the support is 1 Question:
Our lecturer defined support a bit differently than what I've seen on MathExchange.
$$supp_P=\lbrace r \in \Bbb R  \text{ s.t. }   \forall \varepsilon>0, P(r-\varepsilon,r+\varepsilon)>0\rbrace $$
Prove that $\Bbb P(supp_p)=1$ and $\Bbb P((supp_p)^c)=0$.
What I did:
I tried to say that the complement's prob. is 0 because it's an (infinite) union of events with probability of 0 each. I don't know if this is a good direction as I got a hint I need to use Cantor's lemma, but don't really know how.
Thanks for any help.
 A: Probably there are simpler ways than this one:
By definition, for each $r \in \operatorname{supp}^c_p\cap \mathbb Q$ there exists $ε_r>0$ (that can depend on $r$) such that $$P(r-ε,r+ε)=0$$ for all $0<ε<ε_r$. You can write each of these $r$'s as $$r=\bigcap_{0<ε<ε_r}(r-ε,r+ε)\subseteq (r-ε_r,r+ε_r)$$ (by Cantor's lemma if you want). Now since $\mathbb Q$ is dense in $\mathbb R$ you have that $$\operatorname{supp}^c_p\subseteq \bigcup_{r\in\operatorname{supp}^c_p\cap\mathbb Q}(r-ε_r,r+ε_r)$$ and therefore $$P\left(\operatorname{supp}^c_p\right)\le P\left( \bigcup_{r\in\operatorname{supp}^c_p\cap\mathbb Q}(r-ε_r,r+ε_r)\right)\le \sum_{r\in\operatorname{supp}^c_p\cap\mathbb Q}P(r-ε_r,r+ε_r)=0$$

The last equation used that the proability of the union of countably many sets is less than the sum of the probability. This in not possible for uncountably many. That is why we used the intersection with $\mathbb Q$ in the first place.
A: I would like to expand a little more on Jimmy R.'s answer.
A key point in understanding why $$supp_p^c\subseteq\bigcup_{r\in supp_p^c\cap\Bbb Q}(r-\varepsilon_r,r+\varepsilon_r)$$ is the fact that $supp_p$ is a closed group (can be proven without too much effort but belongs in a different question), and therefore $supp_p^c$ is open. That means that for every $r\in supp_p^c$ there is some neighbourhood of it that is fully contained within $supp_p^c$:  $$(r-\delta,r+\delta)\subseteq supp_p^c$$ That is the reason why even if $r$ is irrational, we can choose some rational point arbitrarily close to it, $r_{\Bbb Q}$, that will still be contained in $supp_p^c$ and that will satisfy $$r\in (r-\varepsilon_{r_{\Bbb Q}},r+\varepsilon_{r_{\Bbb Q}})$$
