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My task is to prove that:

Given $E_1 \supset E_2 \supset E_3 \supset ...$ is a decreasing sequence of set with the form $E_k = (-\infty, t_k]$ which is an interval in $\mathbb{R}$ and we have $\lim \limits_{k \to \infty} t_k = -\infty$.
Prove that: $\mu(\bigcap \limits_{k = 1}^{\infty} E_k) = 0$ with $\mu$ is a probability measure.

My attempt was to claim $\bigcap \limits_{k = 1}^{\infty} E_k = \lim \limits_{k \to \infty} E_k = \lim \limits_{k \to \infty}(-\infty, t_k] = (-\infty, \lim \limits_{k \to \infty} t_k] = (-\infty, -\infty]$
and that $\mu((-\infty, -\infty)) = 0$ which leads to the proof.

But it seems to me there isn't notation of limit of intersection of a sequence (the first "$=$" in my proof above), as well as I'm not sure if we can write $\lim \limits_{k \to \infty}(-\infty, t_k] = (-\infty, \lim \limits_{k \to \infty} t_k]$ (it seems wrong).

Could you please have a look at my proof and tell me if it is true or not ? If it is not, could you please give a brief sketch of the proof ?

InTheSearchForKnowledge
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  • https://math.stackexchange.com/questions/902316/proof-of-continuity-from-above-and-continuity-from-below-from-the-axioms-of –  May 22 '21 at 09:22
  • $\bigcap_{k=1}^{\infty}E_k$ is the empty set. There is a notion of limit for nested sequences of sets, but you don't actually need that. The intersection $\bigcap_{k=1}^{\infty}E_k$ is defined simply as the set of points in $\mathbb R$ that are in all of the $E_k$'s. Since there are no such points, the intersection is empty. –  May 22 '21 at 09:23

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Continuity of measures will give you the solution. We have that $E_1 \supset E_2 \supset ...$ and $$E:=\bigcap_{n \in \mathbb{N}}E_n=\bigcap_{n \in \mathbb{N}}(-\infty,t_k]=\emptyset$$ This can be shown by fixing any arbitrary $x \in \mathbb{R}\cup\{-\infty,\infty\}$ and showing that $x \notin E$ (the sets will eventually 'pass it'). Therefore you can use the property of continuity from above, as $E_n \downarrow E$ we have $$\lim_{n \to \infty}\mu(E_n)=\mu(E)=\mu(\emptyset)=0$$

Snoop
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    For decreasing sequences, $\lim_{n\to\infty} \mu(E_n) = \mu(E)$ only holds if one of the $E_n$ has finite measure. That's not the case here. In fact $\lim_{n\to\infty} \mu(E_n) = +\infty$. –  May 22 '21 at 09:25
  • @Bungo $\mu$ is a probability measure – Snoop May 22 '21 at 09:29
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    Ah sorry, I overlooked that. Your argument using continuity is fine. However, note that you don't really need it, since you established on the first line that $E = \emptyset$, of course $\mu(E) = 0$. –  May 22 '21 at 09:30
  • @Snoop: Hi, thanks for your comment. In your proof, i see that you say $\lim \limits_{n \to \infty} \mu(E_n) = \mu(E)$. To have that, i think you use $\lim \limits_{n \to \infty} \mu(E_n) = \mu(\lim \limits_{n \to \infty} E_n) = \mu(E)$. So, indeed, there is a notation of "limit of sequence of set" right ? If so, could you please suggest me a source to read about it ? (I just can't find any source that talks about it). Thank you very much! – InTheSearchForKnowledge May 22 '21 at 09:31
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    @InTheSearchForKnowledge the limit does not pass into the measure. It has to be understood as $\inf_{n \in \mathbb{N}}\mu(E_n)=\mu(E)$, because of measure monotonicity. I suggest this source. – Snoop May 22 '21 at 09:35
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    @InTheSearchForKnowledge This Wikipedia page is enough to get you started. https://en.wikipedia.org/wiki/Set-theoretic_limit –  May 22 '21 at 09:35
  • @Snoop: Hi, after thinking for a while, I see that as the $E_n$ has the form $(-\infty, t_n]$ (i.e. half-closed interval), i think that the intersection should be $(-\infty, -\infty]$ instead of $\emptyset$, right ? – InTheSearchForKnowledge May 22 '21 at 15:36
  • The interval $(-\infty,-\infty]$ is a contradiction @InTheSearchForKnowledge – Snoop May 22 '21 at 16:39
  • @Snoop: I think it does not give contradiction because, as $\mu$ is a probability space, I think we are working in the extended real number, so the value of $-\infty$ can be achieved. What do you think about it ? Thank you very much! – InTheSearchForKnowledge May 22 '21 at 16:42
  • The value of $-\infty$ cannot be achieved in that interval (in this case), because it is excluded by specifying $(-\infty$ on the left @InTheSearchForKnowledge – Snoop May 23 '21 at 00:54