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I know that if $E=\biguplus_{n=1}^{\infty}E_n$ then $\mu(E)=\sum_{n=1}^{\infty}\mu(E)$ else $\mu(E)\leq\sum_{n=1}^{\infty}\mu(E)$

in the case of $E_n=[n,\infty]$ and $E=\cap_{n=1}^{\infty}E_n$ we can say about $\mu(E)?$ I know that $E=\emptyset$, can we just right $\mu(E)=\mu(\emptyset)?$ and then by definition $\mu(\emptyset)=0$ or there is an analogue for sigma additivity for intersection ?

newhere
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    Yes. If $E = \varnothing$, the we may write $\mu(E)= \mu(\varnothing) = 0$. However, you wrote $E_n = [0,\infty]$ including a non-real point called "$\infty$". Did you intend that? If so, we get $E = {\infty} \ne \varnothing$. – GEdgar Dec 26 '19 at 13:20

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Yes, $\mu(E)=0.$ However, you should not say $\mu(\cap_{n=1}^{\infty} E_{n}) \neq \lim_{n \to \infty}\mu(E_{n})$. This is not generally true unless there exits a set of finite measure in your sequence. Yours is a good counter example.

There is nothing like sigma-additivity kinda equality for the intersection. In fact, there should not be such a thing. Lastly, you just can say the following thing which is due to the monotonicity of measure: $\mu(\cap_{n=1}^{\infty} E_{n}) \leq \mu(E_{n})$ for any $n$. This extremely trivial and nothing is interesting here.

I am not sure whether this answers your question or not.

anon
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Matha Mota
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  • the inequality on the intersection is the monotonicity of measure, I wanted to know if any property of the measure is use to show that the measure of $\mu (\cap_{n=1}^{\infty}E_n)=0$ else from $\mu(\emptyset)=0$ – newhere Dec 26 '19 at 13:43
  • Good questions. Sometimes you can do that but this time NO. – Matha Mota Dec 26 '19 at 13:52
  • Are you looking for sth like this? https://math.stackexchange.com/questions/234292/continuity-from-below-and-above – Matha Mota Dec 26 '19 at 13:55
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    Please don't rollback positive edits. – anon Oct 07 '22 at 18:28