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Good day, I have the next theorem: Let {${E_i}$} a collection denumerable of measurable sets such that $E_1\supset E_2\supset... \supset E_n\supset... $ and $m(E_1)$ is finite, then $m(\bigcap_{i=1}^{\infty}E_i)=lim_{n\rightarrow \infty}m(E_n)$.

How is the theorem false if $m(E_1)=\infty$, with a counterexample? I think in intervals $I_k$ in some set A such that $I_k=(a,b)$, $a,b \in \mathbb{Q}$ . Can that perform?

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  • $\begingroup$ every interval $I_k$ in your question @mathreda has a finite measure so you cannot create a counterexample with these $I_k$ $\endgroup$ – Marios Gretsas Aug 19 '17 at 22:12
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The standard counterexample is to set $E_n=(n,\infty)$ for each $n$. Then $m(E_n)=\infty$ for all $n$, but $\bigcap_{n=1}^{\infty}E_n=\emptyset$.

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    $\begingroup$ Hell nooo! 1 second before i pressed <Post your answer> you posted...:P...+1 from me though.(after two hours because im blocked from voting) $\endgroup$ – Marios Gretsas Aug 19 '17 at 21:56
  • $\begingroup$ It happens to all of us... $\endgroup$ – carmichael561 Aug 19 '17 at 21:56
  • $\begingroup$ What about a counterexample where only the first $E_1$ is infinite? $\endgroup$ – mrp Aug 19 '17 at 21:58
  • $\begingroup$ @mrp: The result in the question is true provided that there is some $N$ such that $m(E_n)<\infty$ for $n\geq N$. So there is no such counterexample. $\endgroup$ – carmichael561 Aug 19 '17 at 21:59
  • $\begingroup$ @carmichael561 alright, that makes intuitive sense to me, although that's not what the OP stated. $\endgroup$ – mrp Aug 19 '17 at 22:03

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