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Let $(X,M,m)$ be measurable space with measure $m$ and $(E_n)_n$ sequence in $M$. If $m\left( \bigcup_{k=1}^{+\infty}E_k \right)<+\infty$ prove that $$m\left(\bigcap_{n=1}^{+\infty} \bigcup_{k=n}^{+\infty}E_k\right)\leq{\limsup}_{n \to \infty}m(E_n)$$

I proved that $ \bigcup_{k=n}^{+\infty}E_k\supseteq \bigcup_{k=n}^{+\infty}E_{k+1}$ i.e sequence $(\bigcup_{k=n}^{+\infty}E_k)$ is decreasing.After that i've used continuity of the measure to show that $$m\left(\bigcap_{n=1}^{+\infty} \bigcup_{k=n}^{+\infty}E_k\right)={\lim}_{n\to\infty}m\left(\bigcup_{k=n}^{+\infty}E_k\right)$$ I don't know what next? Any hint would be helpful.

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    $\begingroup$ Wait this is obviously false. I think you want the reverse inequality. $\endgroup$ Mar 22, 2018 at 20:13
  • $\begingroup$ Maybe it's mistake in my paper. $\endgroup$
    – user534900
    Mar 22, 2018 at 20:27
  • $\begingroup$ I do want to refer to you the "Borel-Cantelli Lemma" to give $m(\limsup E_k)$ some meaning $\endgroup$ Mar 22, 2018 at 20:59

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Your inequality is backwards. Just note that $E_n \subseteq \bigcup_{k=n}^{+\infty}E_k$ so

$$ {\lim}_{n\to\infty}m\left(\bigcup_{k=n}^{+\infty}E_k\right) = \limsup_{n\to\infty} m\left(\bigcup_{k=n}^{+\infty}E_k\right) \ge \limsup_{n\to\infty} m(E_n)$$

Also have a look here.

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