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Construct a sequence of measureable sets $E_1\supseteq E_2 \supseteq E_3 \supseteq \cdots$ such that $\mu(E_n)=\infty$ for each $n$ but $$\mu\left(\bigcap_{n=1}^\infty E_n\right)=0$$

Claim: Let \begin{equation*} \begin{aligned} E_1= & \left(\frac{1}{i},1\right]\cup \left(\frac{1}{i+1},2\right] \cup \cdots \\ E_2= & \left(\frac{1}{i},2\right]\cup \left(\frac{1}{i+1},3\right] \cup \cdots \\ \vdots & \vdots \\ E_n= & \left(\frac{1}{i},n\right]\cup \left(\frac{1}{i+1},n+1\right] \cup \cdots \\ \vdots & \vdots \\ \end{aligned} \end{equation*} where $i$ is an arbitrary positive integer.

I believe that this sequence of sets satisfies the conditions above, but I want to formally write it out.

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This is way too complicated. Take $E_i = [i,\infty)$. Then, $\mu(E_i) = \infty$ and $E_j \subset E_i$ for any $j>i$, clearly.

$\bigcap_i E_i = \emptyset$ since if $x \in \bigcap_i E_i$, that means that $x\geq i$ for all $i \in \mathbb{N}$ (and there is no such $x$).

Thus, $\mu\left(\bigcap_i E_i\right) = 0$.

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    $\begingroup$ Beat me to it... $\endgroup$ – user223391 Mar 2 '16 at 21:50
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The question is actually whether the example given works. Notice that $(1/i,1]\subset \bigcap_{n=1}^\infty E_n$ and so it is doesn't work.

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    $\begingroup$ But if you set $i=1$ this argument will no longer hold $\endgroup$ – Korf Mar 2 '16 at 21:53
  • $\begingroup$ @Korf :D Of course it does, simply take $(1/i,2]$, etc. $\endgroup$ – John B Mar 2 '16 at 21:54
  • $\begingroup$ Sure. I do agree that your idea is the right one and applies for any choice $i$. But the argument you have written in your answer actually fails for $i=1$. $\endgroup$ – Korf Mar 2 '16 at 22:00
  • $\begingroup$ @Korf So, we do agree, that what I meant with ":D". And certainly, your comment was very welcome. $\endgroup$ – John B Mar 2 '16 at 22:01
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Take the measure defined by $\mu(X)=\text{Card}(X)$ and $E_n=]0,\frac{1}{n}]$.

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