# Is there a collection of measurable not necessarily nested but converging sets such that we cannot permute measure and limit

We know that for $\{E_n\}_{n\geq0}$ Lebesgue measurable sets s.t. $E_0\subset E_1\subset...$ we have $\lim\limits_{n\rightarrow\infty}E_n=\bigcup\limits_{n=0}^{\infty}E_n$ and $\mu(\lim E_n)=\lim(\mu(E_n))$

Similarily if $E_0\supset E_1\supset...$ and $\mu(E_0)<\infty$ we have $\lim\limits_{n\rightarrow\infty}E_n=\bigcap\limits_{n=0}^{\infty}E_n$ and $\mu(\lim E_n)=\lim(\mu(E_n))$

But what if we lose the property of nestedness? What would be some interesting examples of collections of sets where we cannot permute limit and Lebesgue measure?

An example: $E_n=[n,n+1]$

Then $\mu(E_n)=1$ for every $n$ but $\lim_{n\to\infty}E_n=\varnothing$ so that $\mu(\lim_{n\to\infty}E_n)=\mu(\varnothing)=0$.

edit:

With the lemma of Fatou it can be shown that:$$\mu(\liminf E_n)\leq\liminf\mu(E_n)\tag0$$

If some $c>0$ and some integer $n_0$ exists such that $n>n_0\implies E_n\subseteq[-c,c]$ then this can also be applied on the sets $[-c,c]-E_n$ leading to: $$\mu(\liminf E_n)\leq\liminf \mu(E_n)\leq\limsup\mu(E_n)\leq\mu(\limsup E_n)<\infty\tag1$$

You are dealing with a convergent sequence of measurable sets iff $\liminf E_n=\limsup E_n$ so in that case $(1)$ implies that: $$\mu(\lim E_n)=\lim \mu( E_n)$$

This shows you that it is somehow inevitable to come up with "sets that dissapear to infinity" for an example that you ask for.

• This doesn't look like a collection of converging sets. – Darth Geek Apr 2 '18 at 8:11
• Can we really say that $E_n$ converges though? That would work for a function $\mathbb{1}_{E_n}$ and the integral for sure... – John Cataldo Apr 2 '18 at 8:12
• @DarthGeek We have $\liminf E_n=\limsup E_n=\varnothing$. That means that $\lim E_n$ exists and equals $\varnothing$. – drhab Apr 2 '18 at 8:20
• Ok yes sorry it does converge. But it’s not a very interesting example – John Cataldo Apr 2 '18 at 8:22
• @DarthGeek I just edited on that topic. – drhab Apr 2 '18 at 9:15