I'm trying to prove that if $ \{A_r\; :\: r>0\}$ is a family of measurable subsets of $\mathbb{R^n}$ such that $A_r \subset A_s$, if $r<s$, then uncountable union $\cup_{r>0} A_r$ is measurable.

Now my idea is to use limits to show that $A\subset \bigcup_{r>0}A_r$ and $A \supset\bigcup_{r>0}A_r$, where $A$ is the limit of a subsequence which is measurable as limit of measurable sets. However this relies on the fact that $\bigcup_{r>0}A_r$ is bounded from above for the limit to exists. Now it's clear that the sequence is increasing, but I'm not completely sure that it is bounded from above. My reasoning is that if $A_r\subset\mathbb{R^n}$ for all $r>0$, then $\bigcup_{r>0}A_r$ is bounded.

I'm not completely sure of this reasoning and hope that someone could clarify if the union really is bounded or not.


For each $n \in \mathbb{N}$, $A_{n}$ is measurable, and so $\bigcup \limits_{n=1}^{\infty} A_{n}$ is measurable. But $\bigcup \limits_{n=1}^{\infty} A_{n} = \bigcup \limits_{r > 0} A_{r}$. Why?

Proof that $\bigcup \limits_{n=1}^{\infty} A_{n} = \bigcup \limits_{r > 0} A_{r}$:

$\subseteq$ direction: Let $x \in \bigcup \limits_{n=1}^{\infty} A_{n}$. Then $x \in A_{n}$ for some $n$. But $A_{n} \subseteq A_{r}$ for any $r > n$, which means $x \in A_{r}$ for some $r$, and this implies $x \in \bigcup \limits_{r > 0} A_{r}$.

$\supseteq$ direction: Let $x \in \bigcup \limits_{r > 0} A_{r}$. Then $x \in A_{r}$ for some $r > 0$. But then for $n \in \mathbb{N}$, $n > r$, $x \in A_{n}$ since $A_{r} \subseteq A_{n}$. Then $x \in \bigcup \limits_{n = 1}^{\infty} A_{n}$. Done.

So basically because we have the property that $A_{r} \subseteq A_{s}$ is $r < s$, then we can express the uncountable union as a countable union of measurable sets, which is then measurable.

  • $\begingroup$ This looks good to me, I don't think you're wrong. $\endgroup$
    – 5xum
    Oct 1 '14 at 11:51
  • $\begingroup$ @5xum Cool, thanks! :) $\endgroup$
    – layman
    Oct 1 '14 at 11:53

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