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I'm wondering how to solve this. We have space X, in which there is a sigma-field $M$. We have a sequence of measurable functions $f_n: X \rightarrow \mathbb{R}$. Let $A= \{ x \in X: $ sequence $(f_n(x))$ is bounded from below and unbounded from above $\} $ Prove that $A$ is measurable (which means $A \in M$). I would appreciate any help. Thanks.

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Try to write first $$ B = \{x\in \Bbb R:(f_n(x))\text{ is bounded from below }\} $$ as a union of some sets $B_m$. – Ilya Jan 15 '13 at 17:29

That $(f_n(x))$ is bounded above means, that for some $m$, $f_n(x)<m$ for all $n$, which is equivalent to $x\in{f^{-1}_n(-\infty,m)}$ for all $n$, which is equivalent to $x\in\bigcap_n f^{-1}_n(-\infty,m)$. So $(f_n(x))$ is bounded above if $$x\in\bigcup_m\bigcap_n f^{-1}_n(-\infty,m),$$ a measurable set. Try to write the other condition in a similar way. The intersection of the two sets gives you an answer.

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Thank you for your answer. I've got two more questions: 1) Am i right that when it is unbounded above it means: $x \in \cap_m \cup_n f_n^{-1} (m, \infty) $? And the second question: how do you know that $B = \{ x: x \in \cup \cap f_n^{-1} (-\infty, m) \} $ is measurable? – Anne Jan 15 '13 at 18:02
Err.. I guess I know the answer for my second question: $f_n^{-1} (...)$ is measurable, then we have countable intersection of measurable sets, which is measurable, and then countable sum of measurable. Is it ok? But the first question stays the same. – Anne Jan 15 '13 at 18:10
@Anne 1) It means that for all $q>0$ and for all $n$ there exists $m\geq n$ such that $f_m(x)\in (q,\infty)$. The set of such elements is $\bigcap_q \bigcap_n\bigcup_{m\geq n}f_m^{-1}(q,\infty)$. Generally, you change "for all" to intersections and "exists" to unions. 2) Yes, it is measurable for the reason you gave. – Michael Greinecker Jan 16 '13 at 8:24

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