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psie
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- TL;DR Summary
- Consider
. I've managed to show this set is closed in by showing its complement is open. Now using what I've already showed, how can I argue that is closed in ?
Let be a sequence of measurable functions from into . I'm reading a proof of the fact that the set of all for which converges in as is measurable. The proof goes like this (I'm paraphrasing):
Why is closed in ? As I said, I can show the diagonal with is closed in , but if and the space is , how does one proceed?
Defineby . Then is measurable and if , then So the measurability of follows since is measurable subset of (note that is measurable as a closed subset of ).
Why is