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I saw this post but was still a little confused. My book shows that to prove this it is sufficient to make use of the following identity:

$$\limsup_{i\rightarrow\infty}\phi_i(x) = \inf_i\{\sup_{j\geq i}\phi_j(x)\}$$

And I'm having a bit of trouble understanding what this says. I've gotten a better understanding of $\limsup$ from my post yesterday, but still a bit confused.

So for a sequence of functions $\phi_i$, the $\limsup$ for a given point x is the limit of the supremum of $\phi_i(x)$. This could be increasing as i increases, or decreasing, or staying the same. And this is the same as taking the infimum over i of the set of supremums of $\phi_j$ for $j\geq i$? This last part is confusing me.

So for all i, we look at the $\{\sup_{j\geq i}\phi_j(x)\}$. And then we look at the greatest lower bound of this set? How is this the same as working with the limit?

And finally, how does this help us make it measurable?

Thanks for any help!

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    $\begingroup$ The sequence $\{\sup_{j\ge i} \varphi_j(x) \}_{i\in\mathbb{N}}$ is decreasing in $i$. Basically, the larger a set is, the more chances you have of getting a large value. As you remove things from the set, you potentially remove big things. More formally, $$A\subseteq B \subseteq \mathbb{R} \implies \sup_{x\in A} x \le \sup_{x\in B} x. $$ This can be applied to the sequence of values above by setting $A_i = \{ \varphi_j(x)\}_{i\ge j}$, and noting that $A_i \supseteq A_{i'}$ whenever $i < i'$. $\endgroup$ – Xander Henderson Dec 2 '17 at 20:41
  • $\begingroup$ Could you explain why $\{\sup_{j\geq i}\phi_j(x)\}$ is a sequence? I guess I was confused in the first place about what makes it a set. So I thought $\sup_{j\geq i}\phi_j(x)$ is just the supremum of the sequence of functions, over all functions in the sequence past a certain point. But shouldn't there still only be one value that is a supremum of all those values? $\endgroup$ – letslearnmath Dec 2 '17 at 20:51
  • $\begingroup$ $\sup_{j\ge i} \varphi_j(x)$ is just a number---the supremum of a set is a number. For each $i$, you obtain a different such number. Taken together, these numbers form a sequence, i.e. the sequence $\{ \sup_{j\ge i} \varphi_j(x) \}$, indexed by $i$. $\endgroup$ – Xander Henderson Dec 2 '17 at 20:57
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Observe that when $k\geq j,$ $\sup_{i\geq k}\{\phi_{i}(x)\}\leq\sup_{i\geq j}\{\phi_{i}(x)\},$ since the set $\{\phi_{i}(x):i\geq j\}$ contains $\{\phi_{i}(x):i\geq k\}.$ Then the sequence $(\sup_{i\geq j}\{\phi_{i}(x)\})_{j\geq 1}$ is decreasing, so the limit of this sequence must equal its infimum, which proves the identity in your book.

Now from the answer you linked, $\overline{\phi}_{j}(x):=\sup_{i\geq j}\{\phi_{i}(x)\}$ is a measurable function for every $j\geq 1,$ and by our reasoning above, $(\overline{\phi}_{j})_{j\geq 1}$ is a decreasing sequence of functions. Therefore the pointwise limit is measurable, as we wanted to prove.

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