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There are some statements in something that I'm reading that I don't quite understand:

Consider $c=\sup c_n$ where $c_n$ is a non decreasing sequence of non-negative, uniformly continuous functions. Define the following functional: $$I_j[\pi] = \int c_j \, d\pi$$ Also consider a sequence of measures $(\pi_n)$ which admits a cluster point $\pi_*$. Whenever $n \geq m$ we have $I_n[\pi_n] \geq I_m[\pi_n]$ (because $c_n \geq c_m$ for the same $\pi_n$). By continuity of $I_m$: $$\lim_{n \rightarrow \infty} I_n[\pi_n] \geq \limsup_{n \rightarrow \infty} I_m[\pi_n] \geq I_m[\pi_*]$$ Further: $$\lim_{n \rightarrow \infty}I_n[\pi_n] \geq \lim_{m\rightarrow \infty}I_m[\pi_*]$$

So I'm trying to unpack the string of inequalities and the questions I have are:

  1. Why is this true: $$\lim_{n \rightarrow \infty} I_n[\pi_n] \geq \limsup_{n \rightarrow \infty} I_m[\pi_n]$$

  2. I suspect continuity of $I_m$ is used in the second part of the inequality on the first line: $$\limsup_{n \rightarrow \infty} I_m[\pi_n] \geq I_m[\pi_*]$$ I don't see how its used though.

  3. On the second line: $$\lim_{n \rightarrow \infty}I_n[\pi_n] \geq \lim_{m\rightarrow \infty}I_m[\pi_*]$$ Does this follow from the first string of inequalities? In particular, we showed this statement was true for any particular $m$ on the first line. So we assert it must be true for any $m$ no matter how large?

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    $\begingroup$ Why don't you say what kind of objects $\pi$ and $\pi_n$ are? $\endgroup$
    – amsmath
    Commented Oct 8, 2017 at 18:01
  • $\begingroup$ I assume that the $\pi_n$'s are of course measures and the $I_n's$ act as functionals on your set of measures. Consider adding their definitions though. The statements in question read easier if the indices of the $\pi_j$'s are allowed to vary independently from the limits. Is that what the statement is saying? $\endgroup$ Commented Oct 8, 2017 at 18:31
  • $\begingroup$ @amsmath I didn't think it was important. They are joint distributions. $\endgroup$
    – yoshi
    Commented Oct 8, 2017 at 23:00
  • $\begingroup$ @rt6 Yes, the $\pi$'s are the measure and $I$ is the functional. I'll update this. Could you elaborate on your question? $\endgroup$
    – yoshi
    Commented Oct 8, 2017 at 23:00
  • $\begingroup$ What do you mean by non-decreasing sequence? $c_n(x)\leq c_{n+1}(x)$? $\endgroup$
    – Ice sea
    Commented Oct 13, 2017 at 17:27

1 Answer 1

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It is easy to see that (by fixing $m$ first) $$\lim I_n[\pi_n]=\limsup_{n \rightarrow \infty} I_n[\pi_n] \geq \limsup_{n \rightarrow \infty} I_m[\pi_n]$$ because any sequence $a_n\geq b_n$ implies $\limsup a_n\geq \limsup b_n$. Since $\pi_n$ is a sequence with a cluster $\pi^*$. We know that there exits a subsequence $\pi_{n_k}$ of $\pi_n$ such that $\pi_{n_k}\to \pi^* $ in weak$^*$ topology. $$ \limsup_{n\to\infty} I_m[\pi_n]\geq \lim_{k\to \infty} I_m[\pi_{n_k}]=I_m[\pi^*] $$

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