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I have some trouble understanding the following homework.

With measure space $(\mathbb{N},\mathcal{P}(\mathbb{N}),\mu)$ where $\mu$ is the counting measure, we have $(f_n)_{n\geq1},(g_n)_{n\geq1} \in \mathcal{M}^+$ as

$f_n=\mathbb{1}_{\{n\}}$ and $g_n= \begin{cases} \mathbb{1}_{\{1\}} \text{ if n is odd}\\ \mathbb{1}_{\{2\}} \text{ if n is even} \end{cases} $

(i)

Find $\limsup_{n \rightarrow \infty} f_n, \limsup_{n \rightarrow \infty} g_n$

(ii)

Show that:

$\int_\mathbb{N} \limsup_{n \rightarrow \infty} f_n d\mu < \limsup_{n \rightarrow \infty} \int_\mathbb{N} f_n d\mu$

and

$\int_\mathbb{N} \limsup_{n \rightarrow \infty} g_n d\mu > \limsup_{n \rightarrow \infty} \int_\mathbb{N} g_n d\mu$

I have:

$\limsup_{n \rightarrow \infty} f_n=0, \limsup_{n \rightarrow \infty} g_n=0$

This satisfies the first inequality in (2) as the integral is always 1 so 0<1. However the same approach goes complete wrong in the second inequality where I have 0>2 which is absurd.

Can anyone see what is wrong? I think maybe I am misunderstanding the limsup or liminf concepts :( I am assuming limsup and liminf are pointwise

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You've made two mistakes in dealing with $g_n.$

For the first, note that $$g_n(1) = \begin{cases}1 & \text{ if } n \text{ is odd} \\ 0 & \text{ otherwise}\end{cases}$$ so that $$\left(\limsup g_n\right)(1) = 1$$

Doing the same [point-wise] analysis with $g_n(2),$ we see that $(\limsup g_n)(2) = 1$ and it will be zero for all other inputs. That is, $\limsup g_n = 1_{\{1,2\}}.$ Use this to evaluate $\int_\mathbb{N} \limsup g_n\,d\mu.$

For the second, you should re-visit $\int_\mathbb{N} g_n\,d\mu.$ It's never going to be equal to $2.$

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  • $\begingroup$ Ok, as I understand then $\lim_\mathbb{N} \limsup g_n d\mu=2 > 1=\limsup \int_\mathbb{N} g_n d\mu$. But isn't x chosen first and then limsup is taken afterwards, so it becoms pointwise? $\endgroup$ – Daniel Sep 22 '19 at 8:25
  • $\begingroup$ Yes. At which point do you believe we've done it any other way? $\endgroup$ – Brian Moehring Sep 22 '19 at 8:28

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