Reverse Fatou's lemma on probability space

Question

Let $(\Omega, \mathcal{F},\mathbb{P})$ be probability space and $E_{n \in \mathbb{N}}$ be $\mathcal{F}$-measurable sets.

Show example that reverse Fatou's Lemma, $\mathbb{P}(\limsup_n E_n) \geq \limsup_n \mathbb{P}(E_n)$, meets inequality strictly.

I understand this inequality of inf. However, I cannot solve this. I want to remember the direction of inequality. However, I found that this question's answer on the Internet is only inf version.

My try

I understand $\mathbb{E}[\limsup_{n} E_n] \geq \limsup_n \mathbb{E}[E_n]$

and

$E_{2k-1}(\omega)=1$ if $\omega \in (0,1/2)$

$E_{2k-1}(\omega)=0$ if $\omega \in [1/2,1)$

$E_{2k}(\omega)=0$ if $\omega \in (0,1/2)$

$E_{2k}(\omega)=1$ if $\omega \in [1/2,1)$

so

$1=\mathbb{E}[\limsup_{n} E_n] > \limsup_n \mathbb{E}[E_n]=0$

However, I cannot show this question on probability space. I think that I use the relationship, $\mathbb{P}(A)=\mathbb{E}[1_A]$ , right?

Answer 1

The correct example: $\Omega=[0,1]$, $\mathcal F=\mathcal B(\Omega)$, $P=\mathrm{Leb}$, $E_{2n}=[0,1/2]$ and $E_{2n+1}=(1/2,1]$ for every $n$, then $\limsup E_n=\Omega$ and $P(E_n)=1/2$ for every $n$ hence $$P(\limsup E_n)=1\gt1/2=\limsup P(E_n).$$