Why is this $L^1$-sequence relatively weakly sequentially compact?

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Let $(E,\mathcal E,m)$ be a probability space, $\theta$ be a measurable map on $(E,\mathcal E)$ with $m\circ\theta^{-1}=m$, $s_n$ be a real-valued nonpositive integrable random variable on $(E,\mathcal E,m)$ for $n\in\mathbb N$ with $\lambda:=\inf_{n\in\mathbb N}\int s_n\:{\rm d}m>-\infty$ and $$\varphi_n:=\frac1n\sum_{i=1}^n(s_i-s_{i-1}\circ\theta)\;\;\;\text{for }n\in\mathbb N.$$ It's easy to see that $$\int\varphi_n\:{\rm d}m=\frac{\int s_n\:{\rm d}m}n\xrightarrow{n\to\infty}\lambda\tag1.$$ Moreover, $$Tf:=f\circ\theta\;\;\;\text{for }f\in\mathcal L^1(m)$$ is a linear isometry on $L^p(m)$ for all $p\in[1,\infty]$. In particular, it is continuous with respect to the weak topology on $L^1(m)$.

How can we show that, for all $i\in\mathbb N_0$ and $p\in\mathbb N$, there is an increasing $(n_k)_{k\in\mathbb N}\subseteq\mathbb N$ with $$\max\left(\varphi_{n_k}\circ\theta^i,-p\right)\xrightarrow{k\to\infty}\lambda_{i,\:p}\tag3$$ with respect to the weak topology on $L^1(m)$ for some $\lambda_{i,\:o}\in L^1(m)$? And how can we show that $\lambda_{i,\:p}$ is nondecreasing in $p$ almost surely?

These claims are made in the proof of Theorem 6.7 (in Chapter 4, Paragraph 6) of Revuz' Markov Chains book: proof of theorem 6.7

I don't understand his arguments. For example, I guess he's talking about relative sequential compactness (I'm not sure, but may it be that this is equivalent to relative compactness in the weak topology?). It seems like he's using that $(\varphi_n)_{n\in\mathbb N}$ is contained in a relatively sequentially compact set; but why is that the case? And why do the $\lambda_{i,\:p}$ need to satisfy the claimed monotonicity in $p$? Or does he mean that they can be chosen such that they satisfy this condition?