If $g>0$ is in $L\ln\ln L$, then $\#\{n: g(\theta x)+\cdots+g(\theta^nx)\le t\,g(\theta^nx)\}\le Ct$ when $t\to\infty$ Here are two theorems:

  
*
  
*For every dynamical system $(X, Σ, m, T )$ and function $f \in L \ln \ln L(X,m)$
  (that is, such that  $\int |f| \ln^+ \ln^+ |f|\, {\rm d}m$ is finite), $$N^∗f(x)=\sup_{n\ge 1} \left( \dfrac{1}{n} \# \left\lbrace i\ge 1: \dfrac{f(T^ix)}{i} \ge \dfrac{1}{n} \right\rbrace  \right)$$ is finite for $m$-almost every $x$. 
  
*Let $(S, A, µ)$ be a probability space, and $θ$ a $µ$-measure
  preserving transformation on $S$. Let $g > 0$ such that $\int g \ln^+ \ln^+ g dµ$ is finite. Define $$b_n=\dfrac{\sum_{k=1}^n g(\theta^kx)}{g(\theta^n x)}.$$ If $\theta $ is ergodic then  $$\limsup\limits_{t\to\infty} \dfrac{ \# \left\lbrace n\ge 1: b_n\le t \right\rbrace  }{t}\ \text{is finite}.$$

According to author, to prove the second theorem, we must use the pointwise ergodic theorem and the first theorem. Can you explain this clearly to me?

By the pointwise ergodic theorem, we have (by ergodicity) that $$\frac 1n\sum_{j=1}^ng\circ \theta^k(x)\geqslant \frac{\mathbb E\left[g\right]}2$$
for each $n\geqslant n_0(x)$. Therefore, we have 
$$\left\{n\geqslant n_0(x)\mid b_n\leqslant t\right\}
\subset \left\{n\geqslant n_0(x)\mid \frac{g\left(\theta^nx\right)}n
\geqslant \frac{\mathbb E\left[g\right]}{2t}\right\}=\left\{i\geqslant n_0(x)\mid \frac{g\left(\theta^ix\right)}i
\geqslant \frac{\mathbb E\left[g\right]}{2t}\right\}.$$
Now, assume that $t$ is such that $\mathbb E\left[g\right]/(2t)\in \left[1/u,1/(u-1)\right)$ for some integer $u$. Then 
$$\operatorname{Card}\left\{n\geqslant n_0(x)\mid b_n\leqslant t\right\}
\leqslant \operatorname{Card}\left\{i\geqslant n_0(x)\mid \frac{g\left(\theta^ix\right)}i
\geqslant \frac{\mathbb E\left[g\right]}{2t}\right\}\leqslant \operatorname{Card}\left\{i\geqslant n_0(x)\mid \frac{g\left(\theta^ix\right)}i
\geqslant \frac 1u\right\}.$$
Therefore, 
$$\frac{\operatorname{Card}\left\{n\geqslant n_0(x)\mid b_n\leqslant t\right\}}t\leqslant \operatorname{Card}\left\{i\geqslant n_0(x)\mid \frac{g\left(\theta^ix\right)}i
\geqslant \frac 1u\right\}\cdot \frac 2{\mathbb E\left[g\right] (u-1)}   \leqslant N^*g(x)\frac{2u} {\mathbb E\left[g\right] (u-1)}$$
and we get that 
$$\limsup_{t\to +\infty}\frac{\operatorname{Card}\left\{n\geqslant n_0(x)\mid b_n\leqslant t\right\}}t\leqslant  N^*g(x)\frac{2} {\mathbb E\left[g\right]}.$$
Since $\limsup_{t\to +\infty}\operatorname{Card}\left\{n\lt n_0(x)\mid b_n\leqslant t\right\}/t=0$, we are done.
