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?
 A: 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.
