The proof is from Easy and nearly simultaneous proofs of the ergodic theorem and maximal ergodic theorem (Keane, Petersen, 2006).
Let $(X, \mathcal B, \mu)$ be a probability space, $T:X\to X$ be a measure-preserving map, $f\in L^1(X, \mathcal B, \mu)$. Let $$A_k f = \frac 1 k \sum_{j = 0}^{k-1} f \circ T^j, f^*_N = \sup_{1\le k \le N} A_k f,$$
Take $\lambda \in L^1(X)$, then let $E_N = \{f_N^* > \lambda\}$, then since $f \le f_N^*$, we have that $\forall x \notin \{f_N^* > \lambda\}$, $$[(f - \lambda)\chi_{E_N}](x) = 0 \ge (f - \lambda)(x) $$
Thus $(f - \lambda)\chi_{E_N} \ge (f - \lambda)$
Then we consider a sum, where $m$ is a big but undetermined integer. $$\sum_{k=0}^{m-1} (f - \lambda )\chi_{E_N}(T^kx)$$
This is where the first part of the second page starts, and I don't know what it means.
There is maybe an initial string of $0$’s during which $T^k x \notin E_N$ . Then there is a first time $k$ when $T^k x \in E_N$, which initiates a string of no more than $N$ terms, the sum of which is positive (using on each of these terms the fact that $(f − λ)\chi_{E_N} ≥ (f − λ))$.
From definition of $E_N$, we have that $\forall T^k x \in E_N,$ $\exists 1\le l \le N, $
$$f(T^k x) + f(T^{k+1} x) + \cdots f(T^{k+l-1}x) \ge l \lambda(T^k x)$$ thus $$[(f-\lambda) + (f\circ T - \lambda) + \cdots (f\circ T^{l-1} - \lambda)](T^k x) \ge 0$$
However, Keane seems to want me to show that
$$[(f-\lambda) + (f - \lambda)\circ T + \cdots (f - \lambda)\circ T^{l-1}](T^k x) \ge 0$$
using the fact that $(f − λ)\chi_{E_N} ≥ (f − λ))$.
I can't see how that's possible. Not to mention that $(f − λ)\chi_{E_N} ≥ (f − λ))$ is true over all of $X$, not just over $E_N$, so "using the fact" seems uselessly unspecific.
$[(f-\lambda) + (f\circ T - \lambda) + \cdots (f\circ T^{l-1} - \lambda)](T^k x) \ge 0$ and $[(f-\lambda) + (f - \lambda)\circ T + \cdots (f - \lambda)\circ T^{l-1}](T^k x) \ge 0$ are equivalent (almost everywhere), since in the theorem it's assumed that $\lambda$ is invariant under $T$: ($\lambda\circ T = \lambda$ a.e.) function.