I am reading the specific proof of Birkhoff ergodic theorem given in Katok and Hasselblat's book.
They work with a measure preserving map $T:X\rightarrow X$ on a probability space $(X, \mathcal{B}, \mu)$. They choose $\varphi \in L^1(\mu)$ and as ususal the goal is to show $$ \lim_{n\rightarrow \infty} \sum_{k = 0}^{n - 1}\varphi(T^k(x)) $$ exists for almost every $x$.
The proof starts as something like what follows.
Let $\mathcal{I}:= \{A \in \mathcal{B}:\ T^{-1}(A) = A\}$. This defines a $\sigma$-algebra for itself which they call the invariant $\sigma$-algebra.
They proceed by defining the conditional expectation $f_{\mathcal{I}} := \mathbb{E}[f|\mathcal{I}]$. That is, let $\nu(A) := \int_A f\ d\mu$ be a measure on $\mathcal{I}$. Also restrict $\mu$ on $\mathcal{I}$ and denote it by $\mu'$. Then $f_{\mathcal{I}} := d\nu/d\mu'$ is given by Radon-Nikodym.
Next they define $F_n := \max_{k \le n} \sum_{i = 1}^{k - 1} f\circ T^i$ and claim that $$ \limsup_{n\rightarrow \infty}\sum_{k = 0}^{n - 1}(f \circ T^k)/n \le \limsup_{n\rightarrow \infty} F_n/n \le 0 $$ holds if and only if $A:=\{x \in X: F_n(x) \rightarrow \infty \} \in \mathcal{I}$.
Why is this true?
In general, if you know how to explain the proof in a little bit less messier way, I would highly appreciate it!
Since you ask both about one specific part of the proof and the proof in general, I'll try to rewrite the proof in a more verbose fashion, paying specific attention to the part you asked about. I've bolded the parts that are relevant to your specific "Why is this true?" question. Hopefully this response is somewhat helpful! (It was at least a good exercise for me.)
The main comment I want to make regarding your "Why is this true?" question is that your post has a bit of confusion about the claim KH are making. In particular, KH are making two separate claims:
I'll get to elaborating on both of these. The idea of the proof is to show specifically that the time averages of $\varphi$ converge to the conditional expectation $\varphi_{\mathcal{I}}$. We'll do this by estimating the $\limsup$ and $\liminf$ of these time averages; a very slick lemma allows us to do both of these with the same argument. The lemma is as follows:
Lemma. Let $f \in L^1(\mu)$ satisfy $f_{\mathcal{I}} < 0$. Then $\limsup_{n \to \infty} \tfrac{1}{n} \sum_{k = 0}^{n-1} f \circ T^k(x) \leq 0$ for $\mu$-a.e. $x$.
Proof. [Part 1 of your "Why is this true".] Consider the max-partial-sums of $f$, $$ F_n(x) = \max_{1 \leq k \leq n} \sum_{i = 0}^{k-1} f \circ T^k(x). $$ Note that for every $x$, the sequence $F_n(x)$ is nondecreasing. We let $A = \lbrace x : F_n(x) \to \infty \rbrace$. Then for $x \notin A$ there exists some $C(x)$ such that $F_n(x) \leq C(x)$ for all $n$, and so $$ \tfrac{1}{n} \sum_{k = 0}^{n-1} f \circ T^k(x) \leq \frac{1}{n}F_n(x) \leq \frac{1}{n} C(x) \to 0, $$ from which the conclusion of the lemma follows. [This is the justification for the inequality chain part of your question.]
So it suffices to show $\mu(A) = 0$.
We note, for later use, that the max-partial-sum functions $F_n$ satisfy \begin{equation} \tag{1} F_{n+1}(x) = \begin{cases} f(x) + F_n(Tx) & \text{if } F_n(Tx) \geq 0 \\ f(x) & \text{if } F_n(Tx) < 0. \end{cases} \end{equation} (The second case happens if the max in $F_{n+1}(x)$ happens for $k = 1$, and the first case happens otherwise.)
Subclaim. $A \in \mathcal{I}$. [This is the other part of your direct question.]
Proof of subclaim. We must show $T^{-1}(A) = A$, so show both inclusions:
Back to the proof of the lemma. From our max-partial-sum relation $(1)$ we have $$ F_{n+1} - F_n \circ T = f - \min \lbrace 0, F_n \circ T \rbrace. $$ For $x \in A$, the expression $F_n \circ T(x)$ blows up, so for each $x$ the minimum in this expression is eventually $0$ for large enough $n$. That is, $F_{n+1} - F_n \circ T$ converges pointwise to $f$ from above on $A$. By the monotone convergence theorem (or the dominated convergence theorem), $$ \int_A (F_{n+1} - F_n \circ T ) \, d\mu \to \int_A f \, d \mu. $$ Note though that $$ \int_A (F_n \circ T ) \, d\mu = \int_{T^{-1}(A)} F_n \, d \mu = \int_A F_n \, d\mu $$ since $A \in \mathcal{I}$, and hence (since $F_{n+1} - F_n$ is clearly nonnegative), $$ \int_A (F_{n+1} - F_n \circ T ) \, d\mu \geq 0. $$ On the other hand, again using $A \in \mathcal{I}$ (and by the definition of conditional expectation), $$ \int_A f \, d\mu = \int_A f_{\mathcal{I}} \, d\mu. $$ Putting these together, we conclude $$ \int_A f_{\mathcal{I}} \, d\mu \geq 0, $$ which forces $\mu(A) = 0$ since $f_{\mathcal{I}} < 0$ by hypothesis. This proves the lemma. $\square$
Proof of Birkhoff Ergodic Theorem.
Apply the lemma to $f = \varphi - \varphi_{\mathcal{I}} - \varepsilon$ to conclude that for $\mu$-a.e. $x$, $$ \limsup_{n \to \infty} \frac{1}{n} \sum_{k=0}^{n-1} (\varphi - \varphi_{\mathcal{I}} - \varepsilon) (T^k x) \leq 0. $$ Noting that $\varphi_{\mathcal{I}} \circ T = \varphi_{\mathcal{I}}$, this is $$ \limsup_{n \to \infty} \left( \frac{1}{n} \sum_{k=0}^{n-1} \varphi \circ T^k (x)\right) - \varphi_{\mathcal{I}}(x) - \varepsilon \leq 0, $$ which shows $$\ \limsup_{n \to \infty} \tfrac{1}{n} \sum_{k=0}^{n-1} \varphi \circ T^k \leq \varphi_{\mathcal{I}} + \varepsilon $$ almost everywhere. Similarly, applying the lemma to $f = -\varphi + \varphi_{\mathcal{I}} - \varepsilon$ shows that $$ \liminf_{n \to \infty} \tfrac{1}{n} \sum_{k=0}^{n-1} \varphi \circ T^k \geq \varphi_{\mathcal{I}} - \varepsilon $$ almost everywhere, which proves the theorem.