= Solution
Work with a probability <measure-preserving system>, so $\mu(X)=1$. Put
$$
S_nf=\sum_{j=0}^{n-1}f\circ T^j,\qquad A_nf=\frac{S_nf}{n}.
$$
The \b[maximal ergodic theorem], also called the <maximal ergodic lemma>, says that for real $f\in L^1(\mu)$,
$$
E_N=\{\max_{1\leq n\leq N}S_nf>0\},\qquad E=\{\sup_{n\geq1}S_nf>0\}
\quad\Longrightarrow\quad
\boxed{\int_{E_N}f\,d\mu\geq0,\qquad\int_Ef\,d\mu\geq0.}
$$
No <ergodic transformation> or invertibility assumption is needed for this <maximal ergodic lemma>.
Here is a proof of the <maximal ergodic lemma>. Set $M_0=0$ and $M_N=\max(0,S_1f,\ldots,S_Nf)$. These are integrable: $0\leq M_N\leq\sum_{j=0}^{N-1}|f|\circ T^j$. Splitting off the first summand gives
$$
M_N=\max(0,f+M_{N-1}\circ T),\qquad M_{N-1}\leq M_N.
$$
On $E_N=\{M_N>0\}$, this implies $f=M_N-M_{N-1}\circ T\geq M_N-M_N\circ T$. Outside $E_N$ the right side is $-M_N\circ T\leq0$. Consequently
$$
f\mathbf1_{E_N}\geq M_N-M_N\circ T.
$$
Integrating and using the <measure-preserving transformation> property gives $\int_{E_N}f\,d\mu\geq0$. Since $E_N$ increases to $E$, the <dominated convergence theorem>, with dominating function $|f|$, proves the infinite-supremum assertion of the <maximal ergodic lemma>.
For example, applying the <maximal ergodic lemma> to $|f|-\varepsilon$ yields the useful maximal bound
$$
\mu\{\sup_{n\geq1}|A_nf|>\varepsilon\}
\leq\mu\{\sup_{n\geq1}A_n|f|>\varepsilon\}
\leq\frac{\|f\|_1}{\varepsilon}\qquad(\varepsilon>0).
$$
Indeed on the latter set $G$, the <maximal ergodic lemma> gives $\varepsilon\mu(G)\leq\int_G|f|\,d\mu$.
The \b[<pointwise ergodic theorem>] states that for a probability <measure-preserving system> and $f\in L^1(\mu)$,
$$
\boxed{A_nf\longrightarrow\mathbb E[f\mid\mathcal I]\quad\text{almost everywhere},}
$$
where $\mathcal I$ is the <invariant sigma-algebra>. The <conditional expectation> is invariant and has integral $\int f\,d\mu$. If $T$ is an <ergodic transformation>, this <conditional expectation> is the constant $\int f\,d\mu$. The assertion applies to real or complex integrable functions; it is the <Birkhoff ergodic theorem>.
For a nonnegative measurable $f$, if $\int f\,d\mu<\infty$, the <pointwise ergodic theorem> already proves the required conclusion. For the remaining case, let $f_K=\min(f,K)$, $K=1,2,\ldots$. Each $f_K$ is integrable on the probability <measure-preserving system>. The <pointwise ergodic theorem> and <ergodic transformation> property give, on a single set of full measure obtained by a countable intersection,
$$
\lim_{n\to\infty}A_nf_K(x)=\int f_K\,d\mu\qquad\text{for every }K.
$$
Since $f\geq f_K$, every such $x$ satisfies $\liminf_n A_nf(x)\geq\int f_K\,d\mu$ for all $K$. The <monotone convergence theorem> gives $\int f_K\,d\mu\uparrow\int f\,d\mu=+\infty$. Thus the <nonnegative ergodic averages with infinite integral> tend to $+\infty$, rather than merely having an unbounded subsequence. Combining the two cases,
$$
\boxed{\lim_{n\to\infty}\frac1n\sum_{j=0}^{n-1}f(T^jx)=\int f\,d\mu\in[0,+\infty]\quad\text{almost everywhere}.}
$$
Nonnegativity is what makes truncation a lower bound. The probability normalization is also essential to the displayed value: for a finite measure of total mass $c>0$, the corresponding constant is $c^{-1}\int f\,d\mu$.
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