[proofplan] Shift the function by the threshold and set $g = f - \alpha$, so that the positive maximal set of $g$ is exactly the superlevel set $\{x : M^+ f(x) > \alpha\}$. The Maximal Ergodic Lemma applied to $g$ gives a nonnegative integral of $g$ over this set. Expanding $g$ and bounding the remaining integral by the $L^1$ norm of $f$ gives the weak-type estimate. [/proofplan] [step:Shift the function and identify the superlevel set] Let $g: X \to \mathbb{R}$ be the measurable function defined by \begin{align*} g(x) = f(x) - \alpha . \end{align*} Since $f \in L^1(X,\mu)$ and $\mu(X) < \infty$, the constant function $x \mapsto \alpha$ belongs to $L^1(X,\mu)$, hence $g \in L^1(X,\mu)$. For each integer $n \geq 1$, the one-sided ergodic average satisfies \begin{align*} \frac{1}{n}\sum_{k=0}^{n-1} g(T^k x) &= \frac{1}{n}\sum_{k=0}^{n-1} \bigl(f(T^k x)-\alpha\bigr) \\ &= \frac{1}{n}\sum_{k=0}^{n-1} f(T^k x)-\alpha . \end{align*} Taking the supremum over all $n \geq 1$ gives \begin{align*} M^+g(x)=M^+f(x)-\alpha . \end{align*} Therefore \begin{align*} \{x \in X : M^+g(x)>0\} = \{x \in X : M^+f(x)>\alpha\}. \end{align*} Define this measurable set by \begin{align*} E_\alpha := \{x \in X : M^+f(x)>\alpha\}. \end{align*} [/step] [step:Apply the Maximal Ergodic Lemma to the shifted function] The function $g$ is integrable, and the finite measure-preserving hypotheses on $(X,\mathcal B,\mu,T)$ are exactly the hypotheses required by the [Maximal Ergodic Lemma](/theorems/3432). Applying that lemma to $g$ and using the identity from the previous step, \begin{align*} \int_{E_\alpha} g \, d\mu(x) = \int_{\{x \in X : M^+g(x)>0\}} g \, d\mu(x) \geq 0 . \end{align*} Substituting $g=f-\alpha$ gives \begin{align*} 0 \leq \int_{E_\alpha} (f-\alpha) \, d\mu(x) = \int_{E_\alpha} f \, d\mu(x)-\alpha\,\mu(E_\alpha). \end{align*} Hence \begin{align*} \alpha\,\mu(E_\alpha) \leq \int_{E_\alpha} f \, d\mu(x). \end{align*} [/step] [step:Bound the localized integral by the $L^1$ norm] Taking absolute values inside the integral over $E_\alpha$ gives \begin{align*} \int_{E_\alpha} f \, d\mu(x) \leq \int_{E_\alpha} |f| \, d\mu(x). \end{align*} Since $E_\alpha \subseteq X$, monotonicity of the integral for the nonnegative function $|f|$ gives \begin{align*} \int_{E_\alpha} |f| \, d\mu(x) \leq \int_X |f| \, d\mu(x) = \|f\|_{L^1}. \end{align*} Combining the inequalities, \begin{align*} \alpha\,\mu(E_\alpha) \leq \|f\|_{L^1}. \end{align*} Because $\alpha>0$, division by $\alpha$ yields \begin{align*} \mu\!\left(\left\{x \in X : M^+f(x)>\alpha\right\}\right) = \mu(E_\alpha) \leq \frac{1}{\alpha}\|f\|_{L^1}. \end{align*} This is the desired weak-type $(1,1)$ maximal inequality. [/step]