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There is a simple reduction of the Birkhoff ergodic Theorem for $L^1$ functions to the bounded case using Kakutani-Rokhlin towers, that I learned from H. Furstenberg and B. Weiss decades ago. Given $f \in L^1(X)$ we may assume it is nonnegative, and then (by subtracting the fractional part and adding 1) that it takes values in the positive integers. Consider the subset $$Y=\{(x,k): 1 \le k \le f(x) \}$$ of $X \times {\mathbb N}$, endowed with the product $\sigma$-algebra, the probability measure $\nu$ determined by $$\nu(A \times{k})=\frac{\mu(A)}{\int_X f \, d\mu}$$ for measurable sets $A \subset X$ where $\min_A f \ge k$, and the transformation $S$ defined by $S(x,k)=(x,k+1)$ if $k<f(x)$ and $S(x,k)=(T(x),1)$ if $k=f(x)$.

Then the ergodic theorem in $(Y,\nu,S)$ for the indicator function $$h(x,k):={\mathbf 1}_{\{k=1\}}$$ (along the sequence of return times to the base $\{k=1\}$ of the tower) implies the ergodic theorem for $f$ in $(X,\mu,T)$.

Indeed, consider the Birkhoff sums $$R_n(x):=\sum_{j=0}^{n-1}f(T^jx) \,.$$ For each $x \in X$, we have $$\{\ell \ge 0 : h(S^\ell(x,1))=1 \}= \{R_j(x) \; : \: j \ge 0\} \,$$ whence for $\mu$-almost every $x \in X$, we conclude that $$\lim_{n \to \infty} \frac{n}{R_n(x)}= \lim_{n \to \infty} \frac{ \sum_{\ell=0}^{R_n(x)-1}h(S^\ell(x,a)) }{R_n(x)} =\int_Y h \, d\nu =\frac{\mu(X)}{\int_X f \, d\mu} =\frac{1}{\int_X f \, d\mu}\,.$$

There is a simple reduction of the Birkhoff ergodic Theorem for $L^1$ functions to the bounded case using Kakutani-Rokhlin towers, that I learned from H. Furstenberg and B. Weiss decades ago. We use the notation of the post, and assume that $T$ is ergodic. Given $f \in L^1(X)$ we may assume it is nonnegative, and then (by subtracting the fractional part and adding 1) that it takes values in the positive integers. Consider the subset $$Y=\{(x,k): 1 \le k \le f(x) \}$$ of $X \times {\mathbb N}$, endowed with the product $\sigma$-algebra, the probability measure $\nu$ determined by $$\nu(A \times{k})=\frac{\mu(A)}{\int_X f \, d\mu}$$ for measurable sets $A \subset X$ where $\min_A f \ge k$, and the transformation $S$ defined by $S(x,k)=(x,k+1)$ if $k<f(x)$ and $S(x,k)=(T(x),1)$ if $k=f(x)$.

Then $S$ is ergodic on $(Y,\nu)$, since if a measurable set $E \subset Y$ is invariant under $S$, then $E_1:=\{x \in X \,: (x,1) \in E\}$ is invariant under $T$.

Claim: The ergodic theorem in $(Y,\nu,S)$ for the indicator function $$h(x,k):={\mathbf 1}_{\{k=1\}}$$ (along the sequence of return times to the base $\{k=1\}$ of the tower) implies the ergodic theorem for $f$ in $(X,\mu,T)$.

Proof: consider the Birkhoff sums $$R_n(x):=\sum_{j=0}^{n-1}f(T^jx) \,.$$ For each $x \in X$, we have $$\{\ell \ge 0 : h(S^\ell(x,1))=1 \}= \{R_j(x) \; : \: j \ge 0\} \,$$ whence we conclude that for $\mu$-almost every $x \in X$, $$\lim_{n \to \infty} \frac{n}{R_n(x)}= \lim_{n \to \infty} \frac{ \sum_{\ell=0}^{R_n(x)-1}h(S^\ell(x,a)) }{R_n(x)} =\int_Y h \, d\nu =\frac{\mu(X)}{\int_X f \, d\mu} =\frac{1}{\int_X f \, d\mu}\,.$$

There is a simple reduction of the Birkhoff ergodic Theorem for $L^1$ functions to the bounded case using Kakutani-Rokhlin towers, that I learned from H. Furstenberg and B. Weiss decades ago. Given $f \in L^1(X)$ we may assume it is nonnegative, and then (by subtracting the fractional part and adding 1) that it takes values in the positive integers. Consider the subset $$Y=\{(x,k): 1 \le k \le f(x) \}$$ of $X \times {\mathbb N}$, endowed with the product $\sigma$-algebra, the probability measure $\nu$ determined by $$\nu(A \times{k})=\frac{\mu(A)}{\int_X f \, d\mu}$$ for measurable sets $A \subset X$ where $\min_A f \ge k$, and the transformation $S$ defined by $S(x,k)=(x,k+1)$ if $k<f(x)$ and $S(x,k)=(T(x),1)$ if $k=f(x)$.

Then the ergodic theorem for the indicator function $h(x,k):={\mathbf 1}_{\{k=1\}}$ in $(Y,\nu,S)$ (along the sequence of return times to the base $\{k=1\}$ of the tower) implies the ergodic theorem for $f$ in $(X,\mu,T)$.

Indeed, consider the Birkhoff sums $$R_n(x):=\sum_{j=0}^{n-1}f(T^jx) \,.$$ For each $x \in X$, we have $$\{\ell \ge 0 : h(S^\ell(x,1))=1 \}= \{R_j(x) \; : \: j \ge 0\} \,$$ whence $$\lim_{n \to \infty} \frac{n}{R_n(x)}= \lim_{n \to \infty} \frac{ \sum_{\ell=0}^{R_n(x)-1}h(S^\ell(x,a)) }{R_n(x)}$$

There is a simple reduction of the Birkhoff ergodic Theorem for $L^1$ functions to the bounded case using Kakutani-Rokhlin towers, that I learned from H. Furstenberg and B. Weiss decades ago. Given $f \in L^1(X)$ we may assume it is nonnegative, and then (by subtracting the fractional part and adding 1) that it takes values in the positive integers. Consider the subset $$Y=\{(x,k): 1 \le k \le f(x) \}$$ of $X \times {\mathbb N}$, endowed with the product $\sigma$-algebra, the probability measure $\nu$ determined by $$\nu(A \times{k})=\frac{\mu(A)}{\int_X f \, d\mu}$$ for measurable sets $A \subset X$ where $\min_A f \ge k$, and the transformation $S$ defined by $S(x,k)=(x,k+1)$ if $k<f(x)$ and $S(x,k)=(T(x),1)$ if $k=f(x)$.

Then the ergodic theorem in $(Y,\nu,S)$ for the indicator function $$h(x,k):={\mathbf 1}_{\{k=1\}}$$ (along the sequence of return times to the base $\{k=1\}$ of the tower) implies the ergodic theorem for $f$ in $(X,\mu,T)$.

Indeed, consider the Birkhoff sums $$R_n(x):=\sum_{j=0}^{n-1}f(T^jx) \,.$$ For each $x \in X$, we have $$\{\ell \ge 0 : h(S^\ell(x,1))=1 \}= \{R_j(x) \; : \: j \ge 0\} \,$$ whence for $\mu$-almost every $x \in X$, we conclude that $$\lim_{n \to \infty} \frac{n}{R_n(x)}= \lim_{n \to \infty} \frac{ \sum_{\ell=0}^{R_n(x)-1}h(S^\ell(x,a)) }{R_n(x)} =\int_Y h \, d\nu =\frac{\mu(X)}{\int_X f \, d\mu} =\frac{1}{\int_X f \, d\mu}\,.$$

There is a simple reduction of the Birkhoff ergodic Theorem for $L^1$ functions to the bounded case using Kakutani-Rokhlin towers, that I learned from H. Furstenberg and B. Weiss decades ago. Given $f \in L^1(X)$ we may assume it is nonnegative, and then (by subtracting the fractional part and adding 1) that it takes values in the positive integers. Consider the subset $$Y=\{(x,k): 1 \le k \le f(x) \}$$ of $X \times {\mathbb N}$, endowed with the product $\sigma$-algebra, the probability measure $\nu$ determined by $$\nu(A \times{k})=\frac{\mu(A)}{\int_X f \, d\mu}$$ for measurable sets $A \subset X$ where $\min_A f \ge k$, and the transformation $S$ defined by $S(x,k)=(x,k+1)$ if $k<f(x)$ and $S(x,k)=(T(x),1)$ if $k=f(x)$.

Then the ergodic theorem for $g(x,k):={\mathbf 1}_{\{k=1\}}$ in $(Y,\nu,S)$ (along the sequence of return times to the base $\{k=1\}$ of the tower) implies the ergodic theorem for $f$ in $(X,\mu,T)$.

There is a simple reduction of the Birkhoff ergodic Theorem for $L^1$ functions to the bounded case using Kakutani-Rokhlin towers, that I learned from H. Furstenberg and B. Weiss decades ago. Given $f \in L^1(X)$ we may assume it is nonnegative, and then (by subtracting the fractional part and adding 1) that it takes values in the positive integers. Consider the subset $$Y=\{(x,k): 1 \le k \le f(x) \}$$ of $X \times {\mathbb N}$, endowed with the product $\sigma$-algebra, the probability measure $\nu$ determined by $$\nu(A \times{k})=\frac{\mu(A)}{\int_X f \, d\mu}$$ for measurable sets $A \subset X$ where $\min_A f \ge k$, and the transformation $S$ defined by $S(x,k)=(x,k+1)$ if $k<f(x)$ and $S(x,k)=(T(x),1)$ if $k=f(x)$.

Then the ergodic theorem for the indicator function $h(x,k):={\mathbf 1}_{\{k=1\}}$ in $(Y,\nu,S)$ (along the sequence of return times to the base $\{k=1\}$ of the tower) implies the ergodic theorem for $f$ in $(X,\mu,T)$.

Indeed, consider the Birkhoff sums $$R_n(x):=\sum_{j=0}^{n-1}f(T^jx) \,.$$ For each $x \in X$, we have $$\{\ell \ge 0 : h(S^\ell(x,1))=1 \}= \{R_j(x) \; : \: j \ge 0\} \,$$ whence $$\lim_{n \to \infty} \frac{n}{R_n(x)}= \lim_{n \to \infty} \frac{ \sum_{\ell=0}^{R_n(x)-1}h(S^\ell(x,a)) }{R_n(x)}$$