Finite-measure identification of the Birkhoff limit

First let $g$ be real and integrable, and choose its Birkhoff-limit representative $g^{\ast }$ to be the pointwise limit on the convergence set and zero elsewhere. The shifted-average identity and its rearrangement show that the convergence set is strictly invariant and that $g^{\ast }$ is strictly invariant.

Suppose first that $f$ is real. On $(X,\mathcal{I})$ define $\nu (E)={\int }_{E}fd\mu$. This is a finite signed measure by [F3], is absolutely continuous with respect to $\mu {\mathrm{\mid }}_{\mathcal{I}}$, and has finite total variation. By [F4] there is an integrable $\mathcal{I}$-measurable $h$ such that ${\int }_{E}h={\int }_{E}f$ for every $E\in \mathcal{I}$. This is the unique step spending AC.

Fix a positive integer $m$ and, for $k\in \mathbb{Z}$, put $D_{m,k}=\{k\mathrm{/}m\le g^{\ast }<(k+1)\mathrm{/}m\}\mathrm{.}$ These form a measurable, strict-invariant partition of $X$. For $\epsilon >0$, every point of $D_{m,k}$ belongs to the positive-maximal set of $(g-(k\mathrm{/}m-\epsilon )){\mathbf{1}}_{D_{m,k}}$. The maximal theorem and strict invariance therefore give ${\int }_{D_{m,k}}gd\mu \ge (k\mathrm{/}m-\epsilon )\mu (D_{m,k})\mathrm{.}$ Since $\mu (D_{m,k})<\mathrm{\infty }$, letting $\epsilon \downarrow 0$ gives the same inequality with $k\mathrm{/}m$.

On $D_{m,k}$ one has $g^{\ast }<(k+1)\mathrm{/}m$, so step 2.1 gives ${\int }_{D_{m,k}}{g}^{\ast }d\mu \le {\int }_{D_{m,k}}gd\mu +\frac{1}{m}\mu (D_{m,k})\mathrm{.}$ Countable additivity over the partition is legitimate because $g$ and $g^{\ast }$ are integrable. Summing yields ${\int }_X{g}^{\ast }d\mu \le {\int }_{X}gd\mu +\frac{\mu (X)}{m}\mathrm{.}$ Letting $m\to \mathrm{\infty }$ and then applying the same inequality to $-g$, whose limit is $-g^{\ast }$, proves $\int g^{\ast }=\int g$.

Let $E\in \mathcal{I}$ and take $g=f{\mathbf{1}}_E$. Strict invariance gives $A_{n}g=(A_{n}f){\mathbf{1}}_E$ pointwise, so the Birkhoff limit of $g$ is $f^{\ast }{\mathbf{1}}_E$ a.e. In the real case, step 3.1 therefore gives ${\int }_E{f}^{\ast }d\mu ={\int }_{E}fd\mu \mathrm{.}$

Since $h$ is $\mathcal{I}$-measurable, every rational sublevel set of $h$ is strictly invariant; rational separation therefore gives $h\circ T=h$ pointwise. Thus, because $f^{\ast }$ is invariant almost everywhere, $B_r=\{f^{\ast }-h>r\}$ is invariant modulo null sets for every rational $r>0$. Let $E_r\in \mathcal{I}$ be its strict representative from [F2]. Steps 4.1 and 1.2 give ${\int }_{E_r}(f^{\ast }-h)=0$, while $f^{\ast }-h>r$ a.e. on $E_r$. Monotonicity implies $0\ge r\mu (E_r)$, so $E_r$ is null. Applying the same argument to $h-f^{\ast }$ and taking the countable union over positive rational $r$ proves $f^{\ast }=h$ a.e.

For complex $f$, apply steps 1.2–5.1 to its real and imaginary parts and set $h=h_1+i{h}_2$. This $h$ is integrable and $\mathcal{I}$-measurable, equals $f^{\ast }$ a.e., and has all asserted event-integral identities. Uniqueness follows componentwise from Radon–Nikodym uniqueness. If $\mu (X)=0$, the same proof gives the zero density and all assertions are vacuous off a null set.