We outline the proof of Birkhoff's ergodic theorem, emphasizing key ideas over technical details.

Theorem Statement: For a measure-preserving $(X, \mu, T)$ and $f \in L^1(X, \mu)$, the limit $\lim_{n \to \infty} \frac{1}{n}\sum_{k=0}^{n-1} f(T^k(x)) = f^*(x)$ exists almost everywhere, with $f^*$ invariant and $\int f^* d\mu = \int f d\mu$.

Step 1: Maximal ergodic lemma

Define the maximal function:

$f^+(x) = \sup_{n \geq 1} \frac{1}{n} \sum_{k=0}^{n-1} f(T^k(x))$

The maximal ergodic lemma states: for the set $E = \{x : f^+(x) > 0\}$:

$\int_E f \, d\mu \geq 0$

Proof of lemma: Define $S_n = \sum_{k=0}^{n-1} f \circ T^k$ and note that:

$\max_{1 \leq k \leq n} S_k(x) = \max\{f(x), f(x) + \max_{1 \leq k \leq n-1} S_k(T(x))\}$

Using this recursion and measure-preservation, one shows $\int_E f \, d\mu \geq 0$ through careful estimation. This lemma is the technical heart of the proof.

Step 2: Limsup and liminf

Define:

$\overline{f}(x) = \limsup_{n \to \infty} \frac{1}{n}\sum_{k=0}^{n-1} f(T^k(x))$ $\underline{f}(x) = \liminf_{n \to \infty} \frac{1}{n}\sum_{k=0}^{n-1} f(T^k(x))$

Both $\overline{f}$ and $\underline{f}$ are $T$-invariant (if $T$ moves $x$ to $T(x)$, the time average is the same, just shifted).

Step 3: Showing $\overline{f} = \underline{f}$ almost everywhere

For any $c \in \mathbb{R}$, apply the maximal lemma to $f - c$. The set where $\overline{f} > c$ has:

$\int_{\{\overline{f} > c\}} (f - c) \, d\mu \geq 0$

Similarly, for $\underline{f} < c$:

$\int_{\{\underline{f} < c\}} (f - c) \, d\mu \leq 0$

If $\mu(\{\overline{f} > \underline{f}\}) > 0$, choose $c$ between $\overline{f}$ and $\underline{f}$ on this set (by density of rationals). This yields:

$\int_{\{\overline{f} > c > \underline{f}\}} (f - c) \, d\mu \geq 0 \text{ and } \leq 0$

implying $\mu(\{\overline{f} > c > \underline{f}\}) = 0$. Taking countable union over rationals: $\mu(\{\overline{f} > \underline{f}\}) = 0$.

Thus $\overline{f} = \underline{f} =: f^*$ almost everywhere, so the limit exists.

Step 4: Invariance and integral preservation

$f^*$ is invariant: $f^* \circ T = f^*$ by construction (time averages are shift-invariant).

For integral preservation, use the dominated convergence theorem (or monotone convergence with truncations):

$\int f^* \, d\mu = \int \lim_{n \to \infty} \frac{1}{n}\sum_{k=0}^{n-1} f \circ T^k \, d\mu = \lim_{n \to \infty} \frac{1}{n}\sum_{k=0}^{n-1} \int f \circ T^k \, d\mu$

Since $T$ preserves measure, $\int f \circ T^k d\mu = \int f d\mu$, yielding:

$\int f^* \, d\mu = \lim_{n \to \infty} \frac{1}{n} \cdot n \int f \, d\mu = \int f \, d\mu$

Conclusion: The time average $f^*$ exists almost everywhere, is invariant, and has the same integral as $f$.