Birkhoff ergodic theorem implies Strong Law of Large Number Given $X_n:(\Omega,\mathbb{P}) \rightarrow \mathbb{R}$ a sequence of i.i.d. random variables, we define a path space $(\mathbb{R}^\mathbb{N} = \tilde\Omega, \mathcal P)$ where $X_n:\Omega \rightarrow \mathbb{R}$ and the projection maps $Y_n: \tilde \Omega\rightarrow \mathbb{R}$ where $Y_n(\tilde \omega) = \tilde \omega_n$ have the same law
$$\mathbb{P}(X_1\in B_1, \cdots, X_n \in B_n) = \mathcal{P}(Y_1\in B_1, \cdots Y_n\in B_n)$$
where $B_1, \cdots B_n$ are Borel sets of $\mathbb{R}$. 
Let $T$ be the shift operator on $\tilde \Omega$, it is measure preserving because $X_n$'s have the same distribution, it is ergodic from Kolmogorov $0$-$1$ law.
Apply ergodic theorem, we use $Y_1\in L^1(\tilde \Omega)$, 
$$\frac{1}{n}\sum_{k=1}^{n} Y_k\left(\tilde\omega\right)=\frac{1}{n}\sum_{k=0}^{n-1} Y_1\left(T^k\tilde \omega\right)  \rightarrow \mathcal E[Y_1] \quad \mathcal P \text{ - a.s.}$$
We know $Y_k$ and $X_k$ have the same distribution, and $\mathcal E[Y_1] = \mathbb{E}[X_1]$, but how do we get 
$$\frac{1}{n}\sum_{k=1}^{n} X_k\left(\omega\right)\rightarrow  \mathbb{E}[X_1]\quad \mathbb{P}\text{ -a.s.}$$ 
from 
$$ \frac{1}{n}\sum_{k=1}^{n} Y_k\left(\tilde\omega\right) \rightarrow \mathcal E[Y_1] \quad \mathcal P \text{ - a.s.}?$$

As Shalop said, it is due to the fact that $(X_k)_{k \in \Bbb N}$ has the same distribution as $(Y_k)_{k \in \Bbb N}$. Indeed, let $C\subset\mathbb R^{\mathbb N}$ be the set of the sequences converging to $0$ and let $C'$ be the set of sequences $\left(x_k\right)_{k\geqslant 1}$ such that $\left(n^{-1}\sum_{k=1}^nx_k\right)_{n\geqslant 1}\in C$. Then $C'$ is a Borel subset of $\mathbb R^{\mathbb N}$.
The assertion 
$$\frac{1}{n}\sum_{k=1}^{n} X_k\left(\omega\right)\rightarrow  \mathbb{E}[X_1]\quad \mathbb{P}\text{ -a.s.}$$
means that $\mathbb{P}\left( \left(X_k-\mathbb E\left[X_1\right]    \right)_{k\geqslant 1}\in C'\right)=1$. Using equality of laws, we derive that $\mathcal{P}\left( \left(Y_k-\mathbb E\left[X_1\right]    \right)_{k\geqslant 1}\in C'\right)=1$ and since $\mathbb E\left[X_1\right]=\mathcal E\left[Y_1\right]$, the wanted conclusion follows.
