Ergodic flow has a.e. dense orbits? Clearly this is not true in general (but it should be for some nice spaces). In this case I am interested in the geodesic flow on the modular surface $\mathrm{SL}(2,\mathbb{R})/\mathrm{SL}(2,\mathbb{Z})$. It is known to be ergodic because this surface has finite volume. While reading some lecture notes, I came across the following statement ($g_t$ denotes the geodesic flow, and $z$ is an element of the surface).

Since the flow $g_t$ is ergodic, it follows from the pointwise ergodic theorem that for almost every $z$, the orbit $\left\{z.g_t \right\}$ is dense in $\mathrm{SL}(2,\mathbb{R})/\mathrm{SL}(2,\mathbb{Z})$.

I assume that by "pointwise ergodic theorem" the author means the standard Birkhoff ergodic theorem. But how do we use it to show that the orbits are dense a.e. ?
Thanks in advance. I can provide more details on demand. 
 A: There is a very simple proof here (or in almost every introductory dynamical systems book) that makes no use of the Birkhoff ergodic theorem. This looks like the best way to go since the proof of the Birkhoff theorem is not so easy. But if we really want to we can also cook up a proof that uses the Birkhoff theorem. It looks very convoluted to me (to say the least), but maybe it is not entirely pointless.
Let $T: X \to X$ be ergodic for some probability measure $\mu$ on the Borel sigma algebra of $X$ which is positive on (non-empty) open sets. Assume also that the topology of $X$ is generated by a countable basis $\mathcal{B} = \{ U_1, U_2, \ldots\}$.
Suppose that $x \in X$ does not have a dense orbit. Then there is some basis open set $U_i$ which does not intersect the orbit of $x$. Then the space average of the indicator function of $U_i$
\begin{equation}
\int_X \chi_{U_i} \; d\mu = \mu(U_i)
\end{equation}
is strictly positive, whereas the time average at $x$
\begin{equation}
\lim_{n\to \infty} \quad \frac{1}{n}\sum_{k=0}^{n-1} \,\chi_{U_i}( T^{k}x) =0
\end{equation}
since the orbit of $x$ does not meet $U_i$. Hence $x$ lies in the set $E_i$ of points for which the space average of $\chi_{U_i}$ does not equal the time average. 
This argument shows that the set of points which do not have a dense orbit is included in the union
\begin{equation}
\bigcup_{i=1}^\infty E_i.
\end{equation} 
But the Birkhoff ergodic theorem implies that each $E_i$ has measure $0$ for an ergodic map, which proves the claim.
