probably you already know this example, but in any case...
let $\Omega=\{0,1\}^{\mathbb{N}}$, $\mathcal{F}$ the $\sigma$-algebra generated by the cylinder sets of $\{0,1\}^{\mathbb{N}}$ and
$
\mathbb{P}=\prod_{i\in\mathbb{N}} \mu_i
$
is product measure with $\mu_i:\mathcal{P}(\{0,1\})\to [0,1]$ given by $\mu_i(\{0\})=\frac{1}{2}=\mu_i(\{1\})$ for all $i\in\mathbb{N}$.
We define $\varphi:\Omega\to\Omega$ in the following way:
We think about an element $\omega\in\Omega$ as an infinite sequence of zeros and one, that is, $\omega=(\omega_1,\omega_2,\ldots)$ and then put
$$
\varphi(\omega_1,\omega_2,\ldots)=(\omega_2,\omega_3,\ldots)
$$
this function is known as the left-shift.
You can prove that $\varphi$ is a measure-preserving map. The trick is prove that $\mathbb{P}(\varphi^{-1}\mathscr{C})=\mathbb{P}(\mathscr{C})$ for any cylinder set and then extend this to the whole $\sigma$-algebra using for example the extension measure theorems.
Aside comment: We can prove a more strong fact, this product measure is in fact ergodic for the shift, which implies that the invariant sets has measure zero or one.
To finish the example take $A=\Omega\setminus\{1,1,1,\ldots\}$ which is a measure one set. Note that $\varphi^{-1}A=\Omega$, once for any $\omega\in\Omega$ we have $\mathbb{P}(\{\omega\})=0$ (the proof of this statement is a consequence of the continuity of $\mathbb{P}$), follows that $\mathbb{P}(\varphi^{-1}A\Delta A)=0$.
There is generally no explicit way to write down how the elements of a generated $\sigma$-algebra look like in the way one can do with the generated algebra. There can be sets that can be constructed that need infinitely many steps to be constructed- and then some more. One can "construct" the generated $\sigma$-algebra in $\omega_1$ steps, with $\omega_1$ being the first uncountable ordinal, by transfinite induction.
Best Answer
The notions of ergodic measure and ergodic mapping are more or less interchangeable. If you already have a probability space $(\Omega, \mathcal{F}, \mathbb{P}),$ you would say that a measure-preserving transformation $T: X \rightarrow X$ is ergodic if $\mathbb{P}(A) \in \{0, 1\}$ for all $A \in \mathcal{F}$ for which $T^{-1}(A) = A$ (this is one of many equivalent definitions). You could also say in this scenario that $\mathbb{P}$ is ergodic (with respect to $T$). However, there are situations in which you are only given a mapping $T$ and are asked if there exists an ergodic measure with respect to $T$ (a very well known theorem shows that if $X$ is a complete metric space and $T$ is continuous, then indeed there exists an ergodic measure). So it really depends on the context, really. As for ergodic sequences, I believe that the wikipedia page on ergodic sequences is a good reference. I hope this helps. :)