The answer to both questions is 'no', both for maps and for flows.

For concreteness let $M=\{0,1\}^\mathbb{Z}$ be the set of bi-infinite sequences of $0$s and $1$s, and let $\Phi\colon M\to M$ be the shift map given by $\Phi(x)_j = x_{j+1}$ for $x=(x_j)_{j\in\mathbb{Z}}$.

Q1. Topological transitivity of $\Phi$ only depends on $\Phi$ and $M$, not on the measure $\mu$. In particular the system $(M,\Phi)$ defined above is topologically transitive, but there are many (many!) regular Borel probability measures that are preserved by $\Phi$, and not all of them are ergodic. See this question for some discussion of how intricate this space is. In particular, let $p$ and $q$ be fixed points for $\Phi$, and let $\mu$ be the atomic measure that gives weight $\frac 12$ to each of $p$ and $q$. Then $\mu$ is $\Phi$-invariant but not ergodic.

Q2. The pointwise time averages do not need to exist for every $x$. In fact it is quite typical that they do not exist. Let me make this last statement a little more precise, again using the example of $(M,\Phi)$ from above.

Consider the continuous real valued function $f\colon M\to \mathbb{R}$ defined by $f(x) = x_0$. That is, $f$ is simply the value of the symbol in the $0$ position in the sequence $x$. Then $a_N(x) := \frac 1N \sum_{j=1}^N f(\Phi^j(x))$ is the frequency of the symbol $1$ in the string $x_1 x_2 \cdots x_N$.

The pointwise time averages of $f$ along the orbit of $x$ exist if and only if $a_N(x)$ converges as $N\to \infty$ -- in other words, if and only if the lower and upper asymptotic frequencies of the symbol $1$ are equal. It is straightforward to construct examples of sequences $x\in M$ such that the lower and upper asymptotic frequencies disagree and the limit does not exist.

In fact, one can say some more about how large the set of such points are. Given $x\in M$, let $\lambda(x) = \liminf a_N(x)$ and $\Lambda(x) = \limsup a_N(x)$. Note that $0\leq \lambda(x)\leq \Lambda(x)\leq 1$ for all $x\in M$. Given $0\leq r\leq s\leq 1$, let $K_{r,s}$ be the set of $x\in M$ such that $\lambda(x) = r$ and $\Lambda(x) = s$. The study of the various sets $K_{r,s}$ is called multifractal analysis, and quite a lot is known. I'll state just a few results addressing your question.

Let $K^\neq = \bigcup_{r<s} K_{r,s}$ be the set of points for which $\lambda(x) \neq \Lambda(x)$, so that the limit doesn't exist. Then the following are true (at least for the system I described above -- determining for which general classes of systems these statements hold is a more subtle question):

- $K^\neq$ has zero measure for every $\Phi$-invariant measure.
- $K^\neq$ has Hausdorff dimension equal to the Hausdorff dimension of $M$. (The more honest way of saying this is that they have equal topological entropies, but Hausdorff dimension is a more familiar concept and the statement with dimension is true if you use an appropriate metric.)
- $K^\neq$ is residual -- that is, it is a countable intersection of open and dense subsets of $M$.
- In fact, one can show that $K_{0,1}$ is residual, but that it has Hausdorff dimension $0$. (The fact that $K^\neq$ has full Hausdorff dimension is due to the fact that the Hausdorff dimension of $K_{r,s}$ approaches the Hausdorff dimension of $M$ as $r,s\to \frac 12$.)

So there's an assortment of facts for you illustrating how large the set of points is where convergence fails. In particular the last fact can be interpreted as saying that from a topological point of view, for a generic point $x$ the limit fails to exist as strongly as it can possibly fail. This highlights the fact that ergodic theory is really about measures, not topology. (I will note that the limit exists everywhere if your map is *uniquely* ergodic, that is, if there is only one invariant probability measure. Such systems are quite different from the systems I was describing, which should be thought of as hyperbolic, or informally, chaotic.)

anyinvariant measures. If one of them is fully-supported then $\mu=a\nu_1 + (1-a)\nu_2$ is fully-supported but not ergodic for any $0<a<1$. – Vaughn Climenhaga Nov 22 '12 at 2:28`yes' and can be proved by point- set topological arguments. Pf that Erg. implies top. transitive. By contraposition. Say $\Phi$ is not top. transitive. Then there exist open U, V which are disj., nonempty, and invariant. Since$\mu(U) > 0$ and $\mu(V) > 0$ this proves $\Phi$ not ergodic.`

Fat Cantor sets' [ sets of pos. meas having empty interior] make Top. trans. imply ergodic harder. I warrant it is true though. – Richard Montgomery Nov 22 '12 at 3:22