Can a time series be stationary if the formula for the mean level depends on $t$? I just started working on time series with the book from Brockwell and Davis. I'm still not that familiar with stationary time series. The book says that a time series is stationary if:
1 $\mathbb{E}[X_t] = \mu$ is independent of $t$.
2 $\operatorname{Cov}(X_{t+h},X_t)$ is independent of $t$ for each $h$.
So if I have the following time series:
$X_t=a+bt+Z_t$ with {${Z_t}$} ~ $WN(0,\sigma^2)$
Can you simply say that ${X_t}$ is not a stationary time series because:
$\mu_{x_t} = \mathbb{E}[a+bt+Z_t]=a+\mathbb{E}[bt + 0]=a+b \mathbb{E}[t]=a+ bt$, and this obviously is dependent on $t$?
Edit: Forget to say that $t=1,...,n$ in my time series.
 A: The definition that you claim is from Brockwell and Davis is actually the definition of weakly stationary or wide-sense-stationary time series and not that of stationary series (also called strictly stationary) time series. Strictly stationary series (with finite second moment)
are a subclass
of weakly stationary series and thus also enjoy properties 1. and 2.
(See also this answer of mine on dsp.SE; for the sake of keeping matters as simple as possible,
I had not mentioned finite second moments in the earlier version
of my answer. Also,
those getting ready to nitpick that it is also necessary to assume
that the first moment is finite should bear in mind that $|x| < 1+x^2$ for all $x\in\mathbb R$ and so
$E[|X|] < 1+E[X^2]$ from which it follows that $E[X]$ is finite whenever
$E[X^2]$ is assumed to be finite.
The two conditions that you state make the series strictly stationary 
if we  assume (in addition to 1. and 2.) that all the $X_i$ are jointly normal random variables.
So, yes, the series that you propose is not weakly stationary because
the mean depends on $t$, and so it cannot be strictly stationary
either.
A: There are several different flavours of stationarity. The type described in your definition is weak-sense stationarity, also known as wide-sense stationarity, covariance stationarity, or second-order stationarity.
Your definition is not quite complete: a preliminary condition for weak stationarity is that the mean and covariance must exist and be finite, but this is satisfied here. As you noted, weak stationarity further requires the mean to be constant over time and $\operatorname{Cov}(X_{t+h},X_t)$ to be independent of $t$ for each $h$ i.e. the autocovariance at each lag $h$ is constant over time. The fact that $\mathbb{E}(t)=a+bt$ shows the first of these conditions is not met. Even the mean is not stationary. 
Another form of stationarity is strong stationarity, also called strict stationarity or just stationarity. This requires the joint distribution function of the joint distribution of $X_t$ taken at any $k$ times $t_1, t_2, \dots , t_k$ is the same when lagged by any $\tau$. Technically, for any $k$ and any $\tau$, and for any $t_1, t_2, \dots, t_k$ we require
$$F_X(x_{t_1+\tau}, \dots, x_{t_k+\tau}) = F_X(x_{t_1}, \dots, x_{t_k})$$
This does not just imply that the mean and covariance at any given lag (if either exists) must stay constant over time, but that every conceivable property one can derive from the distribution is invariant under a time shift, in which sense its conditions are "stricter" (though note that if mean or covariance are not finite, we can have a strongly stationary series which does not fulfil the preliminary condition for weak stationarity). Since your mean is not constant over time, then your process can't be strongly stationary either. In general, so long as the mean and covariance exist, we can conclude that a process that is not weakly stationary will not be strongly stationary either. The converse is not true: just because a process is not strongly stationary, doesn't mean it can't be weakly stationary, since it is possible to be weakly stationary yet not strongly stationary.
There is a sense in which your $X_t$ is "stationary": it is trend stationary.
This means that the trend in your time series can be expressed as a function of $t$; if we strip this trend out then what we are left with is a stationary process. In particular, $X_t$ is trend stationary if we can express it as
$$X_t = f(t) + Y_t$$
where $f$ is a deterministic function of time $t$ and $\{Y_t\}$ is a stationary process. In your case we can take $f(t)=a + bt$ and $Y_t = Z_t$; the idea is that by "stripping out" the trend $f(t)$ we would have $X_t - (a + bt) = Z_t$ which is stationary (because it's white noise).
Alternatively we could have taken $f(t)=bt$ and $Y_t= a + Z_t$.
Note that although in this case we could take a process that didn't have a constant mean, then can subtract a deterministic trend to obtain a stationary process, doesn't mean we can do this on any such process. Consider a random walk with drift, for instance. Here the mean is not constant over time, but we can't simply subtract a deterministic trend to obtain a stationary process. So a random walk with drift would not be a trend stationary process. See the question "Difference between series with drift and series with trend" for an illustration.
