What exactly is an elementary outcome corresponding to a sample path of a stochastic process? Suppose a discrete stochastic process $X_n(\omega)$ is defined on the common probability space $(\Omega,P,\mathcal{F})$ and only takes $N$ steps. The state space of each $X_n$ has $K$ elements. Then we at most have $K^N$ distinct sample paths. Since each $\omega$ in the space $\Omega$ of elementary outcomes uniquely corresponds to a sample path, there will be $K^N$ different $\omega$ in $\Omega$, i.e., $|\Omega|=K^N$. But If we add another step to the process, $N\rightarrow N+1$, then using the same logic, $|\Omega|=K^{N+1}$. This is what becomes confusing to me, why $\Omega$ is dependent on the number of steps of the process?
Also, each $X_n(\omega)$ is a measurable function on $\Omega$, assuming they are independent of each other, then adding the extra $X_{N+1}(\omega)$ shouldn't change the measurability and the law of any preceding random variables. But we do see that $\Omega$ is expanded to accommodate the presence of $X_{N+1}(\omega)$, then how do we guarantee that $X_n$ ($n\le N$) is still measurable on the expanded $\Omega$? And since $\Omega$ is expanded, these $X_n$ are  now defined on a new space so they are not the same $X_n$ as before. Can we still say they are the same random variables?
 A: 
"Since each $\omega$ in the space $\Omega$ of elementary outcomes uniquely corresponds to a sample path, there will be $K^N$ different $\omega$ in $\Omega$, i.e., $|\Omega|=K^N$..."

Preassumed that every path can be followed it is correct to state that there will be at least $K^N$ different outcomes in $\Omega$ but it is wrong to state that will be exactly $K^N$ different outcomes in $\Omega$.  This in the first place because the stochastic process will at most provide some constraints on $\Omega$ but is certainly not determining for $\Omega$. On the contrary, I would say. By modeling the situation we have an enormous freedom when it comes to the construction of a suitable probability space.
As an example (already mentioned in the comment of Michael): if we throw just once a fair coin then we can choose for $\Omega=\{H,T\}$ equipped with $\wp(\Omega)$ as $\sigma$-algebra and a probability measure determined by $P(\{H\})=0.5$. 
But nothing stops us from choosing e.g. $\Omega=\mathbb R^{\mathbb 3}$ equipped with $\sigma$-algebra $\mathcal B(\mathbb R^3)$ together with random variable $X:\Omega\to\mathbb R$ prescribed by $(\omega_1,\omega_2,\omega_3)\mapsto\omega_1$. If we interpret a negative result as Tails and a non-negative result as Heads then concerning the probability measure $P$ on $(\mathbb R^3,\mathcal B(\mathbb R^3))$ it is enough to demand that $P\left(X^{-1}([0,\infty))\right)=P([0,\infty)\times\mathbb R\times\mathbb R)=0.5$. Note that without encountering any problems we can go for $3$ fair coin flips by prescribing $Y,Z:\Omega\to\mathbb R$ by $\omega\mapsto\omega_2$ and $\omega\mapsto\omega_3$ respectively, and extra demands $P(\mathbb R\times[0,\infty)\times\mathbb R)=P(\mathbb R\times\mathbb R\times[0,\infty))=0.5$.
So if you are dealing with a discrete stochastic process having $N$ steps then just think of an underlying probability space that makes it possible to do "extra" steps. Actually "expansion of probability space" is rarely needed because we just start with one that has all the benefits we want.  
Also have a look at this question and its answer.
