1 sample space with 2 probability example?

I am a freshman here with the following question, sorry for my gramar I don't know the exact matematical probability terms in english.

So, if I have a sample space, can I define in this space 2 probability?

I know that, the probability function fit 3 condition.

1. $P(a) \ge 0$
2. $P(\Omega) = 1$
3. $P\left(\bigcup_{j\in J}A_j\right) =\sum_{j\in J}P(A_j)$

The following probabilities function fit the conditions: probability as geometrie, probability as counting, probability as algebra. Example: probability as counting P(event) = of successful outcomes / of total outcomes

So can I define a 2 function on the same sample space? If not why, how to prove it? If anyone can give me examples, I would be very grateful.

I want to understand this with 100% percent.

Thank you very much.

-

First, $P(\emptyset)=0$ and the hypothesis at 3. that the $A_j$ sets are pairwise disjoint are missing.
Yes, of course it is possible to define more probability measure on one probability space. The smallest example if $\Omega=\{a,b\}$ and \begin{align} P_1(\{a\}):=P_1(\{b\})&:= 1/2 \\ P_2(\{a\}):=1/3,\quad P_2(\{b\})&:=2/3 \end{align} Or, you can take $\Omega:=[0,1]^2$ and define $P_1$ as the usual Lebesgue measure (ie. area), and take an arbitrary integrable nonnegative function $f$ such that $\displaystyle\int_{[0,1]^2}f(x,y)dxdy=1$, and define for a measurable $A\subseteq [0,1]^2$: $$P_2(A):=\int_A f(x,y)dxdy$$
Moreover, if you have any random variable $\Omega\to X$ to a measure space $X$, then it induces a probability measure on $X$, hence every measurable function $\Omega\to\Omega$ induces such..
I don't think "it is possible to define more probability measure on one probability space" by Berci is very precise statement. Probability space is a tuple $(\Omega, \mathcal{F}, P)$. If you change $P$ keeping $\Omega$ same, it creates different probability spaces but with same sample space. In all fairness, I guess the intention of the question was to know if we can have different $P$ with same $\Omega$ and as Berci pointed out correctly that it is very much possible. You can have several coins each with different probability of head all producing same sample space. – BuckCherry Aug 7 '15 at 16:47