Consider a single period market consisted of the zero-risk asset $S^1$ and two risky assets $S^2,S^3$ with states $$S^1_0 = S^2_0=S^3_0=1, P(S^1_1 = 1 + r, S^2_1 = S^3_1 = u) = \frac{1}{3}, P(S^1_1 = 1 + r, S^2_1 = 1, S^3_1 = d) = \frac{1}{3}, P(S^1_1 = 1 + r, S^2_1 = d, S^3_1 = 1) = \frac{1}{3}$$ with $0 < d < 1 + r < u, r\geq0$.
I need to show that this market is arbitrage-free and then find an EMM on a suitable probability space $(\Omega, \mathcal{A}, \mathbb{Q})$.
I know showing that a market is arbitrage-free is the same as showing that there exists an EMM but since it's the second task I want to use the definition.
A market has arbitrage if the following holds for a portfolio process $V$:
$$V_0 = 0, \quad \mathbb P(V_1 \geq 0) = 1, \quad \mathbb P(V_1 > 0) > 0$$
I want to create a contradiction to the (necessary) condition $0 < d < 1 + r < u$ but I don't know how to use the probabilities given above. And then: how do I find a risk-neutral measure?
It's one-step, so a portfolio has the form $V_i = aS^1_i + bS^2_i + cS^3_i$ for constants $a,b,c$.
Applying $V_0=0$ gives $0=a+b+c$, or $a=-b-c$, so that we may write $$V_i = b(S^2_i-S^1_i) + c(S^3_i-S^1_i).$$
You have three events with positive probabilities. For each one, we must have $V_1 \geq 0$, so $$b(u-(1+r))+c(u-(1+r)) \geq 0 \\ b(1-(1+r))+c(d-(1+r)) \geq 0 \\ b(d-(1+r))+c(1-(1+r)) \geq 0$$ but some of these values are known to be positive, giving us, resp., $$b+c \geq 0 \\ -r'b -c \geq 0 \\ -b - r'c \geq 0$$ where $r' = r / ((1+r)-d) \geq 0.$
Since we assume $\mathbb{P}(V_1 > 0) > 0$, at least one of the inequalities above is strict, so summing all the inequalities gives $-r'(b+c) > 0$, but as $r' \geq 0$ and $b+c \geq 0$, this is a contradiction.
To find an equivalent martingale measure, suppose $$\mathbb{Q}(S^1_1 = 1 + r, S^2_1 = S^3_1 = u) = p_1 > 0 \\ \mathbb{Q}(S^1_1 = 1 + r, S^2_1 = 1, S^3_1 = d) = p_2 > 0 \\ \mathbb{Q}(S^1_1 = 1 + r, S^2_1 = d, S^3_1 = 1) = p_3 > 0 \\ p_1+p_2+p_3 = 1$$
Then we take a discounted portfolio $$V_i = (1+r)^{-i}(aS_i^1 + bS_i^2 + cS_i^3)$$ and note that $$a+b+c = \mathbb{E}^\mathbb{Q}[V_1] = (1+r)^{-1}\left(a\mathbb{E}^\mathbb{Q}[S_1^1] + b\mathbb{E}^\mathbb{Q}[S_1^2] + c\mathbb{E}^\mathbb{Q}[S_1^3]\right)$$ implies that $\mathbb{E}^\mathbb{Q}[S_1^i] = 1+r$ for each $i \in \{1,2,3\}$.
That is, $$up_1 + p_2 + dp_3 = 1+r \\ up_1 + dp_2 + p_3 = 1+r \\ p_1 + p_2 + p_3 = 1$$ and solving this for $p_1,p_2,p_3$ in terms of $d,r,u$ gives $$p_1 = \frac{2r + 1 - d}{2u-d-1}, \qquad p_2=p_3 = \frac{u-(1+r)}{2u-d-1}$$