I am trying to prove the expected utility theorem with three outcomes. The expected utility with $n$ outcomes is rather cumbersome and long in the economics textbook Mas-Colell. But I was hoping that the proof with three outcomes is shorter, however, I am having some difficulty proving it.
Suppose I have three lotteries $x \succeq y \succeq z$. We can think of $x$ as the "best" lottery and $z$ as the "worst". The can set $u(x)=1$ and $u(z)=0$ and then $u(y)=p$ where $p$ is the probability at which a gamble with a $p$ chance of $x$ and a $1-p$ chance of $z$ is indifferent to $y$.
How do I proceed from here to prove the Expected utility theorem?
For $n$ outcomes, the EU states that given $\succeq $ satisfies independence and continuity axioms, there is a utility function $ u:Z\rightarrow \mathbb{R}$ such that if $$p\succeq q\Leftrightarrow \sum_{i=1}^{n}p_{i}u(z_{i})\geq \sum_{i=1}^{n}q_{i}u(z_{i}).$$
Edit: Letting $u(x)= 1$ and $u(z) = 0$ linearly scales the utility function. So we want to show $$u(y) = u\left ( px + (1-p)z \right ) = pu(x) + (1-p)u(z) = p \text{ by the independence axiom}.$$
Since we have assumed $x \succeq y \succeq z$ then it follows that $u(x) \geq u(y) \geq u(z)$, where $p \in [0,1]$. Does this make sense?
I am not sure about the part: $u\left ( px + (1-p)z \right ) = pu(x) + (1-p)u(z)$. Is this a legitimate way of writing the proof for the case of three outcomes?