I want to prove that all Bayes' solutions (those of maximum expected utility) are admissible, meaning that there's no other decision that dominates it.
By definition, a decision $d_a$ dominates $d_b$ if
(i) for all uncertain events, $d_b$ is not prefered by $d_a$
and
(ii) there exists an uncertain event where $d_a$ is strictly prefered to $d_b$.
Let's suppose that we have a finite number of decisions $d_1,...,d_n$ and that each decision has a finite number of consecuences.
Suppose first that each decision has the same uncertain events (each has a strictly positive probability of ocurrance). Let's say that Bayes' solution is $d_k$. Then
$$\sum_{i=1}^n \mathbb{P}(E_{ji})u(c_{ji})<\sum_{i=1}^n \mathbb{P}(E_{ki})u(c_{ki}).$$
We can drop the second index of $E_{ji}$ since all decisions have the same uncertain events, so $E_{ji}=E_j$
Suppose that $d_k$ is dominated by $d_j$. So
(i) for all uncertain events, $d_k$ is not prefered to $d_j$, so $u(c_{ki})\leq u(c_{ji})$ and
(ii) there exists an uncertain event, say $E_h$, such that $u(c_{kh})< u(c_{jh})$
Therefore,
$$\mathbb{P}(E_i)u(c_{ki})\leq \mathbb{P}(E_i)u(c_{ji})$$ and
$$\mathbb{P}(E_h)u(c_{kh})< \mathbb{P}(E_h)u(c_{jh})$$
so
$\sum_{i=1}^n \mathbb{P}(E_{ki})u(c_{ki})<\sum_{i=1}^n \mathbb{P}(E_{ji})u(c_{ji})$. But this is not possible, since $d_k$ is a Bayes' solution.
The problem that I have is the case where the decisions don't have the same number of uncertain events, because in that case we can't drop the second index and must consider $E_{ji}$.
How could I prove that all Bayes' solutions are admissable solutions in that case?