Definite values and Eigenstates

But if you are talking about the usual connection in quantum mechanics that we make between states and vectors and between observables and (self-adjoint) operators, then that includes also the condition that for the observable represented by operator $A$, the distribution of observed values, when it is measured in the state represented by vector $\psi$, is such that the probability of a measurement of $A$ being in any specified interval $L$ is given in terms of the spectral projector $E_L(A)$ by the inner product $\frac{⟨\psi |E_L(A)\psi ⟩}{⟨\psi |\psi ⟩}$ (or just $⟨\psi |E_L(A)\psi ⟩$ if $\psi$ is normalized with $⟨\psi |\psi ⟩=1$).

And if you are requiring that we decide to adopt that representation, then it does follow (essentially immediately from that extra assumption) that a state in which the observable has a definite value is represented by an eigenvector of the corresponding operator.

PS The spectral projectors $E_L(A)$ of any observable $A$ also correspond to question-type observables giving the value 1 when $A$ is observed to give a value in $L$ and 0 otherwise, and the expectation value of any question is just the probability of its value being 1 (ie of the answer being “true”). So the assumption made above actually follows from the simpler version which just says that the expectation values of observables are always given by inner products of the form $⟨\psi |A\psi ⟩$ where $\psi$ is a corresponding normalized vector and $A$ is the corresponding observable (which includes the case of spectral questions about other observables).