Explanation of Dirac's proof of arbitrary ket being expressible with eigenkets of observable In P.A.M. Dirac's The Principles of Quantum Mechanics, Chapter 10 (Observables), pp. 40, at the end of the chapter there is a proof that I don't understand at all.
Here is a pdf link to the book readable online: http://www.fulviofrisone.com/attachments/article/447/Principles%20of%20Quantum%20Mechanics%20-%20Dirac.pdf
The proof in question is on pp. 50 using the pdf reader's numbering and on pp. 40 using the books original numbering. I'm curious about the part starting with "We can now see..." until the end of the chapter.
Would somebody be so kind to explain me what happens there? 
 A: Dirac being opaque and hard to follow? Well I never...
In Chapter 10 Dirac argues on physical grounds that the eigenkets of an observable must form a complete set. His argument goes that say we have an observable with eigenstates $|\varepsilon\rangle$ and some general state $|P\rangle$. Then I can write $|P\rangle$ as \begin{equation}|P\rangle = \sum a_\varepsilon|\varepsilon\rangle + \sum b_\gamma |\gamma\rangle \end{equation} 
Where $|\gamma\rangle$ are some state that cannot be written as a combination of the $|\varepsilon\rangle$s. Unlike Dirac I have normalized my eigenstates so that $\langle \varepsilon|\varepsilon\rangle = 1$ by pulling out a complex factor $a_\gamma$ and have not bothered to write down the integral, as the argument is essentially the same and it just complicates things. Now if we make a measurement of $\varepsilon$ for this state there is a probability $|b_\gamma|^2$ that we will find the system in each of the states $|\gamma\rangle$. But $|\gamma\rangle$ does not correspond to any allowed value of $\varepsilon$, so clearly does not make sense for the result of a measurement of $\varepsilon$! Therefore $b_\gamma = 0$ for all $\gamma$ and the $|\varepsilon\rangle$s form a complete set. 
Edit: It is also worth noting that requiring an operator to have a complete set of real, orthogonal, eigenvalues is equivalent to requiring the operator to be Hermitian, which is how this requirement on observables is normally stated.
In the part you are asking about he is trying to prove that this way of writing $|P\rangle = \sum a_\varepsilon|\varepsilon\rangle$ is unique, i.e. for a given state $|P\rangle$ there is only one possible set of coefficients $a_\varepsilon$. Now we have from orthogonality that \begin{equation}\langle\varepsilon^\prime|\varepsilon\rangle = 
\left\{\begin{array}{lc} 1& \varepsilon = \varepsilon^\prime \\ 0 & \varepsilon\ne\varepsilon^\prime\end{array}\right.\end{equation} Let us say that there is another set of coefficients $a_\varepsilon^\prime$ such that $|P\rangle = \sum a_\varepsilon^\prime|\varepsilon\rangle$. Subtracting these two expressions for $|P\rangle$ we find 
\begin{equation} 0 = \sum (a_\varepsilon - a_\varepsilon^\prime)|\varepsilon\rangle\end{equation} Multiplying by $\langle\varepsilon^\prime|$ we find that all the terms go to 0 except 
\begin{equation} 0 = (a_{\varepsilon^\prime}-a_{\varepsilon^\prime}^\prime)\end{equation} So it terns out $a_{\varepsilon^\prime} = a_{\varepsilon^\prime}^\prime$ and since this must be true for each $\varepsilon$, the expression for $|P\rangle$ was unique. 
With the integral included the discrete coefficients are replaced by functions in the integral $a(\varepsilon)$ and the normalisation is $\langle\varepsilon^\prime|\varepsilon\rangle = \delta(\varepsilon^\prime - \varepsilon)$ but otherwise the argument is unchanged. 
A: Dirac's proof on p. 40 of his book, of the uniqueness of the expansion of an arbitrary ket $|P\rangle$ in terms of the eigenkets of an observable $\xi$, uses several key ideas. It is, however, phrased in a way that does not take advantage of certain concepts and notation that might clarify the discussion. (As these are developed later in the book.)
Nevertheless, the central idea of the proof in question is, paraphrasing from p. 37, that the (most general) condition for the eigenstates of an observable $\xi$ to form a complete set is that any ket $|P\rangle$ can be expressed as an integral plus a sum of eigenkets, that is
$$ |P\rangle = \int |\xi'c\rangle d\xi' + \sum_r |\xi^rd\rangle \, .$$
The integral is over a continuous range of eigenvalues, while the sum is over a discrete set. The sum may also contain eigenkets corresponding to eigenvalues inside the continuous range(!), so that this expression is valid when the expansion of $|P\rangle$ contains eigenkets corresponding to such eigenvalues. The labels $c$ and $d$ are there to distinguish between eigenkets showing up in the integral, and those showing up in the sum, which, it is shown in the book, behave differently.
There are also two axioms/assumptions used in the proof. The first of these is that, for any "finite length" ket $|A\rangle$, the product $\langle A|A\rangle > 0$ unless $|A\rangle=0$. (See p. 21.) However, if we want to be able to express some ket $|Q\rangle$ as, for example,
$$ |Q\rangle = \int |\xi'\rangle d\xi' \, ,$$
then it must follow that $\langle \xi' |\xi'\rangle$ is infinite. (See p. 39.) This leads to the other axiom/assumption used in the proof, which is that
$$ \int \langle \xi' |\xi''\rangle d\xi'' \, > 0, $$
and is finite, unless $|\xi'\rangle = 0$. (See p. 40; the statement about this being finite does not appear here, but later.)
Now comes the proof by contradiction. We assume that the expansion of $|P\rangle$ is not unique, and can therefore be written as, for example
$$ |P\rangle = \int |\xi'a_1\rangle d\xi' + \sum_r |\xi^rb_1\rangle \, ,$$
or
$$ |P\rangle = \int |\xi'a_2\rangle d\xi' + \sum_r |\xi^rb_2\rangle \, ,$$
where the terms in the two integrals and two sums are not, respectively, the same. The difference between these two expansions is zero, that is
$$ |P\rangle - |P\rangle  = \int |\xi'a_1\rangle d\xi' + \sum_r |\xi^rb_1\rangle - \int |\xi'a_2\rangle d\xi' + \sum_r |\xi^rb_2\rangle $$
$$ 0  = \int |\xi'a\rangle d\xi' + \sum_r |\xi^rb\rangle \, ,$$
where on the last line, the remaining integral and sum contain the "leftovers".
We now try to show that there cannot be any eigenvalues appearing in these leftovers. Suppose first that there is an eigenvalue $\xi^t$ appearing in this expression that is not in the continuous range. Then, left multiplying by the bra $\langle \xi^t b|$, we have
$$ 0 = \langle \xi^t b | \xi^t b \rangle \, ,$$
in contradiction with the first of the axioms/assumptions above. Therefore, we cannot have any such $\xi^t$. Suppose instead that there is an another eigenvalue $\xi''$, which is now in the continuous range but does not appear in the sum. (I don't know why Dirac uses the label $t$, and then two primes, but I have tried to stay somewhat consistent with the notation in the book.) Left multiplying the leftovers by the bra $\langle \xi'' a|$ gives
$$ 0 = \int \langle \xi'' a |\xi' a \rangle d\xi' \, , $$
now in contradiction with the second of the axioms/assumptions above. Therefore, we can have no such $\xi''$. Finally, suppose that there is an eigenvalue $\xi^u$ that shows up in both the integral and the sum of this expression. Then we should examine what happens when we left multiply the leftovers by the two different kinds of bras corresponding to this eigenvalue, that is, $\langle \xi^u a|$ and $\langle \xi^u b|$. Examining the former bra first, we have
$$ 0  = \int \langle \xi^u a|\xi'a\rangle d\xi' + \langle \xi^u a |\xi^u b\rangle \, ,$$
where the sum has disappeared because of the orthogonality of the eigenkets. From the second of our axioms/assumptions above, the integral term is non-vanishing and finite, so it must also be true that $\langle \xi^u a |\xi^u b\rangle $ is non-vanishing and finite. Now let's see what happens if we left multiply the leftovers by $\langle \xi^u b|$. We have
$$ 0  = \int \langle \xi^u b |\xi'a\rangle d\xi' + \langle \xi^u b |\xi^u b\rangle \, .$$
However, since $\langle \xi^u a |\xi^u b\rangle $ is finite, the integral term here must vanish, and therefore $\langle \xi^u b |\xi^u b\rangle$ must also vanish, which is contradiction with the first of our axioms/assumptions above. Thus, there can be no such $\xi^u$. This exhausts all of the possible types of eigenvalues that could appear in the leftovers, so that the leftovers must vanish identically. Since this contradicts the assumption with which we started the proof, that the expansion of $|P\rangle$ is not unique, is must be true that the expansion of $|P\rangle$ is unique.
