Quantum mechanics - measuring position I am watching Susskind's Stanford Lectures on quantum mechanics.
The eigenvectors (eigenfunctions) of the position operator are of the form $\delta(x-k)$. But $$\int\delta^{*}(x-k)\delta(x-k)\, \mathrm dx$$ is undefined or infinite. So as far as I can tell it cannot be the state of a system, since $$\int\psi^{*}(x)\psi(x)\,\mathrm dx = 1$$ is required.
I thought that when we measure position, we could say that the wave function is in the state corresponding to the eigenvector of the eigenvalue measured. But this can't be right because of the above. What can we say about the state of a quantum system right after measuring position?
Is this somehow related to measurements of position not being strictly point-like?
 A: Actually, the outcome of the experimental apparatus is an interval $(x_0-\delta, x_0+\delta)$. $\delta>0$ stays for the accuracy of the instrument which can be made smaller and smaller but cannot be removed.  
It is therefore assumed (Luders-von Neumann's axiom) that, if the state immediately before the measurement was determined by the wavefunction $$\psi\:,$$ the one immediately after is, up to the normalisation,  $$\chi_{(x-\delta, x+\delta)}\psi\:.$$ Here $\chi_{(x_0-\delta, x_0+\delta)}(x)=1$ if $x\in (x_0-\delta, x_0+\delta)$ and $\chi_{(x_0-\delta, x_0+\delta)}(x)=0$ if $x\not \in (x_0-\delta, x_0+\delta)$.
Abstractly, if the initial vector state state is $\psi$, the final one is $P_{(x_0-\delta, x_0+\delta)}\psi$. The operator $P_{(x_0-\delta, x_0+\delta)}$ being the orthogonal projector of the PVM of the position operator associated to the interval $(x_0-\delta, x_0+\delta)$. 
Obviously this is the theoretical description of a very ideal measurement procedure.
If, differently from the position observable whose spectrum is continuous, the measurement process regards an observable $A$ with point spectrum and the elements of the spectrum are isolated points, it is always possible to assume the existence of a measurement instrument whose sensibility $\delta>0$ is smaller that the distance of pairs of consecutive values of $A$. This way, even if the measurement is affected by an experimental error represented by $\delta>0$, we can distinguish  between couples of eigenvalues and the Luders-von Neumann's axiom takes the more familiar standard form: The state after the measurement with outcome $a_0$ procedure is represented by the eigenvector with eigenvalue $a_0$. (This is true if the eigenspace has dimension $1$ otherwise the general abstract form of L-vN's axiom holds again.)  
ADDENDUM (After some discussions with  valerio92). 
I prove here in details how the general form of L-vN's postulate leads to the standard postulate of collapse of the wavefunction after an imprecise  measurement of position. 
For a quantum system described over the separable complex Hilbert space $\cal H$, a quantum state is a probability measure $\mu$ over the non-Boolean lattice $\cal L(\cal H)$ of orthogonal projectors in the Hilbert space (see this answer of mine for further details). $P \in \cal L(\cal H)$ has the meaning of an observable whose values are only $0$ or $1$, I mean a YES/NO observable. $\mu(P)$ is the probability that $P$ turns out to be true  if measuring it when the state is $\mu$.
Gleason's theorem proves that, if the Hilbert space is separable with dimension $\neq 2$, there is a one-to-one correspondence between these probability measures and density matrices. These are  unit-trace, trace class, positive operators $\rho : \cal H \to \cal H$. This correspondence is such that $$\mu_\rho (P) = tr(\rho P)$$ for every $P \in \cal L (\cal H)$.
The general form of L-vN's axiom is the following. 
L-vN's axiom. Let  $P$ be an orthogonal projector representing an elementary observable of the physical system and $\rho$ a state. If the outcome of the measurement of $P$ when the state is $\rho$ is $1$ (YES), the post measurement state is 
$$\rho_P = \frac{P\rho P}{tr(\rho P)}\:.\tag{1}$$ 
This postulate has a direct conditional-probabilty interpretation: $\mu_{\rho_P}$ is the unique probability measure such that 
$$\mu_{\rho_P}(Q) = \frac{\mu(Q)}{\mu(P)}$$
for every $Q \in \cal L (\cal H)$ with $Q \leq P$.
Pure states by definition are extremal elements of the convex body of the mentioned measures. In other words a density matrix $\rho$ is a pure state if there are no $p,q\in (0,1)$ with $p+q=1$ and density matrices $\rho_1\neq \rho_2$ such that $\rho = p\rho_1 + q \rho_2$. 
It turns out that $\rho$ is pure if and only if it has the form $$\rho = |\psi \rangle \langle \psi |$$
for some $\psi \in \cal H$ with $||\psi||=1$. 
Evidently unit vectors $\psi$ and $\psi'$ define the same pure state if and only if $\psi = e^{ia}\psi'$ for some $a \in \mathbb R$.
L-vN's postulate applied to pure states specializes to this statement 
L-vN's axiom (pure states). Let $P$ be an orthogonal projector representing an elementary observable of the physical system and $|\psi \rangle \langle \psi |$  a pure state. If the outcome of the measurement of $P$ when the state is $|\psi \rangle \langle \psi |$ is $1$ (YES), the post measurement state is still pure has has the form 
$$|\psi_P \rangle \langle \psi_P | = \frac{P|\psi \rangle \langle \psi | P}{tr(|\psi \rangle \langle \psi | P)}\:.\tag{2}$$ 
Since $tr(|\psi \rangle \langle \psi | P)= ||P \psi||^2$, so that 
$$|\psi_P \rangle \langle \psi_P | = \frac{P|\psi \rangle \langle \psi | P}{||\psi|| \: ||\psi||}\:,$$
the postulate can be rephrased as follows.
L-vN's axiom (pure states 2). Let $P$ be an orthogonal projector representing an elementary observable of the physical system and let the unit vector $\psi \in \cal H$ represent a pure state up to phases. If the outcome of the measurement of $P$ when the state is represented by $\psi$ is $1$ (YES), the post measurement state is still pure and is represented, up to phases, by the unit vector 
$$\psi_P = \frac{P\psi}{||P\psi||}\:.\tag{3}$$ 
Let us eventually come to the position observable for a particle moving along the real axis. 
Here ${\cal H} = L^2(\mathbb R, dx)$ and the position observable is the self-adjoint multiplicative operator $$(X \psi)(x):= x\psi(x)$$ with obvious domain. 
The spectral decomposition of $X$ associates it to the projector-valued measure $\{P_E\}_{E \in B(\mathbb R)}$ where $B(\mathbb R)$ is the class of Borel set of the real axis, for instance $E$ could be an interval $E=(a,b)$. The spectral theorem says that $P_E$ is defined like this
$$(P_E \psi)(x):= \chi_E(x)\psi(x)\quad \forall x \in \mathbb R\:.\tag{4}$$
The meaning of the orthogonal projector $P_E$ is 
$$\mbox{"the position of the particle stays in $E$"}$$
In fact, if  $\psi \in L^2(\mathbb R, dx )$ is a normalized vector representing a pure state, the probability that $P_E$ is true is
$$tr(P_E |\psi\rangle \langle \psi|)= ||P_E\psi||^2 = \int_{\mathbb R} |\chi_E(x) \psi(x)|^2 dx = \int_E |\psi(x)|^2 dx\:,$$
in perfect agreement with more elementary versions of Quantum Mechanics.
With this interpretation, measuring $X$ means measuring every $P_E$ or at least measuring the class of mutually exclusive elementary propositions $P_{(ns, (n+1)s]}$ with $n \in \mathbb Z$ where $s$ is the sensibility of the instrument.
Let us suppose that the initial state of the particle is pure and (up to phases) represented by the (normalized) wavefunction $\psi$. Suppose that we perform a measurement of the position and we find that the particle stays in $E \subset \mathbb R$. What is the post measurement state according to the Postulate by Luders and von Neumann?
We can apply our third version, the one specialized to pure states described in terms of unit vectors. The post-measurement state $\psi_E$, up to phase, is represented by the vector in (3):
$$\psi_E := \frac{P_E \psi}{||P_E \psi||}\:.$$
Taking (4) into account, we find
$$\psi_E(x) = \frac{\chi_E(x) \psi(x)}{\sqrt{\int_E |\psi(z)|^2 dz}}\:.$$
