Eigenstate of position operator after collapse of the wave-function I am studying basic quantum mechanics in undergrad and have hit a wall.
I understand that if a measurement is made for position, the wave function collapses into one of the eigenstates of position, i.e. a Dirac delta distribution, at the point where particle is found. If the particle is found at $x_0$, the wave function collapses into $\psi=\delta(x-x_0)$
The problem is that the delta function does not satisfy the basic requirements of a wave function, it is not finite at $x=x_0$.
The problem also exists for other observables like momentum, where the eigenstate is not square integrable.
 A: It's true that the Dirac delta function is not square-integrable, and so an exact measurement of position takes the position wavefunction outside the Hilbert space. In the same way, an exact measurement of momentum also takes a wavefunction outside the Hilbert space. Note the word "exact".
If you remember the Heisenberg uncertainty principle, this shouldn't surprise you - after all, a state with exactly-determined position has an infinite uncertainty in momentum, and a state with exactly-determined momentum has an infinite uncertainty in position. Clearly, these states aren't physical, and can't be the result of any actual measurement.
So what is the operator corresponding to the position measurements that we actually make? Let's look at the result that we get from such a measurement: "The particle is at position $x_0\pm\sigma_x$". The way that this uncertainty is characterized informs the shape of the eigenstates of this "physical" position operator; for example, when we say "the particle is at position $x_0\pm\sigma_x$", one possible meaning is that the particle is equally likely to be anywhere in the interval $[x_0-\sigma_x,x_0+\sigma_x]$. So the eigenstates of that operator would be:
$$\psi_{x_0}(x)=\begin{cases}\frac{1}{\sqrt{2\sigma_x}}&\text{for }(x_0-\sigma_x)\leq x\leq(x_0+\sigma_x)\\0&\text{otherwise}\end{cases}$$
As you can see, the eigenstates of a "physical" position operator (you might also call it an "approximate position" operator) actually are square-integrable! I won't prove it here, but it should be fairly straightforward that, for most reasonable characterizations of an "approximate position" operator, the eigenstates are square-integrable. (Feel free to try some other characterizations of uncertainty - for example, what if $\sigma_x$ specifies the standard deviation of a Gaussian probability distribution?)
So this prompts the question: If we know that real measurements don't correspond to the action of the position operator, why do we still use the position operator to make predictions? The answer boils down to the fact that working with "approximate position" operators is cumbersome in practice. Their action on a wavefunction is typically much less elegant than the "exact position" operator, which simply multiplies the wavefunction by $x$. Since many parts of the machinery of quantum mechanics don't actually particularly care about square-integrability of wavefunctions (that is, if you handle them carefully enough), we can often get away with using the "exact position" operator as a handy idealization of whatever approximate position measurement we're actually trying to make.
A: 
I understand that if a measurement is made for position, the wave function collapses into one of the eigenstates of position, i.e. a Dirac delta distribution, at the point where particle is found. If the particle is found at $x_0$, the wave function collapses into $\psi=\delta(x-x_0)$

Mathematical aspect:
Continuous position operator $\hat{x}$ does not have eigenfunctions in the usual sense. The problem is there can't be continuous infinity of different functions of $x$ that are square integrable.
Delta distribution $\delta(x-x_0)$ is a non-function concept that helps with remembering and doing formal operations on differential equations and integrals, but it is not a valid psi function in the sense of Born's interpretation. That interpretation requires square integrable function.

The problem is that the delta function does not satisfy the basic requirements of a wave function, it is not finite at $x=x_0$.
Although finiteness is almost always the case with valid square integrable psi functions, this is not a good argument.

Delta distribution is not a function, it does not have values. You can say delta distribution is infinite at 0 in the sense it picks value of other function when integrated, but this different than functions such as $1/x^2$, where it means function values diverge to $+\infty$ as $x$ goes to $0$.
Physical aspect:
No measurement of continuous coordinate can result in exact single real number, there is always some experimental uncertainty involved. All measurements are rational numbers, typically decimals with finite number  of digits. So there isn't really a need for representing state where particle is exactly at $x=2$ m or $x=\pi$ m or any other exact number, because this situation can never be accomplished. All the measurement does is shrink the interval of values that the coordinate can have, but the interval is always non-zero.
