How to understand the measurement on entangled state in the following cases? Assuming an EPR pair AB, event MA is a measurement on A. 

My questions are:
(1) At MB and MB' (depending on where B is located), if we try to describe the state of B (but not measure B yet), what's the difference between MB and MB'?
(2) If MB and MB' are measurements on B, then what's the influences of the operations MB and MB' on the state of B? It seems that before the measurement on B, MB and MB' are different(depending on different observers?), but after the measurement, are they in the same state?
(3) For MB', it seems that MB' is not determined by MA since they are spacelike separated and there is no definite synchronization. But is the state at MB fully determined by MA?  
I am hoping somebody can help to give a complete picture of this problem. There must be somebody has considered this problem through, but I can not find it. 
It might seem trivial for some people, but for me I think it's both related with the measurement problem and the relationship between entanglement/spacetime, and some other topics as well.  I do not agree that it should not be discussed. At least for some people it's a question that need to be checked (See the answer of udrv).
 A: You might be interested in this review on "Quantum Information And Relativity Theory". It isn't very recent anymore, but provides a good grip on the fundamentals.
I am adding some intro. The answer to the questions themselves is at the bottom.
There are two basic concepts that you need to keep in mind relative to quantum measurements:

M1. Quantum measurements are performed on single systems, but the information they produce makes sense only in the context of a statistical ensemble. 

In other words, you always need to repeat the same measurement a large number of times on identically prepared systems in order to extract definitive information. A single measurement on a single system only provides information on the post-measurement state of the system (we always know what it is with 100% certainty), but generally not on its state  before the measurement. For the latter it is necessary to interrogate a large number of identical systems.

M2. For entangled pairs of quantum systems, as far as state statistics is concerned, local measurements on either member of a pair never affect the local state statistics of the other member.   

What this means: 
Alice and Bob share members of a large number $N$ of identically prepared entangled qubit (or any other quantum system) pairs $\{a_i, b_i\}_{i=1}^N$: Alice keeps qubits $\{a_1, a_2, ..., a_N\}$, constituting what we may call ensemble A, while Bob gets qubits $\{b_1, b_2, ..., b_N\}$, or ensemble B. Bob carefully moves his qubits at a distant location from Alice. For convenience let each qubit pair be prepared in a state 
$$
|\Psi\rangle = \alpha \;|\phi_a\rangle\otimes|\phi_b\rangle + \beta\; |\psi_a\rangle\otimes|\psi_b\rangle 
$$
where $|\phi\rangle$ and $|\psi\rangle$ are orthonormal, but otherwise arbitrary single-qubit states, $\langle \phi|\psi\rangle = 0$,$\langle \phi|\phi\rangle = \langle \psi|\psi\rangle = 1$, and $\alpha$, $\beta$ are complex numbers such that $|\alpha|^2 + |\beta|^2 = 1$. 
The entangled nature of state $|\Psi\rangle$ means that:

M1'. If Alice measures qubit $a_i$ and finds it in state $|\phi_a\rangle$ ( $|\psi_a\rangle$ ), then she knows with 100% certainty that when/if Bob measures his corresponding qubit $b_i$ he will find it in state $|\phi_b\rangle$ ( $|\psi_b\rangle$ ). But before the measurement Alice has no idea whether she is going to obtain state $|\phi_a\rangle$ or $|\psi_a\rangle$. Likewise, the fact that Alice knows exactly what she obtained in one individual measurement, and also what Bob will obtain in his corresponding measurement, doesn't tell her anything about what she might obtain in her next measurement, on a different qubit. She cannot reproduce individual outcomes at will.

In order to determine the state of her qubits, Alice needs to measure her entire ensemble,  or at least a statistically significant subensemble. Such an ensemble measurement tells her that the qubits are each in a mixed state described by a density matrix
$$
\rho_a = Tr_b\left[|\Psi\rangle\langle\Psi|\right] \;\;(\;= |\alpha|^2 |\phi_a\rangle\langle \phi_a| + |\beta|^2 |\psi_a\rangle\langle \psi_a|\;)
$$
Even so, she only knows that upon a measurement of any one qubit there is a probability $|\alpha|^2$ to obtain $|\phi_a\rangle$ and a probability $|\beta|^2$ to obtain $|\psi_a\rangle$. Nothing more. Similarly for Bob and his qubits. Their density matrix is
$$
\rho_b = Tr_a\left[|\Psi\rangle\langle\Psi|\right] \;\;(\;= |\alpha|^2 |\phi_b\rangle\langle \phi_b| + |\beta|^2 |\psi_b\rangle\langle \psi_b|\;)
$$

M2'. Local interactions involving only qubits $a_i$ do not affect the state statistics of qubits $b_i$, and conversely. 

This is the critical part: No matter what Alice does to her qubits (and only to her qubits), be it measurements or other operations, the state statistics for Bob's qubits does not change. At least not as long as all interactions are completely positive. Say Alice measures all her qubits and keeps track of the results for each one of them. She finds $N|\alpha|^2$ of them in state $|\phi_a\rangle$ and $N|\beta|^2$ in state $|\psi_a\rangle$. She knows with 100% certainty what Bob is going to obtain for each of his qubits. But unless she tells Bob about it through other channels, he will not be aware of her measurement results in any way. If he can only rely on his local measurements, all he obtains is that $N|\alpha|^2$ of his qubits are in state $|\phi_b\rangle$ and $N|\beta|^2$ are in state $|\psi_b\rangle$. Alice cannot change Bob's ensemble statistics.
Answer to your question:
If your event MA refers to a single measurement, on a single qubit (system), then A knows with certainty the outcome at either B or B'. Space-like or time-like separation makes no difference. On the other hand, neither B nor B' have any idea what they might obtain on their side, unless A informs them of his/her result by independent means, and conversely. If A tries to reproduce the first result, s/he finds that it is impossible to control the outcome in order to build a meaningful signal sequence. All that is produced is a random sequence with a statistics determined by the local mixed state.
If your MA is an ensemble measurement, then again A knows with certainty the outcome at either B or B', but the fact that s/he performed a local measurement is never visible in the ensemble statistics of B or B'. The measurement probabilities for various single-system states at B and B' are not affected. Again, space-like or time-like separation are not important. 
So, assuming B and B' are stationary wrt A, but acted on at different times so as to obtain time-like or space-like separation:
1) Before measurement each individual system at B or B' would be in a mixed state described by a local (reduced) density matrix. There is no difference between B and B'.
2) The reduced state of an individual system is mixed before the measurement and pure after the measurement. Before the measurement there is an ensemble of $N$ systems each in a mixed state described by a density matrix $\rho_B$. Assuming a complete measurement that accounts for all possible outcomes in some orthonormal basis of pure states $\{|\omega_i\rangle\}$, the state of the ensemble is projected on those pure states. That is, after the measurement there is an ensemble of systems each in some pure state $|\omega_i\rangle$ occurring with a frequency $\langle\omega_i|\rho_B|\omega_i\rangle$, such that overall the ensemble represents a mixed state with density matrix
$$
\bar\rho_B = \sum_i{|\omega_i\rangle\langle\omega_i|\rho_B|\omega_i\rangle\langle\omega_i|}
$$ 
If $\{|\omega_i\rangle\}$ is the eigenbasis of $\rho_B$, then the ensemble statistics is not affected by the measurement.       
3) A measurement MA causes the local mixed states of entangled counterparts at B or B' to be reduced to pure states. However, the ensemble statistics embodied by $\rho_B$ (before the B or B' measurement) or $\bar\rho_B$ (after the B / B' measurement) is not affected. The space-like or time-like separation relative to A does not matter. What changes, at most, is the particular realization of the B or B' ensembles: an ensemble of identical individual mixed states (or improper mixtures, as you used in a previous question) before MA vs. a distribution of pure states (a proper mixture) after MA. 
Adding clarifications requested in comments:

(1) For MB', can MA turn an improper mixture to a proper mixture? Since MA and MB' are space-like separated, is the conclusion independent of the reference frame? Can we say that before the measurement operation at MB', the state of B at MB' is always an improper mixture?

The standard argument is often that it doesn't matter whether MB' is in a proper or improper mixture once MA is done, on grounds that if ensemble statistics do not change at MB' then no measurement of a standard observable can distinguish whether B's mixed state is a proper or improper mixture. 
This being said, if MA destroys entanglement between the $a_i$ and $b_i$ it must transform the overall state of ensembles A and B into a separable state (there are only 2 possibilities: entangled or separable). This applies even if A is left entangled with a measuring device M. In the latter case the overall state would be a separable state of (A+M) and B. Then if MA leaves A (or A+M) as a local probabilistic mixture, the post-measurement separable state of A and B must also be a probabilistic mixture, and by definition it must be realized as a superposition of direct product states of A and B. So B would necessarily be a proper mixture too, regardless of whether MB' is performed or not (remote preparation). 
The only alternative is that MA leaves pairs $a_i$, $b_i$ in separable states that are not proper mixtures. This would mean that in general a measurement output may be a separable state, but is not necessarily a proper mixture. 
The argument seems independent of the space-like or time-like separation of A and B.  

(2) If CTC exists, then after MA, if MB' is in a proper mixture, then CTC can be used to achieve non-local signalling. For example, MA is a projective measurement on either basis {|0>,|1>} or {|+>,|->}, then at MB' CTC can distinguish the non-orthogonal states at MB' and MB' can find out which basis MA used. One way to resolve this is to say after MA, MB' is always an improper mixture so that the CTC can not figure out MA's basis and therefore no signaling. Or someone can just argue 'there is no CTC'. 

It doesn't have to be CTCs. According to this very recent article in Nature, it can be causal open timelike curves as well: Replicating the benefits of Deutschian closed timelike curves without breaking causality. The result is rather similar. One way or another CTCs and OTCs bring conflict with the no-signaling clause (finite speed of light), the no-cloning theorem, etc. If the argument that MA necessarily causes B to become a proper mixture holds, then it looks like quantum mechanics provides a very good reason why CTCs and OTCs cannot co-exist with the no-signaling clause, regardless of their compatibility with general relativity. 
This leaves the issue of whether measurements on entangled pairs $a_i$, $b_i$ produce separable states that are not proper mixtures. 
