Probability density of fermions in QFT I was following the book 'Quantum Field theory and the Standard model' written by Mattehw D. Schwartz and I found something unconvincing in p.174 of the book.
In that page, it argues that the zeroth component of the Noether current (to the Dirac Lagrangian) $J_{\mu}=\bar{\psi}\gamma_{\mu}\psi$ corresponding to the global symmetry of the spinors $\psi\rightarrow e^{-i\alpha}\psi$ should be interpreted as the probability density of the fermions. That is, it argues that $J_{0}=\psi^{\dagger}\psi=\psi_{L}^{\dagger}\psi_{L}+ \psi_{R}^{\dagger}\psi_{R}$ is the probability density of the Fermions.
My issue here is, although this $J_{0}$ is a quantity that is conserved according to the continuity equation for the Noether current, it evidently is 'NOT' Lorentz invariant. This in turn means that the probability density should be different depending on what reference frame we choose to be in. How can such a quantity which is not Lorentz invariant be physically interpreted as probability density? My guess is that the 'GLOBAL' amount of charge (or probability) can somehow be Lorentz invariant when we use our usual reasoning that the boundary values should vanish at infinity, but isn't it the 'LOCAL' probability that has to be Lorentz invariant in order to constitute a physically meaningful theory? Probability of an event of observing a particle at a particular spacetime grid has to be reference frame independent at least from my understanding.
 A: 
Probability of an event of observing a particle at a particular spacetime grid has to be reference frame independent

One of your problems is assuming that the particle is observed in a particular spacetime-grid.
The density is not probability divided by spacetime volume. Instead, it's "probability divided by small 3D space volume" at a particular point of time.
So, if you do a boost, the probability density changes, but the 3D space, to which the probability is associated, also changes. The original small 3D space-like hypersurface is no longer a space-like hypersurface after a Lorentz boost.
Anyway, I'm not sure if this resolves the question of probabilistic interpretation of the Dirac equation. Maybe some other answer can clear it up.
In the full quantum field theory, there does not exist a position basis to allow you to talk about the probability of observing a particle at a point in space.
A: More generally, if $J=J^{\mu}\partial_{\mu}$ is a 4-vector current density  that satisfies a continuity equation $$\partial_{\mu}J^{\mu}~=~0,\tag{1}$$
then the corresponding charge defined in one inertial frame as $$\begin{align}Q(t_1)~:=~&\int_{\{t=t_1\}} \!  \mathrm{d}^3r~J^0({\bf r},t)\cr
~=~&\int_{\{t=t_1\}} \!  \mathrm{d}^3r~n_{\mu}J^{\mu}({\bf r},t)\cr
~=~&\int_{\{t= t_1\}} \! J~\lrcorner~ \mathrm{d}^4x~\end{align}\tag{2}$$
is a conserved Lorentz invariant.
To see that the charge (2) is a Lorentz invariant, define a charged in a primed inertial frame in a similar way
$$ Q^{\prime}(t^{\prime}_1)~:=~ \int_{\{t^{\prime}=t^{\prime}_1\}} \!  \mathrm{d}^3r^{\prime}~J^{\prime 0}({\bf r}^{\prime},t^{\prime}). \tag{3}$$
By the divergence theorem, we get
$$ Q^{\prime}(t^{\prime}_1) - Q(t_1)~=~\left(\int_{\{t^{\prime}\leq t^{\prime}_1\}}-\int_{\{t \leq t_1\}} \right)\!  \mathrm{d}^4x~\partial_{\mu}J^{\mu}(x)~=~0, \tag{4}$$
where we implicitly assume that no charge is leaking at spatial infinity.
For a related statement, see this Phys.SE post.
