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My question is regarding a proof given in Greiner's "Relatavistic Quantum Mechanics", 3rd Edition textbook. On pg 148, he proves that the current density $j^{\nu}(x)$ is invariant to a Lorentz transformation so that $$ j'^{\mu}(x')=a_{\nu}^{\mu} j^{\nu}(x) \tag{3.65} $$ where $a_{\nu}^{\mu}$ is the Lorentz transformation matrix. He then writes the continuity equation in Lorentz invariant form: $$ \frac{\partial j^{\mu}(x)}{\partial x^{\mu}}=0 \tag{3.66} $$ He also defines $j^0(x)$ as the probability density: $$ j^{0}(x)=c \varrho(x)=c \psi^{\dagger} \psi(x) \tag{3.66a} $$ The next step is the one I don't understand. He then claims that an invariant probability is guaranteed, because it holds for the Lorentz system of observer A: $$ \frac{\partial}{\partial t} \int j^{0}(x) \mathrm{d}^{3} x=0 \quad \Rightarrow \quad \int j^{0}(x) \mathrm{d}^{3} x=1 \tag{3.66b} $$ and for observer B: $$ \frac{\partial}{\partial t^{\prime}} \int j^{\prime 0}\left(x^{\prime}\right) \mathrm{d}^{3} x^{\prime}=0 \Rightarrow \int j^{\prime 0}\left(x^{\prime}\right) \mathrm{d}^{3} x^{\prime}=1 \tag{3.66c} $$ Please, can someone explain how Eq. (3.66c) follows from the previous equations?

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