I have been thinking about the problem of relativistic path integrals and I encountered several difficulties. Let's assume we have a particle initially a position $x_i$ at $t_i$ in a certain reference frame. In another inertial reference frame the positions are $x_i'$ at $t_i'$. Feynman's path integral method allows us to calculate the conditional problem to observe the particle at $x_f$ at $t_f$ given that it was initially at position $x_i$ at $t_i$: $$P(x_f,t_f|x_i, t_i)$$ In another frame of reference the probability is $$P'(x_f',t_f'|x_i', t_i')$$ Common sense would tell us that $$P(x_f,t_f|x_i, t_i)=P'(x_f',t_f'|x_i', t_i')$$ Since we are observing the same event. However now the problem comes: $$\int_{-\infty}^{+\infty} dx_f P(x_f,t_f|x_i, t_i) = 1$$ Which basically means that the probabilities have to add up to 1. In the different frame of reference the same must be true, but according to special relativity the simultaneity hypersurface is different, but still the probabilities along that hypersurface have to add up to 1. But these probabilities are in general not the same as the probabilities in a different reference frame. If my arguemtn is correct then it seems to me that we are left with three choices:

1. The probabilities are different in each frame of reference. This is absurd.
2. Not all the paths are allowed such that this condition is indeed satisfied. This is also absurd.
3. There is no free will, meaning that an observer can't arbitrarily choose his velocity.

Is my argument correct?