I have been studying the Feynman path integral and its various derivations, and I've run into a bit of a problem. The standard Feynman path integral appears as follows: $$ \int \mathcal{D}[x(t)]\exp\left(\frac i \hbar \int_0^t d\tau\ \left( \frac m2 \dot{x}(\tau) - V(x(\tau) \right) \right) $$ Where the path integral is defined by passing the following into the continuum limit: $$ \lim_{n \to \infty}\idotsint d^{n-1}x\ \mathcal{N} \exp\left( \frac i \hbar \sum_{j=1}^n \left( \frac m2 \left( \frac{x_{j+1} - x_j}{\Delta t} \right)^2 - V(x_j) \right) \Delta t \right) $$ where $\Delta t = t/n$. In this form, we independently integrate each $x_j$ from $-\infty$ to $+\infty$, and hence there is no innate notion that the paths considered must be continuous; $x_{j+1}$ and $x_j$ can be as far apart as we please.
In doing readings, I have found multiple sources that claim that the paths we consider are continuous, for example in this paper. In transforming to imaginary time ($-t \to -i\tau$), this claim makes sense, since discontinuities in the path will blow up the the sum and they will be exponentially suppressed. However, in the form presented above, I am not quite sure where and how we can make this claim.
I have seen the argument that paths that deviate greatly from the classical path will lead to extreme oscillatory behavior from our phase, and hence will be suppressed since they will on-average integrate to zero. However, I don't think I am satisfied with this argument to claim we must have continuity. This shows which paths contribute the most, but I don't think it rules out discontinuous paths. Does anyone have insight into this? Any help would be greatly appreciated.