In deriving the path integral, one often finds an expression of this type:
$$\langle x_N|e^{-iH\Delta t} e^{-iH \Delta t} \ldots e^{-iH\Delta t}|x_0\rangle =\tag{1}$$ $$ \langle x_N|e^{-iH\Delta t}\left(\int dx_{N-1}|x_{N-1}\rangle\langle x_{N-1}|\right)e^{-iH\Delta t}\left(\int dx_{N-2}|x_{N-2}\rangle\langle x_{N-2}|\right)e^{-iH\Delta t}\ldots |x_{0}\rangle \tag{2}$$
which is claimed to be writable, equivalently, as follows:
$$\int dx_{N-1}\int dx_{N-2}...\int dx_1 \langle x_N|e^{-iH\Delta t}|x_{N-1}\rangle\langle x_{N-1}|e^{-iH\Delta t}|x_{N-2}\rangle\langle x_{N-2}|\ldots|x_1\rangle\langle x_1|e^{-iH\Delta t}|x_0\rangle \tag{3}$$
But, I am confused as to the use of the notation. I am probably wrong, but all I see are violations of the properties of integrals.
In (3), what are we integrating exactly? Is (3) a claim that we integrate the braket terms $N$ times, once for each $dx_i$?
I am familiar with the integrated term being between the $int$ and the $dx$ signs. However, here it seems the notation is $\int dx_{N-1}$ but the integral is perform edfor the entire expression that follows it. For example, these are not the same: $\int xdx \neq (\int dx) x$?
In (2) why is one allowed to break the $\int dx_{N-1}$ to the front of the expression, leaving its friends $|x_{N-1}\rangle\langle x_{N-1}|$ behind? Doesn't that changes the whole integral? For instance if I normalize $\int \psi(x) dx=\mathbb{I}$, then inject it into an expression $x^2 \mathbb{I}\to x^2\int \psi(x) dx$ on the grounds that the integral is equal to unity, but then I play with the integral moving terms around as follows $\int dx x^2 \psi(x)$, then I do not see how I still injected the identity into my expression. Clearly, the result for $\int dx x^2 \psi(x)$ are not the same as those for $x^2\int \psi(x) dx$. But I have made the same changes as going from (2) to (3).
