If I understand correctly the definition of a Riemann integral is with the following sum:
$$ \int_a^b f(x)dx=\lim_{n\rightarrow\infty}\sum_{i=1}^n f(x+i\Delta x)\cdot \frac{b-a}{n}$$
If we substitute an improper integral I get the following:
$$ \int_{-\infty}^{\infty} f(x)dx=\lim_{n\rightarrow\infty}\sum_{i=1}^n f(x+i\Delta x)\cdot \frac{(\infty)-(-\infty)}{n}$$
Where the last fraction turns into
$$ \frac{\infty}{\infty} (?)$$
You could also do multiple limits but I have never seen this:
$$ \lim_{b\rightarrow\infty} \lim_{a\rightarrow-\infty}\int_{a}^{b} f(x)dx=\lim_{b\rightarrow\infty} \lim_{a\rightarrow-\infty}\lim_{n\rightarrow\infty}\sum_{i=1}^n f(x+i\Delta x)\cdot \frac{b-a}{n}$$
But I have never seen a sequence of limits like:
$$ \lim_{b\rightarrow\infty} \lim_{a\rightarrow-\infty}\lim_{n\rightarrow\infty}$$
How do you properly convert a riemann sum into an improper integral?
You could also just use one limit:
$$ \lim_{n\rightarrow\infty} \int_{-n}^{n} f(x)dx=\lim_{n\rightarrow\infty}\sum_{i=1}^n f(x+i\Delta x)\cdot \frac{n-(-n)}{n}$$
This would give:
$$ \lim_{n\rightarrow\infty}\sum_{i=1}^n f(x+i\Delta x)\cdot \frac{2n}{n}=2\lim_{n\rightarrow\infty}\sum_{i=1}^n f(x+i\Delta x)$$
Here we lose the '$\Delta x$' term.
As a side note, the formal definition of the Riemann integral is slightly different to the one you're using because it allows for the partitions to be of unequal width, as long as their size goes to zero in the limit.
But beyond that, when looking at the informal integral, we would look at a double limit:
$$\begin{eqnarray} \int_{-\infty}^\infty f(x)\ dx & := & \lim_{a \rightarrow -\infty \\ b \rightarrow \infty} \int_a^b f(x)\ dx \\ & = & \lim_{a \rightarrow -\infty \\ b \rightarrow \infty} \lim_{\|P\| \rightarrow 0} \sum_i f(x_i) \Delta x_i \end{eqnarray}$$
It is absolutely necessary to let the two limits of integration move independently, because it is easy to construct examples where $\int_{-n}^n f(x) dx$ will give inconsistent results under transformations because the integral diverges at each end.