Let us consider the usual derivation of the gaussian integral: \begin{align} \left(\int dx e^{-ax^2}\right)\left(\int dy e^{-ay^2}\right) =\int\int dx dy e^{-a(x^2 + y^2)} = \sqrt{\frac{\pi}{a}}. \end{align} If we extract the "diagonal part" of the double integral, which is $y = x$, \begin{align} \int dx \int_{y=x} dy e^{-2ax^2} \end{align} is zero, because the integral measure is zero.
However, how can I understand this result via the continuum limit of finite sum?
I mean, I consider \begin{align} \int dx f(x) = \lim_{L \rightarrow \infty}\left(\lim_{N \rightarrow \infty}\frac{L}{N}\sum_{k = -\frac{N}{2}}^{k = +\frac{N}{2}} f\left(\frac{L}{N}k\right) \right). \end{align} Once I forget detail treatments of limits, then \begin{align} \int dx f(x) \int dy f(y) \supset \lim_{L \rightarrow \infty}\lim_{L' \rightarrow \infty} \left(\lim_{N \rightarrow \infty}\frac{L}{N}\sum_{k = -\frac{N}{2}}^{k = +\frac{N}{2}} f\left(\frac{L}{N}k\right) \right)\left(\lim_{N' \rightarrow \infty}\frac{L'}{N'}\sum_{l = -\frac{N'}{2}}^{l = +\frac{N'}{2}} f\left(\frac{L'}{N'}l\right) \right), \end{align} which I do not understand how the diagonal part $k = l$ is suppressed to be zero.
In physics, sometimes the right hand side comes first, and we should take its continuum limit. Often, such the diagonal part is ignored. However, I think there is some mathematical reasons to get it. If we start from the left hand side, then the integral will be a surface integral by definition. However, what I need to understand when the right hand side comes first?