Consider a triangle in the 1st quadrant defined by
\begin{align} &x \geq 0 \\ &y \geq 0 \\ &y = 1-x \end{align}
To calculate the area using double integrals we could for example take \begin{align} \int_{0}^1 \int_{0}^{1-y} dxdy = [y-\frac{y^2}{2}]_{0}^1 = \frac{1}{2} \end{align} as expected. I am wondering how to reach this answer using Riemann sums ie defining the area as \begin{align} A = \lim_{n \to \infty} \sum_{j=0}^n \sum_{i=0}^m \Delta x_i \Delta y_j \end{align} In particular how do we set $m$ to be the discretized limit $x \leq 1-y$ ?
Define Let $x_i$, $y_j$ be the points in the domain associated with the Riemann summands. Then:
\begin{align} A = \lim_{n,m \to \infty} \sum_{j=0}^n \sum_{i=0}^m \Delta x_i \Delta y_j \mathbf{1}[1-{y}_j \geq {x}_i], \end{align} where $\mathbf{1}[B]= 1$, if $B$ is true and $=0$ otherwise.
When using the same Riemann-partition for $x$ and $y$, one can simplify $$A = \lim_{n \to \infty}\sum_{j=0}^n\sum_{i=0}^{n(1-y_j)} \Delta x_i \Delta y_j $$