Understanding what ij mean in a Double Riemann Sum (Double Integral) I am having trouble understanding what the ($x^*_{ij} $, $y^*_{ij}$) in this diagram (circled in blue) is explaining. What I do know is that $i$ is the iteration of the $x$ Riemann Sum and the $j$ is the iteration for the $y$ Riemann Sum. What I am having an issue with is understanding why $x$ needs the $j$ and $y$ needs the $i$. Couldn't the $P$ simply be represented as ($x^*_i$, $y^*_j$)?

 A: The double index is needed to specify an intermediate point in a partition subrectangle in the most general way for constructing a Riemann sum.
To define the double integral over the rectangle $[a,b]\times [c,d]$, we must first specify a partition $P = (P_1, P_2)$ where $P_1$ is a partition of $[a,b]$ defined by points $a = x_0 < x_1 < \ldots < x_n = b$ and  $P_2$ is a partition of $[c,d]$ defined by points $c = y_0 < y_1 < \ldots < y_m = d$.  The partition $P$ can then be viewed as the collection of surectangles $R_{ij} =[x_{i-1}, x_i] \times [y_{j-1}, y_j] = \{(x,y): x_{i-1} \leqslant x \leqslant x_i, \, y_{j-1} \leqslant y \leqslant y_j\}$ for $i = 0,1,\ldots n$ and $j = 0,1,\ldots,m$.
A Riemann sum is then defined by
$$S(P,f) = \sum_{i=1}^n\sum_{j=1}^mf(x^*_{ij},y^*_{ij})\,(x_i-x_{i-1})\,(y_j - y_{j-1}),$$
where the intermediate point or tag $(x^*_{ij},y^*_{ij})$ can be any point in the closed subrectangle $R_{ij}$. Recall that the theory is that Riemann sums converge to the integral as the mesh of the partition (area of the largest subrectangle in the partition) tends to $0$, regardless of how the intermediate points are selected.
If we had specified the intermediate point for $R_{ij}$ as $(x^*_{i},y^*_{j})$ then the points would not be completely arbitrary in that all $m$ subrectangles with fixed index $i$ and $j$ ranging from $1$ to $m$ would have intermediate points with the same $x-$component $x^*_i$ (as well as a similar constraint on $y^*_j$ for fixed $j$). 
Such a constrained specification of intermediate points is not unacceptable -  Riemann sums defiend this way will still converge to the integral.  Nevertheless the Riemann integral can be defined in a much more general way using completely arbitrary points $(x^*_{ij},y^*_{ij})$.   
