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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 ($x^*_i$, $y^*_j$)?

Pulled from here https://math.libretexts.org/Bookshelves/Calculus/Book%3A_Calculus_(OpenStax)/15%3A_Multiple_Integration/15.1%3A_Double_Integrals_over_Rectangular_Regions

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  • $\begingroup$ Are you sure it's not a typo? $\endgroup$ – clathratus Feb 5 at 2:47
  • $\begingroup$ It's not. It's written like this in multiple textbooks and in multiple examples throughout the internet. Even my Prof. used this notation. $\endgroup$ – Doxyy Feb 5 at 3:06
  • $\begingroup$ Have you asked your Prof.? $\endgroup$ – clathratus Feb 5 at 4:16
  • $\begingroup$ Not to be mean, but that question is obvious. If I had access to my professor I would have asked him instead of coming here. Instead of questioning the question could you please be more productive? $\endgroup$ – Doxyy Feb 5 at 16:10
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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 $(x^*_{ij},y^*_{ij})$.

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  • $\begingroup$ Ok, so by specifying $x$ with $ij$ instead of just $i$ gives us the ability to make it a point that is arbitrary within that subrectangle? $\endgroup$ – Doxyy Feb 5 at 15:54
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    $\begingroup$ @Doxyy: Yes that is correct. It must be free to be any point within the subrectangle and so two indices are required to distinguish it from other intermediate points. That is a requirement for the function to be Riemann integrable. $\endgroup$ – RRL Feb 5 at 16:39
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    $\begingroup$ Just recall the definition of the Riemann integral in one dimension. A Riemann sum approximates the area as a sum of rectangles with heights specified by the function values at any points within the subintervals. All such sums must converge. It is not enough for only special sums, say using the left endpoints, to converge. $\endgroup$ – RRL Feb 5 at 16:43

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