Visualizing the 2d Riemann sums

MAT236H5F, University of Toronto, Mississauga. Instructor: Ke Zhang.

Let $$ R = \{(x, y) : a \le x \le b, \, c \le y \le d\} =: [a, b] \times [c, d] $$ and $f = f(x, y): R \to \mathbb{R}$, the double integral of $f$ over $R$ is the signed volume between the graph of $f$ and the rectangle $R$ in the $xy$ plane. Again it's helpful to think of $f(x)$ as a weight on $R$ since we will run out of dimensions when it comes to triple integral.

Given integers $m, n$, set:

• $\Delta x = (b-a)/n$, $x_0 = a$, $x_i = x_0 + i\Delta x$, $i = 1, \ldots, n-1$, $x_n = b$.
• $\Delta y = (d-c)/m$, $y_0 = c$, $y_j = y_0 + j\Delta y$, $j = 1, \ldots, m-1$, $y_m = d$.
• Choose arbitrary $(x_{ij}^*, y_{ij}^*) \in [x_{i-1}, x_i] \times [y_{i -1}, y_i]$.

Definition. We say that $f(x,y)$ is integrable over $R$ if the following limit exists $$ \lim_{m, n \to \infty} \sum_{i = 1}^n \sum_{j = 1}^m f(x_{ij}^*, y_{ij}^*) \Delta A, $$ where $\Delta A = (\Delta x)(\Delta y)$. Denote it $\iint_R f(x, y) dA$.

Theorem. Suppose $f(x,y)$ is continuous over $R = [a, b] \times [c, d]$, then $f$ is integrable over $R$.

Difference between the lower and upper sums

Similar to the single variable case, one can show that as $n, m \to \infty$, the limit of the Riemann sums is independent of the choice of the sample points. To this send, define:

• $M_{ij} = \max \{f(x, y): x \in [x_{i-1}, x_i], y \in [y_{j-1}, y_j]\}$, and the corresponding sums $$ \sum_{i = 1}^n \sum_{j = 1}^m M_{ij}(\Delta x)(\Delta y) $$ is the Darboux upper sums.
• Similarly define $m_{ij} = \min \{f(x, y): x \in [x_{i-1}, x_i], y \in [y_{j-1}, y_j]\}$ and the lower sums by $$ \sum_{i = 1}^n \sum_{j = 1}^m m_{ij}(\Delta x)(\Delta y) .$$

Similarly to the single variable case, one can prove that the difference of the upper sums and the lower sums converges to $0$ as $n, m \to \infty$.

Proposition Suppose $f(x, y)$ is continous on $[a, b] \times [c, d]$, then $$ \lim_{n, m \to \infty} \left( \sum_{i = 1}^n \sum_{j = 1}^m M_{ij}(\Delta x)(\Delta y) - \sum_{i = 1}^n \sum_{j = 1}^m m_{ij}(\Delta x)(\Delta y) \right) = 0. $$

The idea of the proof is the following:

• For every $\epsilon >0$, there is $N \in \mathbb{N}$ such that if $n > N$, $M_{ij} - m_{ij} < \epsilon$ for all $i = 1, \ldots, n$, $j = 1, \ldots, m$. (This uses continuity, and more precisely uniform continuity)
• We have $$ \lim_{n, m \to \infty} \left( \sum_{i = 1}^n \sum_{j = 1}^m M_{ij}(\Delta x)(\Delta y) - \sum_{i = 1}^n \sum_{j = 1}^m m_{ij}(\Delta x)(\Delta y) \right) < \epsilon \sum_{i = 1}^n \sum_{j = 1}^m (\Delta x)(\Delta y) = \epsilon(b-a)(d -c). $$ Since $\epsilon$ is arbitrary, we obtain what we need.

Visually, this means the difference between the upper and lower sums is represented by the difference between the volumes, which converges to $0$ as seen in the picture.