Fubini's theorem for iterated integrals

Partition $[a,b]$ into $m$ subintervals $[x_{i-1}, x_i]$ of width $\Delta x$ and $[c,d]$ into $n$ subintervals $[y_{j-1}, y_j]$ of width $\Delta y$. For each fixed $x$, define the cross-sectional integral $A(x) = \int_c^d f(x,y)\,dy$. By the Mean Value Theorem for integrals, on each strip $[y_{j-1}, y_j]$ there is $y_{ij}^* \in [y_{j-1}, y_j]$ such that $\int_{y_{j-1}}^{y_j} f(x_i, y)\,dy = f(x_i, y_{ij}^*)\,\Delta y$.

Summing over $j = 1, \dots, n$ gives $A(x_i) = \sum_{j=1}^n f(x_i, y_{ij}^*)\,\Delta y$. Multiplying by $\Delta x$ and summing over $i = 1, \dots, m$ yields $\sum_{i=1}^m A(x_i)\,\Delta x = \sum_{i=1}^m \sum_{j=1}^n f(x_i, y_{ij}^*)\,\Delta x\,\Delta y$. Because $f$ is uniformly continuous on the compact rectangle $R$, letting $\Delta x, \Delta y \to 0$ makes the left-hand side converge to $\int_a^b A(x)\,dx = \int_a^b \left(\int_c^d f(x,y)\,dy\right) dx$ while the right-hand side converges to $\iint_R f(x,y)\,dA$. Repeating the argument with $x$ and $y$ swapped establishes equality with the other order of integration.