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If I have a Riemann integrable function $f: [a,b] \to R$, then the following is true:

$$\frac{b-a}n \sum_{r=1}^n f(a + \frac{r(b-a)}n) \longrightarrow \int_a^b f(x)dx$$

is there an analogue for this when considering double integrals? Say I have a function $f$ Riemann integrable on a rectangle $[a,b] \times [c,d]$, can this be expressed as the limit of some sum?

$$\iint_{[a,b] \times [c,d]} f(x,y)dxdy = \lim_{n \to \infty} (?)$$

Maybe something like this?

$$\lim_{n \to \infty} \frac{(d-c)(b - a)}{n^2} \sum_{r,t=1}^n f(a + \frac{r(b-a)}n, c + \frac{t(d-c)}n)$$

using say Fubini's theorem? I am not very sure

Topological cat
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1 Answers1

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Yes, if $ \iint_{[a,b]\times [c,d]} f(x,y) dA$ exists. This is just a hint for you to work. Let $\mathscr{A}_n=\{a=x_0<x_1<\ldots<x_n=b\}$ and $\mathscr{C}_n=\{c=x_0<y_1<\ldots<y_n=d\}$. Let $x_i^\ast\in I_i=[x_{i-1},x_{i}]$ and $y_j^\ast\in J_j=[y_{j-1},x_{j}]$.

Fix \begin{align} B_1=&I_1\times J_1, \ldots, B_n=I_1\times J_n; \\ B_{1\cdot n+1}=&I_2\times J_1,\ldots,B_{1\cdot n+n}=I_2\times J_n; \\ \vdots\quad & \quad \quad\vdots \\ B_{(n-1)\cdot n+1}=&I_{n}\times J_1,\ldots,B_{(n-1)\cdot n+n}=I_n\times J_n \end{align} Realize that $$ \mathscr{P}_{n^2}=\left\{ B_{k}\left| B_{k}=[x_{i-1},x_{i}]\times [y_{j-1},x_{j}], \begin{array}{c}1\leq k\leq n^2 \\ 1\leq i\leq n\\ 1\leq j\leq n\end{array}\right., k=(j-1)n+i\right\}. $$ is a partition of $B=[a,b]\times [c,d]$. Now set $x_i^\ast=x_i=a + \frac{i(b-a)}{n}$, $y_j^\ast= y_j=c + \frac{j(d-c)}{n}$ and $$ p_k^\ast=p^\ast_{(j-1)n+i}=(x_i^\ast,y_j^\ast)=(a + \frac{i(b-a)}n, c + \frac{j(d-c)}n) $$

Now note that: \begin{align} \iint_{B} f(x,y)\;\mathrm{d} A =& \lim_{\|\mathscr{P}_{n^2}\|\to 0} \sum_{k=1}^{n^2} f(p_k^\ast)\cdot \mathrm{area}(B_k) \\ = & \lim_{n^2\to \infty } \sum_{k=0}^{n^2} f(p_k^\ast)\cdot \mathrm{area}(B_k) \\ =& \lim_{n \to \infty} \frac{(d-c)(b - a)}{n^2} \sum_{i=1}^n\sum_{j=1}^n f(a + \frac{i(b-a)}n, c + \frac{j(d-c)}n) \end{align} What can you say about the last equality?

Elias Costa
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