5.7 Integration in the plane  (Page 2/5)

As in the proof of [link] , we use the fact that a continuous function on a compact set is uniformly continuous.

For each positive integer $n,$ let ${\delta }_n$ be a positive number satisfying $\left|f\right(z)-f(w\left)\right|<1/n$ if $|z-w|<{\delta }_n.$ Such a ${\delta }_n$ exists by the uniform continuity of $f$ on $S.$ Because $S$ is compact, it is bounded, and we let $R=[a,b]\times [c,d]$ be a closed rectangle that contains $S.$ We construct a partition $\left\{S_i^n\right\}$ of $S$ as follows. In a checkerboard fashion, we write $R$ as the union $\cup R_i^n$ of small, closed rectangles satisfying

1. If $z$ and $w$ are in $R_i^n,$ then $|z-w|<{\delta }_n.$ (The rectangles are that small.)
2.   ${R_i^n}^0\cap {R_j^n}^0=\varnothing .$ (The interiors of these small rectangles are disjoint.)

Now define $S_i^n=S\cap R_i^n.$ Then ${S_i^n}^0\cap {S_j^n}^0=\varnothing ,$ and $S=\cup S_i^n.$ Hence, $\left\{S_i^n\right\}$ is a partition of $S.$

For each $i,$ choose a point $z_i^n$ in $S_i^n,$ and set $a_i^n=f\left(z_i^n\right).$ We define a step function $h_n$ as follows: If $z$ belongs to one (and of course only one) of the open geometric sets ${S_i^n}^0,$ set $h_n\left(z\right)=a_i^n.$ And, if $z$ does not belong to any of the open geometric sets ${S_i^n}^0,$ set $h_n\left(z\right)=f\left(z\right).$ It follows immediately that $h_n$ is a step function.

Now, we claim that $|f\left(z\right)-h_n\left(z\right)|<1/n$ for all $z\in S.$ For any $z$ in one of the ${S_i^n}^0$ 's, we have

because $|z-z_i^n|<{\delta }_n.$ And, for any $z$ not in any of the ${S_i^n}^0$ 's, $f\left(z\right)-h_n\left(z\right)=0.$ So, we have defined a sequence $\left\{h_n\right\}$ of step functions on $S,$ and the sequence $\left\{h_n\right\}$ converges uniformly to $f$ by [link] .

What follows now should be expected. We will define the integral of a step function $h$ over a geometric set $S$ by

We will define a function $f$ on $S$ to be integrable if it is the uniform limit of a sequence $\left\{h_n\right\}$ of step functions,and we will then define the integral of $f$ by

Everything should work out nicely. Of course, we have to check the same two consistency questions we had for the definition of the integral on $[a,b],$ i.e., the analogs of [link] and [link] .

Let $S$ be a closed geometric set, and let $h$ be a step function on $S.$ Suppose $P=\{S_1,...,S_n\}$ and $Q=\{T_1,...,T_m\}$ are two partitions of $S$ for which $h\left(z\right)$ is the constant $a_i$ on $S_i^0$ and $h\left(z\right)$ is the constant $b_j$ on $T_j^0.$ Then

We know by part (d) of [link] that the intersection of two geometric sets is itself a geometric set.Also, for each fixed index $j,$ we know that the sets $\{T_j\cap S_i^0\}$ are pairwise disjoint. Then, by [link] , we have that $A\left(T_j\right)={\sum }_{i=1}^{n}A(T_j\cap S_i).$ Similarly, for each fixed $i,$ we have that $A(S_i={\sum }_{j=1}^{m}A(T_j\cap S_i).$ Finally, for each pair $i$ and $j,$ for which the set $T_j^0\cap S_i^0$ is not empty, choose a point $z_{i,j}\in T_j^0\cap S_i^0,$ and note that $a_i=h\left(z_{i,j}\right)=b_j,$ because $z_{i,j}$ belongs to both $S_i^0$ and $T_j^0.$

With these observations, we then have that

which completes the proof.

OK, the first consistency condition is satisfied. Moving right along:

Let $h$ be a step function on a closed geometric set $S.$ Define the integral of $h$ over the geometric set $S$ by the formula

$${\int }_{S}h={\int }_{S}H\left(z\right)dz=\sum _{i=1}^n{a}_{i}A\left(S_i\right),$$

where $S_1,...,S_n$ is a partition of $S$ for which $h$ is the constant $a_i$ on the interior $S_i^0$ of the set $S_i.$

Just as in the case of integration on an interval, before checking the second consistency result, weneed to establish the following properties of the integral of step functions.