Step Functions in Two Dimensions

References are to Salas & Hille, Calculus, 7th Edition

Some functions are so easy to integrate that we do not need calculus at all!

Step functions In the plane, let R be the rectangle

R: a ≤ x ≤ b, c ≤ y ≤ d, whose area is area(R) = (b − a)(d − c).

Definition 1 The function s(x, y), defined on R, is a step function on R if there exists a partition P = P_{1} × P_{2} of R, consisting of subdivision points

P_{1} = {x_{0}, x_{1}, …, x_{m}}, where a = x_{0} < x_{1} < x_{2} < … < x_{m−1} < x_{m} = b,

P_{2} = {y_{0}, y_{1}, …, y_{n}}, where c = y_{0} < y_{1} < y_{2} < … < y_{n−1} < y_{n} = d, (2)

such that s satisfies:

(i) For each y and i (1 ≤ i ≤ m), s(x, y) is constant for x_{i−1} < x < x_{i}; (ii) For each x and j (1 ≤ j ≤ n), s(x, y) is constant for y_{j−1} < y < y_{j}. (3)

The effect is to subdivide R into mn small rectangles R_{ij}, namely

R_{ij} : x_{i−1} ≤ x ≤ x_{i}, y_{j−1} ≤ y ≤ y_{j}, for each 1 ≤ i ≤ m and 1 ≤ j ≤ n.

Then s is constant, s(x, y) = s_{ij}, say, on the interior of each R_{ij}. (More accurately, we subdivide R into mn open rectangles, many open intervals, and (m+1)(n+1) grid points.) We ignore the values of s on the points and lines because these have area zero, and define the integral of s over R as

∫∫_{R} s(x, y)dx dy = ∑_{i=1}^{m} ∑_{j=1}^{n} s_{ij} area(R_{ij}) = ∑_{i=1}^{m} ∑_{j=1}^{n} s_{ij}(x_{i} − x_{i−1})(y_{j} − y_{j−1}). (4)

It is important to note that the definition of a step function does not specify which partition to use. There are many choices; the simplest is not always best.

Lemma 5 The integral (4) is independent of the choice of partition P.

Proof If the partition P works for s, we can always insert a new subdivision point x_{∗} between x_{k−1} and x_{k} to create a new partition P′ that also works for s. In effect, this breaks up the rectangle R_{kj}, for each j, into two smaller rectangles R′_{kj} and R′′_{kj}, say. Then in equation (4) for P, the term s_{kj} area(R_{kj}) is replaced for P′ by two terms s_{kj} area(R′_{kj}) + s_{kj} area(R′′_{kj}). Because area(R_{kj}) = area(R′_{kj}) + area(R′′_{kj}), the sum (4) is unaffected. We can do the same for the y-coordinate.

Given any two partitions P and P′ that work for s, we can make them coincide simply by adding enough new subdivision points to each. It follows that they yield the same value for the integral (4).

Remark Moreover, given two step functions s and t, there is a partition that works for both of them. We simply choose partitions that work for s and t separately, and form a new partition that includes all the subdivision points of either of these.

Elementary Properties These all follow easily from equation (4) (cf. p. 1046).

Proposition 6 Let s and t be step functions on R.

(a) For any constant k, ks is a step function on R, and

∫∫_{R} ks(x, y)dx dy = k ∫∫_{R} s(x, y)dx dy.  (7)

(b) The sum s + t is another step function on R, and

∫∫_{R} s(x, y) + t(x, y) dx dy = ∫∫_{R} s(x, y)dx dy + ∫∫_{R} t(x, y)dx dy.  (8)

(c) If s ≤ t on R, then

∫∫_{R} s(x, y)dx dy ≤ ∫∫_{R} t(x, y)dx dy.  (9)

(d) If we cut the rectangle R into two rectangles R′ and R′′ by the line x = e, where a < e < b (or the line y = f, where c < f < d), then

∫∫_{R} s(x, y)dx dy = ∫∫_{R′} s(x, y)dx dy + ∫∫_{R′′} s(x, y)dx dy.  (10)

Proof In (a), if we replace s by ks, each term in (4) is multiplied by k. Parts (b) and (c) are clear from (4) if we use the same partition for both s and t; according to the above Remark, we can always do this. In (d), we can make sure that e = x_{k} is one of the subdivision points; then the terms with i ≤ k in (4) for R form the sum (4) for R′, and the terms with i > k form the sum (4) for R′′.

Remark Equation (4) is a consequence of equation (10) and the basic statement that if s(x, y) = k whenever a < x < b and c < y < d, then

∫∫_{R} s(x, y)dx dy = k area(R) = k(b − a)(d − c).  (11)

Iterated integrals We express the double integral of s as an iterated integral. If we fix y, we note from (3) that s(x, y) is a step function in x. We put

S(y) = ∫_{a}^{b} s(x, y)dx = ∑_{i=1}^{m} s_{ij}(x_{i} − x_{i−1}),  if y_{j−1} < y < y_{j}.  (12)

This shows that S(y) is a step function in y.

Lemma 13 Given any step function s(x, y) on R, we define the step function S(y) on [c, d] by equation (12). Then

∫_{c}^{d} S(y)dy = ∫∫_{R} s(x, y)dx dy.  (14)

Proof We rewrite equation (4) as

∫∫_{R} s(x, y)dx dy = ∑_{j=1}^{n} {∑_{i=1}^{m} s_{ij}(x_{i} − x_{i−1})} (y_{j} − y_{j−1}) = ∫_{c}^{d} S(y)dy.

Alternatively, we could equally well integrate y first.