The Riemann Integral in Two Dimensions

See also Step Functions in Two Dimensions, in this series. References are to Salas-Hille’s Calculus, 7th Edition.

Two separate tasks: we wish to define and compute the definite integral

∫∫_{R} f(x, y)dx dy

of a function f over the rectangle R given by a ≤ x ≤ b, c ≤ y ≤ d. The Riemann integral is based on two simple non-negotiable axioms:

(i) If f ≤ g on R, then ∫∫_{R} f(x, y)dx dy ≤ ∫∫_{R} g(x, y)dx dy; (ii) Some functions we already know how to integrate, namely step func- tions. (1)

If s and t are step functions on R such that

s(x, y) ≤ f(x, y) ≤ t(x, y) for all (x, y) ∈ R,  (2)

axiom (i) requires

∫∫_{R} s(x, y)dx dy ≤ ∫∫_{R} f(x, y)dx dy ≤ ∫∫_{R} t(x, y)dx dy,  (3)

and axiom (ii) specifies the two outer integrals. Moreover, we know that because s ≤ t, we have ∫∫_{R} s(x, y)dx dy ≤ ∫∫_{R} t(x, y)dx dy. The idea is that for favorable f, the inequality (3) is sufficient to determine the integral of f completely.

The Riemann integral

Definition 4 (cf. Defn. 16.2.3) Given a function f on R, we call f Riemannintegrable on R if there exists a unique number I such that

∫∫_{R} s(x, y)dx dy ≤ I ≤ ∫∫_{R} t(x, y)dx dy  (5)

whenever s and t are step functions that satisfy (2). If this is the case, we define ∫∫_{R} f(x, y)dx dy = I and call it the Riemann integral of f over R.

Note that f must be bounded or the definition breaks down; unless f is bounded below, s does not exist, and unless f is bounded above, t does not exist.

We have a squeeze principle: f is (Riemann-) integrable if and only if the difference

∫∫_{R} t(x, y)dx dy − ∫∫_{R} s(x, y)dx dy = ∑_{i=1}^{m} ∑_{j=1}^{n}(t_{ij} − s_{ij}) area(R_{ij})  (6)

can be made arbitrarily small for suitable choices of s and t. Here we find it convenient to use (as we may) the same partition P for both s and t. Moreover, it is not necessary to check all step functions.

Lemma 7 Suppose given a function f on R and a number I. Suppose there are step functions s and t that satisfy equations (2) and (5) and make the difference (6) arbitrarily small. Then f is integrable and ∫∫_{R} f(x, y)dx dy = I.

Elementary properties (cf. p. 1046) These all follow directly from the corresponding statements for step functions, with the help of Lemma 7.

Theorem 8 Let f and g be functions on the rectangle R.

(a) If f is integrable on R and k is constant, then kf is integrable on R and

∫∫_{R} kf(x, y)dx dy = k ∫∫_{R} f(x, y)dx dy.  (9)

(b) If f and g are integrable on R, so is f + g, and

∫∫_{R} f(x, y) + g(x, y) dx dy = ∫∫_{R} f(x, y)dx dy + ∫∫_{R} g(x, y)dx dy;  (10)

(c) If f and g are integrable on R and f(x, y) ≤ g(x, y) for all (x, y) ∈ R, then axiom (1)(i) holds,

∫∫_{R} f(x, y)dx dy ≤ ∫∫_{R} g(x, y)dx dy.  (11)

(d) If we slice the rectangle R into two rectangles R′ and R′′ by the line x = e or y = e, then f is integrable on R if and only if it is integrable on both R′ and R′′, and we have

∫∫_{R} f(x, y)dx dy = ∫∫_{R′} f(x, y)dx dy + ∫∫_{R′′} f(x, y)dx dy.  (12)

Evaluation Let us fix y and consider the integral with respect to x,

F (y) = ∫_{a}^{b} f(x, y)dx.  (13)

Theorem 14 (cf. (16.4.1)) Suppose f is integrable on R and that the integral (13) exists for all y (i. e. c ≤ y ≤ d). Then F is integrable on [c, d], and

∫_{c}^{d} F (y)dy = ∫∫_{R} f(x, y)dx dy.  (15)

Proof Choose step functions s and t such that s ≤ f ≤ t, and define

S(y) = ∫∫_{R} s(x, y)dx, T (y) = ∫∫_{R} t(x, y)dx.

Then S(y) ≤ F (y) ≤ T (y) for all y. But S and T are again step functions, and

∫_{c}^{d} S(y)dy = ∫∫_{R} s ≤ ∫∫_{R} f ≤ ∫∫_{R} t = ∫_{c}^{d} T (y)dy,

in abbreviated notation. But the difference

∫_{c}^{d} T (y)dy − ∫_{c}^{d} S(y)dy = ∫∫_{R}(t(x, y) − s(x, y)) dx dy

is arbitrarily small, and the squeeze principle (in one dimension) applies, to show that F is integrable and that equation (15) holds.