Evaluation of Double Integrals

The following Fubini-type theorem is fundamental to the evaluation of any Riemann double integral. Yet I have found it stated in this form only in the advanced text by Apostol (and even there, only with a complicated proof of a special case). It is in fact quite easy, and is a result in pure integration (or measure) theory, with no mention of continuity; it is obvious for step functions. We sketch a direct proof below. Everything here generalizes (with a little care) to three dimensions, or even n dimensions.

Theorem 1 Assume that the function f(x, y) is integrable over the rectangle R defined by a ≤ x ≤ b, c ≤ y ≤ d in the xy-plane.

(a) We have

∫∫_{R} f(x, y)dxdy = ∫_{c}^{d} F (y)dy = ∫_{c}^{d} {∫_{a}^{b} f(x, y)dx} dy,

provided that the inner integral

F (y) = ∫_{a}^{b} f(x, y)dx

exists for all y (c ≤ y ≤ d). In particular, the function F is integrable.

(b) We have

∫∫_{R} f(x, y)dxdy = ∫_{a}^{b} G(x)dx = ∫_{a}^{b} {∫_{c}^{d} f(x, y)dy} dx,

provided that the inner integral

G(x) = ∫_{c}^{d} f(x, y)dy

exists for all x (a ≤ x ≤ b). In particular, the function G is integrable.

Examples show that the integrability hypothesis on f is essential. For this result to be useful, we need a supply of integrable functions.

Theorem 2 Suppose that the function f(x, y) is bounded on the rectangle R, and is continuous except on a set E with area zero. Then f is integrable over R.

We do not give the proof, which is quite advanced even if we assume that f is continuous everywhere. And this special case fails to cover many essential applications, such as the following corollary.

Corollary 3 Let Ω be a bounded region in the plane. If the set of boundary points of Ω has area zero, then the area of Ω is defined.

Proof We apply Theorem 2 to the characteristic function χ_{Ω} of Ω.

In view of this corollary, Theorem 2 does look rather circular. To make it useful, we need a supply of sets that are known to have area zero. Of course, the union of any two (or finitely many) sets of area zero also has area zero.

Theorem 4 The following sets have area zero:

(i) The graph y = f(x) of an integrable function on the interval a ≤ x ≤ b; (ii) The graph x = g(y) of an integrable function on the interval c ≤ y ≤ d; (iii) The image curve of a function r(t) that is continuously differentiable on the

interval a ≤ t ≤ b.

We omit the proofs. Parts (i) and (ii) are again pure integration theory, but somewhat messier, and are closely related to the following result.

Theorem 5 Suppose that f(x) is integrable on the interval a ≤ x ≤ b, and that f(x) ≥ 0 for all x. Then the region Ω under the graph of f, defined by 0 ≤ y ≤ f(x) and a ≤ x ≤ b, has area ∫_{a}^{b} f(x)dx.

Once we know that the area exists, it is readily computed by Theorem 1.

Sketch proof of Theorem 1(a) We have to show that F is integrable, and that its integral is I = ∫ ∫_{R} f(x, y)dxdy.

So take any Riemann sum

∑_{j=1}^{n} F (y_{j}^{∗})(y_{j} − y_{j−1})  (6)

for the integral of F, where c = y_{0} < y_{1} < … < y_{n} = d is a partition (or subdivision) of [c, d] and y_{j−1} ≤ y_{j}^{∗} ≤ y_{j} for each j. We also choose a partition a = x_{0} < x_{1} < … < x_{m} = b of [a, b], thus dividing the rectangle R into little rectangles R_{ij} as usual.

For each j, we choose points P_{ij} ∈ R_{ij} having y_{j}^{∗} as the y-coordinate; then

∑_{i=1}^{m} f(P_{ij})(x_{i} − x_{i−1})  (7)

is a Riemann sum for the integral

∫_{a}^{b} f(x, y_{j}^{∗})dx = F (y_{j}^{∗}),

and will therefore be close to F (y_{j}^{∗}) if the x-partition is fine enough. This holds for all j simultaneously, if the x-partition is fine enough to handle all the integrals F (y_{j}^{∗}); this works because we are considering here only finitely many integrals.

Now multiply (7) by y_{j} − y_{j−1} and add. Then

∑_{j=1}^{n} ∑_{i=1}^{m} f(P_{ij})(x_{i} − x_{i−1})(y_{j} − y_{j−1})  (8)

will be close to (6). But (8) is a Riemann sum for the double integral I, and is therefore close to I if the partitions of [a, b] and [c, d] are fine enough. So (6) is close to (8) which is close to I, as required.

As it stands, this is not a formal proof, but can easily be made so. One has to insert δ and ϵ in all the right places, and check that the various choices are made in a legitimate order. The method also works well if the definite integral is defined in terms of upper and lower sums instead of Riemann sums.