Double Integrals and Fubini's Theorem

A single integral adds up contributions along an interval. Replace the interval with a region of the plane and the construction is unchanged: chop, multiply each piece by a height, add, take a limit. Area elements stand where widths stood.

That definition is exact and almost useless for computing. A two-dimensional limit over shrinking rectangles is not something to evaluate by hand.

What makes it practical is a theorem. Fubini's theorem says the two-dimensional limit equals two ordinary one-dimensional integrals performed in sequence, and over a rectangle with a continuous integrand, in either order. So ${\int }_0^2{\int }_0^3(2x+y)dydx=21$, and reversing the order gives 21 as well.

The freedom to choose is a property of the rectangle. Constant limits are what let the orders be exchanged as written. Over a region whose boundary bends, the inner limits depend on the outer variable, and the two orders sweep the region differently, in vertical strips or in horizontal ones. Exchanging the written limits there describes a different region and produces a different number.

So this unit has two things to get right: the mechanics of holding one variable fixed while integrating the other, and the question of when the order is actually free.

The canonical definition states the double integral as a limit and Fubini's theorem as its evaluation. What follows is why the theorem carries a hypothesis and what the hypothesis is doing.

The two sides are different objects. ${\iint }_{R}fdA$ is a single limit over shrinking two-dimensional pieces, taken without reference to any order. An iterated integral ${\int }_c^d{\int }_a^{b}fdxdy$ is two one-dimensional limits performed in sequence. That these coincide is a theorem, not a restatement, and it can fail.

Why continuity suffices. On a closed bounded rectangle a continuous function is uniformly continuous and bounded, so the Riemann sums converge and the order in which the two limits are taken cannot change the value. Absolute integrability, , is the more general condition, and it is what the theorem actually needs; continuity is the convenient sufficient case.

What goes wrong without it. The standard counterexample takes $f(x,y)={\displaystyle \frac{x^2-y^2}{(x^2+y^2{)}^2}}$ on $[0,1]\times [0,1]$, where the two iterated integrals are $\pi /4$ and $-\pi /4$. Both exist, they differ in sign, and the double integral does not exist at all, because $\iint |f|dA$ diverges near the origin. Nothing in the arithmetic of either iterated integral signals the problem; only checking the hypothesis does.

Why the inner integration treats the outer variable as a constant. Fixing $y$ reduces $f(x,y)$ to a function of $x$ alone, and ${\int }_a^{b}f(x,y)dx$ is then an ordinary integral whose value depends on the $y$ that was fixed. Carrying out that integration for every $y$ produces a function of $y$, which the outer integral then integrates. The outer variable's symbols pass through untouched precisely because they are constants for the duration.

Why a non-rectangular region breaks the symmetry. Constant limits make the region a product of two intervals, so describing it by $x$ then $y$ or by $y$ then $x$ gives the same set. When the inner limits are functions of the outer variable the region is not a product, and the two descriptions are genuinely different decompositions. The order may then be forced: one may require splitting the region into pieces where the other does not.

A double integral, both orders: ${\displaystyle {\int }_0^2{\int }_0^3(2x+y)dydx}$.

Inner first, over $y$ with $x$ held:

Then over $x$:

Reversing the order: ${\int }_0^2(2x+y)dx={[x^2+xy]}_0^2=4+2y$, then ${\int }_0^3(4+2y)dy={[4y+y^2]}_0^3=12+9=21$.

Check: a $1500\times 1500$ midpoint sum. Fubini's theorem is what guarantees the two orders agree, and the integrand is continuous on the rectangle, so its hypothesis holds.

Four cases, chosen because each one settles a different question about what the order and the region cost.

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1. A separable integrand, where the double integral is a product. Over $R=[0,2]\times [0,3]$ with $f(x,y)=x{y}^2$, the inner integration carries the $x$ through as a constant:

But because the integrand factors as $g(x)h(y)$ and the limits are constants, the whole thing factors too:

The factorisation needs both conditions. ${\int }_0^2{\int }_0^3(x+y^2)dydx=24$, and the corresponding product of separate integrals is $(2)(9)=18$, which is not it: a sum does not factor. And over a non-rectangular region the limits carry the outer variable, so even a product integrand does not separate.

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2. A region where the order decides how much work it is. Consider ${\displaystyle {\int }_0^1{\int }_y^1{e}^{x^2}dxdy}$.

As written, the inner integral $\int e^{x^2}dx$ has no elementary antiderivative, and the calculation stops. Reversing the order rescues it. The region is $\{(x,y):0\le y\le 1,y\le x\le 1\}$, which is the triangle below the line $y=x$; described the other way, $x$ runs over $[0,1]$ and for each $x$, $y$ runs from $0$ to $x$:

The inner integration produced the factor $x$ that the substitution needed. Both orders are valid and give the same number; only one is computable in closed form. Check: a $3000\times 3000$ midpoint sum over the triangle returns $0.85914\dots$.

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3. A region needing two pieces in one order and one in the other. Let $S$ be bounded by $y=x$, $y=2-x$ and $y=0$: a triangle with vertices $(0,0)$, $(1,1)$, $(2,0)$.

Sweeping in vertical strips, the upper boundary changes at $x=1$, so $dydx$ needs two integrals:

Sweeping in horizontal strips, each strip at height $y$ runs from $x=y$ to $x=2-y$ throughout, so $dxdy$ needs one:

Both give the triangle's area of 1. The choice of order is not about correctness; it is about how many pieces the region breaks into.

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4. Where Fubini has nothing to say. Over $[0,1]\times [0,1]$ take

The two iterated integrals both exist and come to $\pi /4$ and $-\pi /4$. They disagree, and the double integral does not exist at all, because $\iint |f|dA$ diverges near the origin where $f$ is unbounded.

Nothing in the arithmetic of either iterated integral signals this. Each is a perfectly ordinary calculation returning a finite number. Only checking the hypothesis does, and the hypothesis fails because $f$ is not continuous on the closed rectangle and not absolutely integrable over it. This is the case that makes Fubini a theorem rather than a convention.

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What the four have in common. The region and the integrand together decide what is possible: whether the integral factors, whether either order is computable, how many pieces are needed, and whether the orders agree at all. None of that is visible from the notation ${\iint }_{R}fdA$, which is why the first step is always to describe the region and check the integrand rather than to start integrating.

To evaluate a double integral over a rectangle.

1. Choose an order. Over a rectangle either works, by Fubini.
2. Integrate the inner variable, holding the outer one fixed as a constant, its symbols pass through untouched.
3. Substitute the inner limits, leaving an expression in the outer variable alone.
4. Integrate that over the outer limits.

Check: redo it in the opposite order. Agreement is a strong check on both, and disagreement over a rectangle with a continuous integrand means an arithmetic slip rather than a failure of Fubini.

Over a non-rectangular region the inner limits are functions of the outer variable, and swapping the order means redescribing the region, not exchanging the limits as written.

Swapping the limits of a non-rectangular double integral. Over a rectangle the limits are constants and either order works. Over a region where the inner limits depend on the outer variable, exchanging the order requires describing the region the other way round, reversing the written limits produces a different region and a different number.

Treating Fubini as unconditional. Over a rectangle with a continuous integrand the two orders agree, and disagreement means an arithmetic slip. That is a consequence of the hypothesis, not a general fact about integration: without continuity or absolute integrability the two iterated integrals can genuinely differ, and the two-dimensional integral may not exist at all. Quoting "either order works" without the rectangle and the continuity is quoting a conclusion without its premise.

Reading the inner limits as fixed when they are not. Over a region bounded by $y=x^2$ and $y=x$, the inner limits in $\int \int dydx$ run from $x^2$ to $x$, both functions of the outer variable. Substituting them leaves an expression in $x$ alone, which is the point. Writing constant inner limits there integrates over a rectangle the region does not fill.

Mass, probability and marginalisation. Mass is ${\iint }_R\rho dA$ for a density $\rho$; probability for a joint distribution is ${\iint }_{R}f(x,y)dA$; and the requirement that a joint density integrate to 1 over the plane is a double improper integral. Marginalising one variable out is the inner integration of an iterated integral, so Fubini is what licenses computing a marginal distribution by integrating the joint one.

Why the order matters in practice. Marginalising a joint density integrates out one variable, and which one is integrated first is the difference between obtaining $f_X$ and $f_Y$. Over a rectangle either is available; over a region whose boundary bends, one order may need the region split into pieces while the other does not, which is the usual reason for choosing.

The through-line. A single integral accumulates along an interval. A double integral accumulates over a region, and Fubini is what makes that accumulation computable by two ordinary integrations rather than a limit nobody could evaluate directly.