Double-integral depth: Darboux construction and a complete continuous Fubini proof

Learning goals and exact theorem class

Define a proper double integral through upper and lower rectangular sums, prove continuous integrability, and prove Fubini's theorem for continuous functions on a compact rectangle. Then examine the extra boundary argument needed for ordinary nonrectangular regions.

Let $R=[a,b]\times[c,d]$ with $a<b,c<d$. A grid consists of partitions of each coordinate interval. Its cells are closed subrectangles with disjoint interiors. For bounded $f:R\to\mathbb R$, let $m_{ij},M_{ij}$ be its infimum and supremum on cell $R_{ij}$, of area $\Delta A_{ij}$. Define $L(f,P)=\sum m_{ij}\Delta A_{ij}$ and $U(f,P)=\sum M_{ij}\Delta A_{ij}$.

1. The definition and its useful criterion

Refining a grid raises lower sums and lowers upper sums. Given any two grids, their common refinement proves every lower sum is at most every upper sum. Thus the lower integral $\underline I=\sup_P L(f,P)$ is at most the upper integral $\overline I=\inf_P U(f,P)$, and both are finite by boundedness. The function is Riemann integrable when these numbers agree, with common value $\iint_R f$.

Theorem: rectangular Darboux criterion

The bounded function is integrable if and only if for every $\varepsilon>0$ some grid has $U(f,P)-L(f,P)<\varepsilon$.

Proof. If both integrals equal $I$, choose a lower sum above $I-\varepsilon/2$ and an upper sum below $I+\varepsilon/2$, then refine both grids. The new gap is below epsilon. Conversely any such grid bounds $0\le\overline I-\underline I\le U-L$. Arbitrarily small gaps force equality. The proof is the supremum argument in two dimensions, not an assumption of Fubini's theorem.

Theorem: continuity implies proper integrability

If $f$ is continuous on $R$, it is bounded and uniformly continuous there. Write $A=(b-a)(d-c)>0$. Choose a distance scale making output differences less than $\varepsilon/(2A)$. A sufficiently fine grid has every cell diameter smaller than that scale, so each oscillation $M_{ij}-m_{ij}$ is at most $\varepsilon/(2A)$. Therefore $U-L\le\varepsilon/2<\varepsilon$, proving integrability.

For any tags $\xi_{ij}\in R_{ij}$, the corresponding finite sum lies between the grid's lower and upper sums. Thus whenever the grid gap is small, its tagged approximation is close to the integral. This statement is a consequence of the definition, not a new limiting convention.

2. Prove the continuous rectangle Fubini theorem

Theorem

For continuous $f$ on $R$,

$$\iint_R f(x,y)\,dA =\int_a^b\left(\int_c^df(x,y)\,dy\right)dx =\int_c^d\left(\int_a^bf(x,y)\,dx\right)dy.$$

First, the iterated integrals exist. Each one-variable slice is continuous and integrable. Let $g(x)=\int_c^df(x,y)dy$. Uniform continuity of $f$ gives

$$|g(x)-g(x_0)|\le(d-c)\sup_{y\in[c,d]}|f(x,y)-f(x_0,y)|\longrightarrow0.$$

Hence $g$ is continuous and integrable. The reversed inner-integral function is continuous by the same argument.

Next, squeeze them between the two-dimensional sums. On every cell, $m_{ij}\le f(x,y)\le M_{ij}$. For a fixed $x$ in its $i$th strip, integrate these bounds over each $y$ interval and add:

$$\sum_jm_{ij}\Delta y_j\le g(x)\le\sum_jM_{ij}\Delta y_j.$$

Integrate over that $x$ strip and sum over $i$. The result is $L(f,P)\le\int_a^bg(x)dx\le U(f,P)$. The double integral lies between the same bounds. Grids with arbitrarily small gap therefore force equality. Repeating with coordinates reversed proves the second equality. No order of an unproved infinite double sum was exchanged; only finite sums, ordinary integral inequalities and the Darboux criterion were used.

A separable continuous integrand $f(x,y)=p(x)q(y)$ consequently has integral $(\int p)(\int q)$. This is an integration identity under the stated hypotheses, not a claim about probabilistic independence in an unspecified experiment.

3. Why ordinary curved boundaries do not destroy proper integrability

Let $D=\{a\le x\le b,\ell(x)\le y\le u(x)\}$ with continuous $\ell\le u$, and let $f$ be continuous on $D$. This is a compact set. Choose a bounding rectangle $R=[a,b]\times[c,d]$ with $c\le\ell(x)\le u(x)\le d$. The clamping map

$$p(x,y)=\bigl(x,\min(u(x),\max(\ell(x),y))\bigr)$$

is continuous, maps $R$ into $D$ and fixes every point of $D$. Consequently $\widetilde f=f\circ p$ is a continuous extension to $R$. In the estimates below write $f$ for this extension. Now set it to zero outside $D$ within $R$. This zero extension may jump across the boundary, so one cannot cite continuity everywhere.

Instead control boundary cells. In a vertical grid strip, uniform continuity makes the oscillation of either boundary graph as small as desired. The cells meeting that graph can be covered by a rectangle whose height is at most that oscillation plus two vertical mesh widths. Summing strip widths shows the total area of these graph-boundary cells tends to zero as both mesh widths tend to zero. The two vertical end boundaries can likewise be covered by strips of arbitrarily small total area.

If $|f|\le M$, boundary cells contribute at most $2M$ times their total area to the upper-lower gap. Cells away from the boundary lie wholly inside or wholly outside the region; inside, uniform continuity of $f$ makes their oscillations small, and outside the function is zero. Choose the boundary area small first and then the ordinary cell oscillation small. The total gap becomes arbitrarily small, proving integrability of the zero extension and therefore of $f$ over $D$.

The same bounds show that its integral is $\int_a^b\int_{\ell(x)}^{u(x)}f(x,y)dydx$: on each strip, changing the vertical endpoints and the $x$ input changes the inner integral by uniformly small amounts, while the vanishing boundary-cell area controls the endpoints. Explicitly, using the bounding rectangle's vertical span $H$,

$$|g(x)-g(x_0)|\le H\sup_y|f(x,y)-f(x_0,y)|+M(|\ell(x)-\ell(x_0)|+|u(x)-u(x_0)|),$$

so the inner-integral function is continuous, and the finite-strip Darboux squeeze applies after excluding the already controlled boundary cells. This establishes the common continuous-simple-region case. More general region or improper versions require their own hypotheses and are not inferred from this proof.

4. Quantitative grids and bounded discontinuities

If $f$ is Lipschitz with constant $K$ on the unit square, each cell of an $n$-by-$n$ grid has oscillation at most $K\sqrt2/n$. Since total area is one, $U-L\le K\sqrt2/n$. Thus a mesh can be selected from a desired error rather than chosen by trial. For $f=x+y$, the exact gap is $2/n$, agreeing with $K=\sqrt2$.

A bounded function changed at finitely many points has the same integral: cover those points by grid cells of arbitrarily small total area and bound their contribution by the finite uniform magnitude. But unbounded singular behavior near a single point is not covered by that argument. The rational improper counterexample in the coordinate/slicing extension has unequal iterated values precisely because the required bounded/absolute-integrability control fails.

Graded construction and proof tasks

1. Definition. Prove a common grid refinement makes every lower sum less than or equal to every upper sum. Why is boundedness needed before taking the real supremum and infimum?
2. Existence before interchange. In the Fubini proof, explain why slice integrability alone would not establish the existence of the outer integral. Supply the uniform-continuity estimate that does.
3. Error certificate. For a Lipschitz function on the unit square, give a sufficient integer mesh size making the upper-lower gap below epsilon. Check the exact gap for $f=x+y$.
4. Counterexample. Let $f=1$ when both coordinates are rational and zero otherwise on the unit square. Prove it is bounded but not Riemann integrable. Do the continuous Fubini hypotheses hold?
5. Boundary logic. Why is extending by zero across a curved boundary not generally continuous? Explain how the boundary-cell area estimate repairs the proper-integrability proof.
6. Finite alteration. Prove changing a continuous rectangular integrand at finitely many points leaves its double integral unchanged. Explain why the same sentence cannot be used to dismiss an unbounded puncture-neighborhood singularity.

Full solutions

1. Add all coordinate partition points to get a common grid. Refinement raises the first lower sum and lowers the second upper sum, while on that grid every cell infimum is at most its supremum. Thus $L(f,P)\le U(f,Q)$ for arbitrary grids. A finite bound on $|f|$ bounds every sum in magnitude by that bound times rectangle area, ensuring the extrema used to define the lower and upper integrals are finite real numbers.
2. The outer integrand is the slice integral $g(x)$, and arbitrary collections of integrable slices need not yield an integrable outer function. Here $|g(x)-g(x_0)|\le(d-c)\sup_y|f(x,y)-f(x_0,y)|$, and compact uniform continuity makes that supremum tend to zero. Thus the outer integrand is continuous, supplying its integrability before an equality is asserted.
3. Choose an integer $n>K\sqrt2/\varepsilon$; if $K=0$, any grid has zero gap. For $x+y$, each uniform cell has range width $2/n$ and area $1/n^2$. Summing $n^2$ cells gives exact gap $2/n$, matching the displayed bound with $K=\sqrt2$.
4. Every nondegenerate cell contains a pair of rational coordinates, giving supremum one, and also a point with an irrational coordinate, giving infimum zero. Hence every grid has upper sum one and lower sum zero. The criterion fails. The function is not continuous; at rational fixed $x$, its $y$ slice is the nonintegrable rational indicator, so the existence issue is visible even before attempting the outer integral.
5. A nonzero boundary value inside can approach a zero exterior value across the boundary, causing a jump. The proof does not pretend otherwise: it confines all such jumps to boundary cells of arbitrarily small total area, whose gap contribution is bounded by $2M$ times that area. Everywhere else ordinary uniform continuity or the zero value controls oscillation. Both ingredients are needed.
6. The difference is bounded and supported on finitely many points. For any positive area target, choose sufficiently fine coordinate intervals around those points; the finite union of affected grid cells has total area below that target. Its upper and lower integral magnitudes are at most the difference bound times that area, forcing integral zero. An unbounded singularity has no common bound $M$ on those shrinking neighborhoods, so that estimate cannot be applied and no conclusion follows from the exceptional point's area alone.

Summary and terms

A rectangular Darboux integral is defined by coinciding upper and lower sums. A common refinement compares unrelated grids. A boundary-cell estimate controls bounded jumps near ordinary region boundaries. The continuous Fubini theorem follows after proving both double and iterated integrals exist. Improper singularities and more general domains need additional hypotheses, not an automatic extension of a finite-rectangle proof.