Double integrals and iterated integration

Fix an integer $n\ge1$ and the Euclidean norm on $\mathbb R^n$. A set is bounded if it lies in some finite-radius ball, closed if its complement is open, and compact if every cover by open subsets of $\mathbb R^n$ has a finite subcover. We use real completeness, including the nested closed interval property.

Heine–Borel in finite-dimensional Euclidean space

Theorem. A set $K\subseteq\mathbb R^n$ is compact if and only if it is closed and bounded. The empty set satisfies all three properties.

Compact implies bounded. The open balls $B(0,m)$, $m\ge1$ an integer, cover $K$. A finite subcover puts $K$ inside the largest of these balls.

Compact implies closed. Suppose $K\ne\varnothing$ and $y\notin K$. The balls $B(x,\|x-y\|/2)$, $x\in K$, cover $K$. Choose a finite subcover with centers $x_1,\ldots,x_r$ and set $\delta=\min_i\|x_i-y\|/2>0$. If $z$ lies in the ball centered at $x_i$, the triangle inequality gives $\|z-y\|>\|x_i-y\|/2\ge\delta$. Hence $B(y,\delta)$ misses $K$, so the complement is open.

Closed and bounded implies compact. Let nonempty $K$ be closed and bounded. Enclose it in a closed cube $Q_0=[-M,M]^n$ with $M>0$. Suppose an open cover of $K$ has no finite subcover. Divide each coordinate interval into two closed halves. The resulting $2^n$ closed subcubes cover $Q_0$. At least one subcube has intersection with $K$ that is not finitely coverable by the given family; otherwise the finite union of the finite subcovers would cover $K$. Select such a subcube and repeat. This produces nested closed cubes $Q_j$ with side length $2M/2^j$ and $K\cap Q_j$ not finitely coverable, hence nonempty.

For each coordinate, the nested interval property gives a common coordinate value. Together these give $p\in\bigcap_jQ_j$. Every pair of points in $Q_j$ has distance at most $2M\sqrt n/2^j$, so the intersection contains only $p$. Choose $x_j\in K\cap Q_j$. Then $\|x_j-p\|\le2M\sqrt n/2^j\to0$. A closed set contains such a limit: if $p$ were in its open complement, a ball around $p$ would eventually contain $x_j$ outside $K$, a contradiction. Thus $p\in K$.

One member $O$ of the cover contains $p$, and openness gives $B(p,\varepsilon)\subseteq O$ for some $\varepsilon>0$. For sufficiently large $j$, every point of $Q_j$ is within $\varepsilon$ of $p$, so $O$ alone covers $K\cap Q_j$. This contradicts the selection of $Q_j$. Therefore every open cover has a finite subcover. The number $2^n$ is finite; replacing it by infinitely many subpieces would invalidate this argument.

Continuous images and extreme values

Theorem. If $K\subseteq\mathbb R^n$ is compact and $F:K\to\mathbb R^m$ is continuous relative to $K$, then $F(K)$ is compact. In particular, a continuous $f:K\to\mathbb R$ on nonempty compact $K$ is bounded and attains its maximum and minimum.

Proof. An open cover $(O_i)$ of $F(K)$ pulls back to relatively open sets $F^{-1}(O_i)=K\cap U_i$, with $U_i$ open in $\mathbb R^n$. These $U_i$ cover $K$, so finitely many suffice. Their corresponding $O_i$ cover $F(K)$, proving compactness. For scalar $f$, the image is nonempty, closed and bounded by the theorem already proved in dimension one. Let $a=\sup f(K)$. For each $j\ge1$, the definition of supremum supplies $y_j\in f(K)$ with $a-1/j<y_j\le a$. Then $y_j\to a$, and closedness places $a$ in the image. Thus the maximum is attained. Applying the same argument to $-f$ gives the minimum. This proves existence, not a formula for the maximizing input.

Compactness gives boundedness and uniform continuity

Let $K\subset\mathbb R^n$, $n\ge1$, be compact, and let $u:K\to\mathbb R$ be continuous relative to $K$. Compactness means the finite-subcover property. The empty set is harmless. At each $p\in K$, continuity supplies $r_p>0$ such that $|u(q)-u(p)|<1$ for $q\in K\cap B(p,r_p)$. A finite selection of these balls covers $K$; the largest of the finitely many numbers $|u(p)|+1$ bounds $|u|$.

For $\varepsilon>0$, instead choose $r_p>0$ such that $|u(q)-u(p)|<\varepsilon/2$ whenever $q\in K\cap B(p,r_p)$. Finitely many balls $B(p_i,r_{p_i}/2)$ cover $K$. Put $\delta=\min_i r_{p_i}/2>0$. If $x,y\in K$ and $\|x-y\|<\delta$, choose $i$ with $\|x-p_i\|<r_{p_i}/2$. Then both $x,y$ belong to $B(p_i,r_{p_i})$, so $|u(x)-u(y)|<\varepsilon$. This proves uniform continuity using the relative topology; no extension off $K$ was used. Closed bounded subsets of finite-dimensional Euclidean space are compact by the separately included Heine–Borel proof. This argument neither uses integration nor assumes a Fubini theorem.

Product-grid Darboux criterion and arbitrary rectangular tags

Let $R=[a,b]\times[c,d]$, with finite $a<b$, $c<d$, and let $f:R\to\mathbb R$ satisfy $|f|\le M<\infty$. A product grid consists of partitions $a=x_0<\cdots<x_r=b$, $c=y_0<\cdots<y_s=d$. Its cells $C_{ij}$ have areas $A_{ij}=\Delta x_i\Delta y_j$. Define $m_{ij}=\inf_{C_{ij}}f$, $M_{ij}=\sup_{C_{ij}}f$, and

Each tagged sum lies between these two numbers. Refinement increases $L$ and decreases $U$, because the subcell areas sum to the parent area. Two grids have a common refinement obtained by taking both sets of coordinate cuts. Consequently $L(f,P)\le U(f,Q)$ for all grids. Real completeness defines $I_-=\sup_P L(f,P)\le I_+=\inf_P U(f,P)$. These are finite. They coincide with a number $I$ if and only if, for every $\varepsilon>0$, some grid has $U-L<\varepsilon$. Indeed that gap bounds $I_+-I_-$; conversely choose an upper sum within $\varepsilon/2$ of $I_+$, a lower sum within $\varepsilon/2$ of $I_-$, and refine both.

Here is the bridge to arbitrary subrectangles in the original definition. A rectangular tiling is a finite cover of $R$ by closed axis-parallel positive-area rectangles with disjoint interiors; common edges may overlap. Its mesh is the largest cell diameter. Its lower and upper sums are defined in the same way. Taking every cell-edge coordinate gives a product-grid refinement, so its lower sum is at most $I$, and its upper sum is at least $I$.

Fix a product grid $P$ with gap less than $\varepsilon/2$. Let it have $v$ internal vertical lines and $h$ internal horizontal lines. In any tiling $Q$ of mesh less than $\delta$, call a cell bad if it meets one of these lines. A bad cell lies in a strip of width $2\delta$ about a line. Thus the sum of the bad-cell areas is at most

This is an elementary finite-area estimate: refining all rectangle and strip edges reduces it to addition of disjoint cell areas, without using an integral. Every other cell is contained in one cell of $P$. Its contribution to the oscillation sum is bounded by that parent-cell oscillation times its area; summing the good cells gives at most $U(f,P)-L(f,P)$. On bad cells the oscillation is at most $2M$. Hence

Choose $\delta$ so the last term is less than $\varepsilon/2$; when its coefficient is zero there is no restriction from that term. Every tag choice then gives a sum within $\varepsilon$ of $I$. This proves convergence for every sufficiently fine rectangular tiling, not only for grid refinements.

Conversely, if all tagged sums on sufficiently fine rectangular tilings approach $I$, this holds for product grids. On a fixed fine grid choose each tag within $\eta/|R|$ of its cell supremum in function value; the resulting sum differs from $U$ by at most $\eta$. Such tags exist by the definition of supremum; use infima for $L$. The assumed uniform control of every tag choice, followed by $\eta\downarrow0$, makes $U-L$ arbitrarily small. Thus the bounded-function Darboux criterion and the original arbitrary-diameter tagged definition agree.

The one-dimensional version is proved by the same construction with intervals and lengths. Explicitly, refine at the union of cuts; a fine interval that meets one of $q$ fixed interior cuts contributes to bad total length at most $2q\delta$. The same $2M$-oscillation estimate proves equivalence with all fine interval tags, and near-supremum/infimum tags give the converse. We use this proper one-dimensional Riemann integral below. Sums of tagged sums prove linearity; pointwise comparison proves monotonicity; refinement at a fixed cut proves interval additivity. In detail, if a function is integrable on the whole interval, refine a small-gap partition at the cut: each subinterval gap is at most the total gap, proving integrability of both restrictions. Conversely, join small-gap partitions of the two restrictions; lower and upper sums add, so the integral of the whole is the sum of the two limits. Constants have integral equal to value times length. These statements also follow directly by taking the just-proved tagged limits. A bounded change supported on finitely many points has integral zero: isolate those points in finitely many intervals of arbitrarily small total length, and bound the upper sum of its absolute value by its bound times that length. Therefore endpoint values do not change a proper integral.

If either side length of $R$ is zero, define every area sum and the double integral to be zero. An integral over a zero-length interval is also zero. The positive-area argument above is not applied after dividing by a zero area.

Fubini with an integrable function and all integrable sections

Assume $a<b$, $c<d$, that bounded $f:R\to\mathbb R$ is Riemann integrable, and that for every $x\in[a,b]$ the section $y\mapsto f(x,y)$ is properly Riemann integrable on $[c,d]$. Put $F(x)=\int_c^d f(x,y)\,dy$. Then $|F(x)|\le M(d-c)$. We prove that $F$ is properly Riemann integrable and

For any product grid use its $m_{ij},M_{ij}$ above and set $A_i=\sum_jm_{ij}\Delta y_j$, $B_i=\sum_jM_{ij}\Delta y_j$. For every $x\in[x_{i-1},x_i]$, one-dimensional additivity and comparison give $A_i\le F(x)\le B_i$. Taking infima and suprema in that entire closed $x$-interval yields

Given $\varepsilon>0$, integrability of $f$ supplies a grid with $U(f,P)-L(f,P)<\varepsilon$. The displayed inequality makes the outer Darboux gap less than $\varepsilon$, proving outer integrability rather than assuming it. Both $\iint_R f$ and $\int_a^b F$ lie in $[L(f,P),U(f,P)]$; their difference is at most $\varepsilon$, so they are equal. If instead all horizontal sections are properly integrable, the identical argument with the coordinate names exchanged gives the horizontal formula. Both orders are asserted only when both sets of sections exist. For a zero-width rectangle, the same equality holds whenever the displayed section integrals exist, since an outer or inner interval has zero length.

Continuous rectangle and interval integrability

Let $f$ be continuous relative to the closed rectangle $R$. If $|R|>0$, compactness and the finite-cover proof give boundedness and uniform continuity. For any $\varepsilon>0$, choose $\delta>0$ such that points less than $\delta$ apart have function values differing by less than $\varepsilon/(2|R|)$. A grid whose cell diameters are less than $\delta$ has

The Darboux criterion proves integrability, and its tiling bridge proves the original arbitrary-rectangle tagged definition. On a compact interval the identical proof uses length instead of area, with zero-length intervals handled separately. Each fixed-coordinate section of continuous $f$ is therefore properly integrable. The proved section theorem gives both iterated formulas. Moreover $F(x)=\int_c^d f(x,y)\,dy$ is continuous: its difference at $x,x_0$ is bounded by

which tends to zero by uniform continuity. If a side of $R$ has zero length, continuous sections still have their one-dimensional proper integrals, and all relevant double/iterated integrals are zero. No nonsingular-change-of-variables or measure-theoretic theorem is involved.

Continuous graphs have arbitrarily small rectangle covers

For continuous $g:[a,b]\to\mathbb R$, $a<b$, its graph can be covered by finitely many open axis-parallel rectangles with total area less than any prescribed $\tau>0$. To prove this, write $A=b-a>0$. Choose positive $\omega,\eta$ with $A(\omega+2\eta)<\tau/2$. Uniform continuity permits a partition $a=x_0<\cdots<x_N=b$ on each interval of which the oscillation is at most $\omega$. Let $m_i,M_i$ be the infimum and supremum on that interval. The open rectangles

cover the graph, including the cut points. Their total area is at most $(A+2Nr)(\omega+2\eta)$, which is less than $\tau$ when $r>0$ is sufficiently small. A finite vertical or horizontal line segment has such covers as well, by putting it in a thin open rectangle; a single point is covered by a small square. Finite unions of these graphs and vertical or horizontal segments inherit the property by splitting the allowed total area among their finitely many members.

We will also need control of cells that touch such a compact set $E$, not just of the points in $E$. Choose a finite open-rectangle cover of $E$ with total area below $\tau$. For each $p\in E$, choose a ball $B(p,r_p)$ contained in one rectangle. Finitely many $B(p_i,r_{p_i}/2)$ cover $E$. For $\delta<\min_i r_{p_i}/2$, every cell of diameter less than $\delta$ meeting $E$ is contained in one of the chosen rectangles: use a meeting point and the triangle inequality. Disjoint-interior cell areas therefore sum to less than $\tau$, by finite rectangle-area addition. This is the precise zero-area property used below; no assertion about arbitrary irregular boundaries is made.

Continuous Type-I regions without an ambient extension

Let $a\le b$ be finite, let continuous $g_1,g_2:[a,b]\to\mathbb R$ satisfy $g_1\le g_2$ everywhere, and let

Assume only that $f:D\to\mathbb R$ is continuous relative to $D$. The set $D$ is bounded, because $g_1,g_2$ are bounded. It is closed: limits of points in $D$ retain $x\in[a,b]$ and both inequalities by continuity. Equivalently, a point outside the interval or strictly violating an inequality has a small open ball missing $D$. Hence $D$ is compact. Thus $|f|\le M$ and $f$ is uniformly continuous on $D$, even though it need not extend to any surrounding rectangle.

First suppose $a<b$, and choose $c<d$ strictly below all $g_1$-values and above all $g_2$-values, respectively. On $R=[a,b]\times[c,d]$ define $h=f$ on $D$ and $h=0$ off $D$. Boundary points of $D$ lie on the two graphs or on the two vertical edge segments. Indeed a point with $a<x<b$ and $g_1(x)<y<g_2(x)$ is interior by continuity, and a point outside the closed set $D$ is not a boundary point unless it is a limit of $D$. The finite graph/segment union $E$ is compact and has the small-cover property just proved.

For a fine product grid, call cells meeting $E$ bad. Their total area is below any chosen $\tau>0$. Every other cell lies entirely in $D$ or entirely outside it: otherwise the straight segment joining an inside to an outside point has a first exit from the closed set $D$, hence a boundary point in the cell, a contradiction. Uniform continuity bounds the oscillation on each good inside cell by any chosen $\eta>0$; on a good outside cell it is zero. The oscillation on every bad cell is at most $2M$. Taking a mesh small enough for both requirements gives

For a requested error $\varepsilon>0$, choose $\eta<\varepsilon/(2|R|)$ and $\tau<\varepsilon/(4(M+1))$. This proves that the bounded zero extension $h$ is properly Riemann integrable. Only differences of values of $f$ at two points of $D$ were used; continuity of an ambient extension was never assumed.

For fixed $x$, the restriction of $f$ to the closed interval $[g_1(x),g_2(x)]$ is continuous. Splitting $[c,d]$ at its endpoints, and ignoring the finitely many endpoint values, proves that every section of $h$ is properly integrable and

When the two endpoints coincide, the section of $h$ is supported at one point and both integrals are zero. The bounded-function section theorem now proves existence of the outer proper integral and the Type-I formula

The same good/bad-cell proof works on any positive-area axis-parallel bounding rectangle containing $D$: its boundary-control set $E$ is unchanged, and outside $D$ the extension is still zero. Every vertical section with $x$ outside $[a,b]$ is zero. At every $x\in[a,b]$, including $x=a,b$, the preceding proper-section splitting applies. Thus the proved section theorem on that bounding rectangle, followed by one-dimensional interval additivity and endpoint invariance, gives the same displayed right side. The definition is therefore independent of the bounding rectangle; this assertion does not assume a continuous ambient extension. Bounded changes confined to $E$ also leave the double integral unchanged, since their absolute upper sums are bounded by their bound times the bad-cell area.

In fact the inner integral $G(x)$ is continuous. For $x,x_0$, compare the two integration intervals on their overlap; there the values $f(x,y)$, $f(x_0,y)$ are both defined in $D$. Uniform continuity bounds their difference by $\eta$ when $|x-x_0|$ is small. The remaining pieces have total length at most $|g_1(x)-g_1(x_0)|+|g_2(x)-g_2(x_0)|$. Hence

This tends to zero; it also covers empty overlaps and collapsed intervals. If $g_1=g_2$ everywhere, the whole region is a graph: both sides of the Type-I formula are zero. If $a=b$, $D$ is a vertical segment, its double integral is zero by a thin-rectangle cover, its one section is properly integrable by continuity, and its outer zero-length integral is zero. Interchanging the coordinate names proves the horizontal Type-II assertion under continuous ordered horizontal bounds and relative continuity on that same region. Nothing here asserts Fubini for arbitrary Jordan regions, nonintegrable sections, unbounded functions, or improper integrals.

For a rectangle, group the two-dimensional Riemann sums by vertical columns. Within each column, the inner sum approximates the integral in $y$ for a fixed $x$, and the remaining sum approximates the outer integral. Continuity on the compact rectangle supplies the uniform control needed for the limiting argument. For a simple region, integrate the indicator-extended function over a rectangle; slices outside the lower and upper boundaries contribute zero. This explains the formulas; the theorem supplies the full convergence justification.