Improper and Parameter-Dependent Multiple Integrals

If $D=\mathrm{\varnothing }$, take $K_j=\mathrm{\varnothing }$ for all $j$; [L9] verifies the Jordan clause. If $D={\mathbb{R}}^n$, take $K_j=[-(j+1),j+1{]}^n$; [L7], [L9], and [L10] make these compact Jordan boxes. Both sequences are exhaustions, and the radii begin at $1$.

Suppose $D$ is proper and nonempty, put $F:={\mathbb{R}}^n\setminus D$, and define $C_j:=\{x\in {\mathbb{R}}^n:\mathrm{\mid }{x}_k\mathrm{\mid }\le j+1\text{ for every }k<n,d(x,F)\ge 1\mathrm{/}(j+1)\}$. By [L2], each $C_j$ is closed and bounded, hence compact by [L7], lies in $D$, and satisfies $C_j\subseteq C_{j+1}$.

Fix a bijection from $\mathbb{Q}$ to $\mathbb{N}$ using [L3]. Iterating the explicit pairing in [L11] codes every finite rational endpoint list by one natural, with its length included in the code; [L4] verifies each fixed-length stage. Thus all finite unions of closed rational grid rectangles admit one fixed enumeration by natural-number codes without Countable Choice.

Apply [L1] to $C_0\subseteq D$. Its finite grid union has a positive margin between the compact core and the complement of its interior and between the union and ${\mathbb{R}}^n\setminus D$. By [L8], move each of its finitely many grid endpoints by less than that margin to rational endpoints, preserving $C_0\subseteq \mathrm{int}{K}_0\subseteq K_0\subseteq D$. Thus the candidate codes of step 1.3 are nonempty, and [L5] selects their least member. The same argument applied to the compact set $C_{j+1}\cup K$ defines a single-valued successor $K^{\mathrm{\prime }}$ with $C_{j+1}\cup K\subseteq \mathrm{int}{K}^{\mathrm{\prime }}\subseteq K^{\mathrm{\prime }}\subseteq D$.

Apply [L6] to the seed $(0,K_0)$ and the successor rule of step 2.1. The second coordinates form compact Jordan sets with $C_j\subseteq K_j\subseteq \mathrm{int}{K}_{j+1}\subseteq D$.

If $C\subseteq D$ is compact, [L7] bounds all of its coordinates. Also, open balls contained in $D$ cover $C$; a finite subcover and the minimum of their halved radii give a positive lower bound for $d(x,F)$ on $C$. Hence $C\subseteq C_j\subseteq K_j$ for some $j$; in particular every point of $D$ is eventually included, and $(K_j)$ is an exhaustion.

Since $K_j\subseteq K_{j+1}$, extend both restricted functions by zero to one common bounding rectangle. Their zero extensions are ordered pointwise, so [L2] and [L3] make the numbers ${\int }_{K_j}f$ increasing, and every one is bounded above by the defining supremum ${\int }_{D}f$ of [L1].

Every compact Jordan set $K\subseteq D$ lies in some $K_j$ by [L4]. Extending the two restrictions by zero to one bounding rectangle and applying [L2] and [L3] gives ${\int }_{K}f\le {\int }_{K_j}f\le {\sup }_i{\int }_{K_i}f$.

Taking the supremum over all compact Jordan $K$ in step 1.2 gives ${\int }_{D}f\le {\sup }_i{\int }_{K_i}f$, while step 1.1 gives the reverse inequality; equality follows, including when the value is $+\mathrm{\infty }$.

Since $0\le f^{+},f^{-}\le \mathrm{\mid }f\mathrm{\mid }$, extend the restrictions to every compact Jordan set by zero on one bounding rectangle. Monotonicity in [L3], the Jordan-set definition [L4], and the defining suprema in [L2] make both nonnegative improper integrals finite; [L1] then gives ${\int }_{K_j}{f}^{+}\to {\int }_D{f}^{+}$ and ${\int }_{K_j}{f}^{-}\to {\int }_D{f}^{-}$ along every exhaustion.

On each compact $K_j$, extend the three restrictions by zero to one bounding rectangle. The identity $f=f^{+}-f^{-}$ and linearity in [L3], interpreted through [L4], give ${\int }_{K_j}f={\int }_{K_j}{f}^{+}-{\int }_{K_j}{f}^{-}$.

Subtracting the two finite limits in step 1.1 and using step 2.1 yields ${\int }_{K_j}f\to {\int }_D{f}^{+}-{\int }_D{f}^{-}={\int }_{D}f$, independently of the exhaustion; the converse is the defining condition in [L2].

On every compact Jordan $K\subseteq D$, extend $f{\mathrm{\mid }}_K$ and $g{\mathrm{\mid }}_K$ by zero to one bounding rectangle. Their zero extensions satisfy the same pointwise inequalities, so [L3] and [L4] give $0\le {\int }_{K}f\le {\int }_{K}g$; taking the defining suprema, equivalently using [L1] on any exhaustion, gives $0\le {\int }_{D}f\le {\int }_{D}g$.

If $\mathrm{\mid }f\mathrm{\mid }\le g$ and ${\int }_{D}g$ is finite, step 1.1 applied to $\mathrm{\mid }f\mathrm{\mid }$ gives ${\int }_D\mathrm{\mid }f\mathrm{\mid }\le {\int }_{D}g<+\mathrm{\infty }$, so [L2] gives absolute improper integrability of $f$.

Choose a compact Jordan exhaustion $(K_j)$ by [L4]. Then [L1] gives ${\int }_{K_j}g\to {\int }_{D}g<+\mathrm{\infty }$, so choose $K=K_j$ with $0\le {\int }_{D}g-{\int }_{K}g<\epsilon$.

For every $t\in C$, [L2] first makes $f_t$ absolutely improperly integrable. For every later exhaustion member $K_i\supseteq K$, [L3] applied to the zero extensions gives $\mathrm{\mid }{\int }_{K_i}{f}_t-{\int }_K{f}_t\mathrm{\mid }\le {\int }_{K_i}g-{\int }_{K}g$. Passing to the exhaustion limits by [L1] and [L5] gives $\mathrm{\mid }{\int }_D{f}_t-{\int }_K{f}_t\mathrm{\mid }\le {\int }_{D}g-{\int }_{K}g<\epsilon$.

Since $\mathrm{\mid }{f}_t-f_s\mathrm{\mid }\le \mathrm{\mid }{f}_t\mathrm{\mid }+\mathrm{\mid }{f}_s\mathrm{\mid }\le 2g$, [L2] makes every difference absolutely improperly integrable, and the same [L1], [L3], and [L5] argument gives the second estimate with $2\epsilon$, uniformly for $s,t\in C$.

Choose a compact relative parameter neighborhood $C\subseteq I$ of $t_0$ and an integrable $g$ dominating $\mathrm{\mid }{f}_t\mathrm{\mid }$ there. Continuity gives local integrability and [L5] gives absolute improper integrability of the slices. Given $\epsilon >0$, [L1] supplies a compact Jordan core $K\subseteq D$ on which both tail errors are below $\epsilon \mathrm{/}3$.

The restriction of $f$ to the compact set $K\times C$ is uniformly continuous by [L2], while [L6] supplies all proper core integrals. Hence, for $t\in C$ sufficiently close to $t_0$, $\mathrm{\mid }f(x,t)-f(x,t_0)\mathrm{\mid }<\epsilon \mathrm{/}(3(1+\mathrm{cont}K))$ for every $x\in K$, and [L3] with [L4] makes the compact-core integral difference smaller than $\epsilon \mathrm{/}3$.

Split $F(t)-F(t_0)$ into the two tail errors and the core-integral difference. Steps 1.1 and 1.2 make its absolute value smaller than $\epsilon$, proving relative continuity at $t_0$, including a one-sided parameter endpoint.

On the compact parameter interval between $t_{\ast }$ and any $t\in I$, [L1] gives $\mathrm{\mid }f(x,t)-f(x,t_{\ast })\mathrm{\mid }\le \mathrm{\mid }t-t_{\ast }\mathrm{\mid }{g}_C(x)$ pointwise. Fact [L4] makes the difference absolutely improperly integrable. Proper linearity in [L6] along one exhaustion and convergence in [L7] show that a sum of two absolutely improperly integrable functions is again absolutely improperly integrable, so $f_t=f_{t_{\ast }}+(f_t-f_{t_{\ast }})$ is absolutely improperly integrable.

For nonzero $h$ with $t_0+h\in I$, [L1] bounds the difference quotient $q_h(x):=(f(x,t_0+h)-f(x,t_0))\mathrm{/}h$ by one integrable dominator. On a compact Jordan core, [L5] and [L1] make $q_h\to {\mathrm{\partial }}_{t}f(\cdot ,t_0)$ uniformly as $h\to 0$.

Use [L2] to make the tails of both $q_h$ and ${\mathrm{\partial }}_{t}f(\cdot ,t_0)$ uniformly small, then use the uniform convergence from step 1.2 on the core and [L6]. It follows that ${\int }_D{q}_h\to {\int }_D{\mathrm{\partial }}_{t}f(x,t_0)dx$. On every compact member of one exhaustion, proper linearity in [L6] gives $\int q_h=(\int f_{t_0+h}-\int f_{t_0})\mathrm{/}h$; applying [L7] to the three absolutely integrable functions passes this identity to $D$. Hence the left side is the difference quotient of $F$, proving the asserted derivative formula.

Apply [L4] to each derivative slice and [L3] to the continuous integrand ${\mathrm{\partial }}_{t}f$, using the same local dominators, to see that the derivative integral is continuous in $t$. Thus $F$ is $C^1$ on $I$.

Put $h:=\tilde{g}-\tilde{f}$. It is bounded, vanishes on $Q\setminus N$, and has some bound $\mathrm{\mid }h\mathrm{\mid }\le M$.

In the case $M=0$, the function $h$ is identically zero, hence integrable with integral zero.

In the case $M>0$, given $\epsilon >0$, [L2] covers $N$ by finitely many cubes of total volume below $\epsilon \mathrm{/}(4M)$, and [L3] gives a grid whose cells meeting $N$ have total volume below $\epsilon \mathrm{/}(2M)$. On all other cells $h=0$, while on a cell meeting $N$ its oscillation is at most $2M$, so the total Darboux gap is below $\epsilon$. Thus [L4] makes $h$ integrable; the bounds $-M{1}_N\le h\le M{1}_N$ with the same grids force its integral to be zero.

The two cases exhaust $M\ge 0$, so $h$ is integrable with integral zero. If $f$ is integrable, then $\tilde{g}=\tilde{f}+h$ is integrable and has the same integral by [L5].

Interchanging $f$ and $g$ applies step 3.1 to $-h$, proving the reverse integrability implication and the same equality of values.

If $\mathrm{\mid }x\mathrm{\mid }\ge 1$, then [L1] and [L3] give $0<e^{-x^2}\le 1\mathrm{/}(1+x^2)\le x^{-2}$. The $p$-test [L2] and comparison [L5] make the positive tail converge, and the substitution $u=-x$ in [L9] gives the identical negative-tail estimate.

On $[-1,1]$, the integrand is continuous and hence integrable by [L8]; [L6] gives $e^{-x^2}\ge e^{-1}>0$, so [L7] gives ${\int }_{-1}^1{e}^{-x^2}dx\ge 2e^{-1}>0$.

By [L4], the two finite tails and the proper middle integral combine to a finite mixed improper integral, and step 1.2 makes the total strictly positive.

For $R>0$, [L4] gives $e^{-(x^2+y^2)}=e^{-x^2}{e}^{-y^2}$, so [L1] yields ${\int }_{[-R,R{]}^2}{e}^{-(x^2+y^2)}d(x,y)={\left({\int }_{-R}^R{e}^{-x^2}dx\right)}^2$.

Put $R=j+1$. The squares form a compact Jordan exhaustion of ${\mathbb{R}}^2$, so [L2] makes their plane integrals tend to the nonnegative improper plane integral; [L3] makes each one-dimensional factor tend to $I$.

Passing to the limit in the product identity of step 1.1 gives the plane integral equal to $I^2$.

On each compact rectangle $[\epsilon ,R]\times [0,\pi ]$ and $[\epsilon ,R]\times [-\pi ,0]$, the polar map has Jacobian determinant $r({\cos }^2\theta +{\sin }^2\theta )=r>0$ by [L12], [L13], and [L17]. Enlarge the radial interval inside $(0,\mathrm{\infty })$ and each angular interval by less than $\pi \mathrm{/}2$. If two points in one enlarged rectangle have the same polar image, [L13] gives equal positive radii; reducing both angles modulo $2\pi$ by [L23] and using its injective half-open parametrization shows that their difference is an integer multiple of $2\pi$. The enlarged angular interval has length below $2\pi$, so the angles are equal. Thus the map is injective on an open neighbourhood of each compact rectangle, and their images are the closed upper and lower half-annuli. Each half-annulus is closed and bounded, hence compact by [L6]; its boundary is contained in two continuous semicircle graphs and two radial segments, so [L4] and [L5] make it Jordan measurable.

The Gaussian is continuous on each compact half-annulus and the pulled-back function $e^{-r^2}r$ is continuous on each parameter rectangle, so [L22] supplies both integrability conditions in [L1]. Apply [L1] to each half-annulus. By [L18], the parameter-rectangle integral is $\pi {\int }_{\epsilon }^R{e}^{-r^2}rdr$. Facts [L10], [L19], and [L20] give $(-\frac{1}{2}{e}^{-r^2}{)}^{\mathrm{\prime }}=r{e}^{-r^2}$, so [L14] evaluates each half as $\frac{\pi }{2}(e^{-{\epsilon }^2}-e^{-R^2})$; the lower half has the same angular length.

The half-annuli overlap only in the two radial boundary segments, which have content zero by [L4]. On a common bounding rectangle, $1_{A_{+}}+1_{A_{-}}$ differs from $1_{A_{+}\cup A_{-}}$ only on that overlap, so [L3], [L7], and [L8] combine the two values from step 2.1 into the full-annulus integral $\pi (e^{-{\epsilon }^2}-e^{-R^2})$.

Every circle is the union of its upper and lower continuous semicircle graphs, hence has content zero by [L4]; [L5] makes every closed disc Jordan measurable, and [L6] makes it compact.

The omitted inner disc lies in $[-\epsilon ,\epsilon {]}^2$, whose content is $4{\epsilon }^2$ by [L7] and [L9]. Since $-(x^2+y^2)\le 0$, strict increase in [L16] and the normalization/positivity in [L21] give $0<e^{-(x^2+y^2)}\le 1$. Thus [L8] bounds its Gaussian integral by $4{\epsilon }^2$. The annulus and inner disc overlap only on a content-zero circle, so [L3] recombines them. Letting $\epsilon \downarrow 0$ in step 3.1 gives the radius-$R$ disc integral $\pi (1-e^{-R^2})$.

The closed discs of radii $j+1$ form a compact Jordan exhaustion of ${\mathbb{R}}^2$. The Gaussian is nonnegative by [L21], and it is locally Riemann integrable because [L10] and elementary algebra make it continuous and [L22] makes its restriction to every compact Jordan set integrable. Thus [L11], step 5.1, and [L15] show that the disc integrals tend to $\pi$, which is the improper plane integral.

By [L1], $I>0$, and [L2] with [L3] gives $I^2=\pi$.

Since $I$ is nonnegative, uniqueness in [L4] identifies it with $\sqrt{\pi }$.