9 Change of variables

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1 Tel Aviv University, 2014/15 Analysis-III,IV Change of variables 9a What is the problem b Examples c Rotation invariance d Linear transformation e Differentiating set functions (again) f Set function induced by mapping g Change of variable in general h Change of variable for a diffeomorphism Change of variables is the most powerful tool for calculating multidimensional integrals. Two kinds of differentiation are instrumental: of mappings (treated in Sections 2 5) and of set functions (treated in Sect. 8c). 9a What is the problem The area of a disk {(x, y) : x 2 + y 2 < 1} R 2 may be calculated by iterated integral, 1 1 x 2 1 dx dy = 2 1 x 1 2 dx =... 1 x 2 1 or alternatively, in polar coordinates, 1 0 r dr 2π 0 dϕ = 1 0 2πr dr = π ; the latter way is much easier! Note rdr rather than dr (otherwise we would get 2π instead of π). Why the factor r? In analogy to the one-dimensional theory we may dx dy expect something like ; is it r? Well, basically, it is r because an infinitesimal rectangle [r, r + dr] [ϕ, ϕ + dϕ] of area dr dϕ on the (r, ϕ)-plane dr dϕ corresponds to an infinitesimal rectangle or area dr rdϕ on the (x, y)-plane. dr rdφ

2 Tel Aviv University, 2014/15 Analysis-III,IV 144 9a1 Theorem. Let U, V R n be Jordan measurable open sets, ϕ : U V a diffeomorphism, and f : V R a bounded function such that the function (f ϕ) det Dϕ : U R is also bounded. Then (a) f is integrable on V if and only if (f ϕ) det Dϕ is integrable on U; and (b) if they are integrable, then f = (f ϕ) det Dϕ. V U 9a2 Exercise. Prove that the following is an equivalent reformulation of Theorem 9a1. 1 Let U, V R n be Jordan measurable open sets, ϕ : V U a diffeomorphism, and f : U R a bounded function such that the function f det D(ϕ 1 ) : U R is also bounded. Then (a) f ϕ is integrable on V if and only if f det D(ϕ 1 ) is integrable on U; and (b) if they are integrable, then f ϕ = f det D(ϕ 1 ). 9b Examples V U In this section we take for granted Theorem 9a1 (to be proved in Sect. 9h). 9b1 Exercise. Show that 7d4 and 7d5 are special cases of 9a1. 9b2 Exercise (polar coordinates in R 2 ). (a) Prove that f(x, y) dxdy = f(r cos θ, r sin θ) r drdθ x 2 +y 2 <R 2 0<r<R,0<θ<2π for every integrable function f on the disk x 2 + y 2 < R 2 ; 2 1 Hint: ϕ U V f f ϕ R ; redraw: V ϕ 1 U f f ϕ R ; relabel: ϕ V f ϕ U f R 2 Do you use a diffeomorphism between (0, R) (0, 2π) and the disk? (Look closely!)

3 Tel Aviv University, 2014/15 Analysis-III,IV 145 (b) it can happen that the function (r, θ) rf(r cos θ, r sin θ) is integrable on (0, R) (0, 2π), but f is not integrable on the disk; find a counterexample; (c) however, (b) cannot happen if f is bounded on the disk; prove it. 1 In particular, we have now the curvilinear Cavalieri principle for concentric circles promised in 7d13. 9b3 Exercise (spherical coordinates in R 3 ). Consider the mapping Ψ : R 3 R 3, Ψ(r, ϕ, θ) = (r cos ϕ sin θ, r sin ϕ sin θ, r cos θ). (a) Draw the images of the planes r = const, ϕ = const, θ = const, and of the lines (ϕ, θ) = const, (r, θ) = const, (r, ϕ) = const. (b) Show that Ψ is surjective but not injective. (c) Show that det DΨ = r 2 sin θ. Find the points (r, ϕ, θ), where the operator DΨ is invertible. (d) Let V = (0, ) ( π, π) (0, π). Prove that Ψ V is injective. Find U = Ψ(V ). 9b4 Exercise. Compute the integral x 2 +y 2 +(z 2) 2 1 Answer: π ( 2 3 log 3). 2 2 over one loop of the lem- 9b5 Exercise. Compute the integral niscate (x 2 + y 2 ) 2 = x 2 y 2. 3 dxdy (1+x 2 +y 2 ) 2 dxdydz x 2 +y 2 +z 2. 9b6 Exercise. Compute the integral over the four-dimensional unit ball: x 2 +y 2 +u 2 +v 2 1 ex2 +y 2 u 2 v 2 dxdydudv. 4 9b7 Exercise. Compute the integral xyz dxdydz over the ellipsoid {x 2 /a 2 + y 2 /b 2 + z 2 /c 2 1}. Answer: a2 b 2 c Do not forget: Theorem 9a1 is taken for granted. 2 Hint: 1 < r < 3; cos θ > r2 +3 4r. 3 Hints: use polar coordinates; π 4 < ϕ < π 4 ; 0 < r < cos 2ϕ; 1 + cos 2ϕ = 2 cos 2 ϕ; dϕ cos 2 ϕ = tan ϕ. 4 Hint: The integral equals x 2 +y 2 1 ex2 +y 2 (u 2 +v 2 1 (x 2 +y 2 ) e (u2 +v 2) dudv ) dxdy. Now use the polar coordinates. 5 Hint: 6d17 can help.

4 Tel Aviv University, 2014/15 Analysis-III,IV 146 The centroid 1 of a Jordan measurable set E R n of non-zero volume is the point C E R n such that for every linear (or affine) f : R n R the mean of f on E (recall (6g18)) is equal to f(c E ). That is, C E = 1 ( ) x 1 dx,..., x n dx, v(e) which is often abbreviated to C E = 1 v(e) E E E x dx. 9b8 Exercise. Find the centroids of the following bodies in R 3 : (a) The cone {(x, y, z) : h x 2 + y 2 < z < h} for a given h > 0. (b) The tetrahedron bounded by the three coordinate planes and the plane x a + y b + z c = 1. (c) The hemispherical shell {a 2 x 2 + y 2 + z 2 b 2, z 0}. (d) The octant of the ellipsoid {x 2 /a 2 + y 2 /b 2 + z 2 /c 2 1, x, y, z 0}. The solid torus in R 3 with minor radius r and major radius R (for 0 < r < R < ) is the set Ω = {(x, y, z) : ( x2 + y 2 R )2 + z 2 r 2 } R 3 generated by rotating the disk Ω = {(x, z) : (x R) 2 + z 2 r 2 } R 2 on the (x, z) plane (with the center (R, 0) and radius r) about the z axis. Interestingly, the volume 2π 2 Rr 2 of Ω is equal to the area πr 2 of Ω multiplied by the distance 2πR traveled by the center of Ω. (Thus, it is also equal to the volume of the cylinder {(x, y, z) : (x, z) Ω, y [0, 2πR].) Moreover, this is a special case of a general property of all solids of revolution. 1 In other words, the barycenter of (the uniform distribution on) E.

5 Tel Aviv University, 2014/15 Analysis-III,IV 147 9b9 Proposition (the second Pappus s centroid theorem). 1 2 Let Ω (0, ) R R 2 be a Jordan measurable set and Ω = {(x, y, z) : ( x 2 + y 2, z ) Ω} R 3. Then Ω is Jordan measurable, and v 3 ( Ω) = v 2 (Ω) 2πx CΩ ; here C Ω = (x CΩ, z CΩ ) is the centroid of Ω. 9b10 Exercise. Prove Prop. 9b9. 3 9c Rotation invariance 9c1 Theorem. Let T : R n R n be a linear isometry, 4 and f : R n R a bounded function with bounded support. Then (a) f T = R n f, R n f T = R n f ; R n (b) f T is integrable if and only if f is integrable, and in this case f T = f. R n R n 9c2 Corollary. (a) v ( T (E) ) = v (E) and v ( T (E) ) = v (E) for all bounded E R n ; (b) T (E) is Jordan measurable if and only if E is, and then v ( T (E) ) = v(e). In order to prove Th. 9c1 we need v ( T () ) = v() for every box R n. ut first we need Jordan measurability of T (). There are several ways to prove it. Here is one of them (not the most elementary). 5 9c3 Lemma. For every norm on R n, the set {x : x = 1} is of volume zero, and the sets {x : x < 1}, {x : x 1} are Jordan measurable. 1 Pappus of Alexandria ( ) was one of the last great Greek mathematicians of Antiquity. 2 The first Pappus s centroid theorem, about the surface area, has to wait for Analysis 4. 3 Hint: use cylindrical coordinates: Ψ(r, ϕ, z) = (r cos ϕ, r sin ϕ, z). 4 That is, a linear bijection satisfying x T x = x. Note that T 1 is also a linear isometry. 5 Do you see a more elementary way?

6 Tel Aviv University, 2014/15 Analysis-III,IV 148 Proof. We denote E = {x : x 1}. It follows easily from 1e7 that E = {x : x < 1}, E = E, and E = {x : x = 1} = (E ). y (8e3), v ( (1 ε)e ) v (E ) for all ε (0, 1). y 6g12, v ( (1 ε)e ) = (1 ε) n v (E). y 8b7(b), v (E ) = v (E). We get (1 ε) n v (E) v (E) for all ε (0, 1), therefore v (E) v (E), that is, E is Jordan measurable. y 8d4(b), = ( ) is of volume zero, and E is Jordan measurable. 9c4 Exercise. (a) For every box R n there exist x 0 R n and a norm on R n such that = {x : x x 0 < 1} and = {x : x x 0 1}. (b) For every box R n and every invertible linear T : R n R n, the set T () is Jordan measurable. Prove it. We return to a linear isometry T. 1 9c5 Lemma. v ( T () ) = v() for every box R n. Proof. WLOG, the box is centered: = {x : x 1} where x = 2 max ( x 1 c 1,..., xn ) c n ; v() = c1... c n. For arbitrary R (0, ) we introduce the set J R Z n of all (i 1,..., i n ) such that the shifted box + ic intersects the ball RD = {x : x R}; here ic stands for (i 1 c 1,..., i n c n ). We have RD i JR ( + ic), therefore v(rd) #(J R )v(). Also, T (RD) T ( i JR ( + ic) ) ; taking into account that T is isometric we have RD i JR (T () + T (ic)), therefore v(rd) #(J R )v ( T () ). On the other hand, taking r (0, ) such that rd, we have i JR ( + ic) (R + 2r)D (think, why); and the interiors + ic are pairwise disjoint; therefore #(J R )v() v ( (R + 2r)D ). Applying T as before we get 1 See also 7d14. #(J R )v ( T () ) v ( (R + 2r)D ).

7 Tel Aviv University, 2014/15 Analysis-III,IV 149 It follows that R n (R + 2r) v ( ) T () n v() (R + 2r)n R n for all R (while r is constant); it remains to take R. Proof of Th. 9c1. Item (b) for the indicator f = 1l of a box follows from 9c5, since f T = 1l T 1 () (and T 1 also is a linear isometry). y linearity, Item (b) holds for all step functions f. In general, given f and ε > 0, we take a step function h such that h f and h f + ε, note that h T f T and get f T h T = h T = h f + ε for all ε > 0; that is, f T f. Also, f = (f T ) T 1 f T ; thus, f T = f. Similarly (or taking ( f)), f T = f, which completes the proof of Item (a). Item (b) follows immediately. Given an n-dimensional Euclidean vector space E, we choose a linear isometry E R n and transfer the Riemann integral (and the Jordan measure) from R n to E. y Theorem 9c1 the result does not depend on the choice of the linear isometry E R n. y translation invariance, the same holds for Euclidean affine spaces. Riemann integral and Jordan measure are well-defined on every n-dimensional Euclidean affine space, and preserved by affine isometries between these spaces. 9c6 Exercise. Find the volume cut off from the unit ball by the plane lx + my + nz = p. 9c7 Exercise. Reformulate the result of 9b8(a) as a geometric theorem about the centroid of an arbitrary right circular cone in a classical Euclidean space. 9d Linear transformation 9d1 Theorem. Let T : R n R n be an invertible linear operator. Then the image T (E) of an arbitrary E R n is Jordan measurable if and only if E is Jordan measurable, and in this case v ( T (E) ) = det T v(e).

8 Tel Aviv University, 2014/15 Analysis-III,IV 150 Also, for every bounded function f : R n R with bounded support, det T f T = f and det T f T = f. Thus, f T is integrable if and only if f is integrable, and in this case det T f T = f. Proof. The Singular Value Decomposition 3i1 gives an orthonormal basis (a 1,..., a n ) in R n such that vectors T (a 1 ),..., T (a n ) are orthogonal. Invertibility of T ensures that the numbers s k = T (a k ) do not vanish. Taking b k = (1/s k )T (a k ) we get an orthonormal basis (b 1,..., b n ) such that T (a 1 ) = s 1 b 1,..., T (a n ) = s n b n. We have s 1... s n = det T, since the singular values s k are square roots of the eigenvalues of T T (thus, s 1... s n = det(t T ) = (det T ) 2 = det T ). Similarly to the proof of 3c4 we downgrade the two copies of R n into a pair of Euclidean vector spaces, choose new bases and upgrade back to two copies of R n. In other words, we treat T : R n R n as T : X Y where X, Y are Euclidean vector spaces, and upgrade X, Y to Cartesian spaces via the bases (a 1,..., a n ), (b 1,..., b n ). The (matrix of the) operator becomes diagonal: T (x 1,..., x n ) = (s 1 x 1,..., s n x n ). It remains to apply 6d17 (and 6g12). If 1, 2 are two Euclidean norms on an n-dimensional vector space, then the ratio of norms 1 2 varies between min(s 1,..., s n ) and max(s 1,..., s n ) (here s 1,..., s n are the singular values), depending on the direction of a vector; but the ratio of volumes v 1( ) is s v 2 ( ) 1... s n, invariably. On an n-dimensional vector or affine space the volume is ill-defined, but Jordan measurability is well-defined, and the ratio v(e 1) v(e 2 of volumes is ) well-defined. That is, the volume is well-defined up to a coefficient. 9e Differentiating set functions (again) Recall Sect. 8c: the derivative F of a box function F is basically F (x 0 ) = lim x 0 F () v() (more generally, we use lim inf and lim sup for F and F ); the idea of x 0 was formalized as x 0 and sup x x 0 < δ x

9 Tel Aviv University, 2014/15 Analysis-III,IV 151 (and ultimately δ 0+). Equivalently, (9e1) F (x 0 ) = a if and only if F ( i ) v( i ) a whenever i x 0 ; here i x 0 means i i x 0 and sup x i x x 0 0 (as i ). More generally, F F ( (x 0 ) is the supremum (in fact, maximum) of lim i ) i v( i over all ) sequences ( i ) i such that i x 0 and F ( i) converges. v( i ) As noted in 8c5, the restriction x 0, unnecessary in Sect. 8, 1 is stipulated in order to conform to the one-dimensional case, and enlarges the class of differentiable box functions. Here (in Sect. 9) we introduce one more restriction (with no one-dimensional counterpart), enlarging further the class of differentiable box functions. First, we define the aspect ratio α() of a box, = [s 1, t 1 ] [s n, t n ] R n, by α() = max(t 1 s 1,..., t n s n ) min(t 1 s 1,..., t n s n ). Clearly, α() = 1 if and only if is a cube. Second, we redefine the relation i x 0 as follows: (9e2) i i x 0 ; sup x i x x 0 0 ; sup α( i ) <. i Third, we redefine accordingly the relation F (x 0 ) = a (still (9e1) but with the new interpretation of i x 0 ). Similarly, we redefine F (x 0 ) and F (x 0 ) as (respectively) the infimum and supremum 2 F ( of lim i ) i v( i over all ) sequences ( i ) i such that i x 0 (in the new sense) and F ( i) converges. v( i ) Still, 8c4 holds; (F + G) = F + G etc. Lemma 8c7 holds, again. The only change needed in the proof is, when taking a partition P i of a box i, to require additionally that C P i α(c) α( i ), which is easy to satisfy; and then sup i α( i ) α( 0 ) <. 3 Still, 8c9, (8c10) and 8c11 hold; F () = F, etc. Why bother about the aspect ratio? The ultimate answer will be given in Sect. 9h, but here is a partial answer. Recall 9c4: = {x : x x 0 < 1} and = {x : x x 0 1} for some norm. As every norm, it is equivalent to the Euclidean norm, c C, but these c, C depend on ; if is small, then C must be large. 1 And still unnecessary here in Sect Not necessarily maximum. 3 It is easy to get even more: α( i ) 1; we could include such requirement into the definition of i x 0.

10 Tel Aviv University, 2014/15 Analysis-III,IV 152 9e3 Lemma. For every box R n there exist x 0 R n, c (0, ) and a norm on R n such that = {x : x x 0 < c}, = {x : x x 0 c}, and 1 α() n. Proof. We have = [s 1, t 1 ] [s n, t n ] R n. As noted in the proof of 9c5, x 0 = { x : 2 max ( x 1 ) } c 1,..., xn c n 1 where c1 = t 1 s 1,..., c n = t n s n ; note that α() = max(c 1,..., c n ) min(c 1,..., c n ). We take c = 1 min(c 2 1,..., c n ), C = max(c 1,..., c n ) n, and x = 2c max ( x 1 ) ( c 1,..., xn c n, then x c 2 max x1 ) c 1,..., xn c n 1 x + x 0, thus = {x : x x 0 c}; similarly, = {x : x x 0 < c}. On one hand, k c k 2c, therefore x 2c max ( x 1,..., xn ) 2c 2c = max( x 1,..., x n ) x. On the other hand, k c k 2cα(), therefore x 2c max ( x 1,..., x n ) 2cα() 2cα() = 1 max( x α() 1,..., x n ) 1 α() x. n 9f Set function induced by mapping Recall (7d9): v m+n (E) = v R m n (E x ) dx for a Jordan set E R m+n and its sections E x = {y : (x, y) E} R n for x R m. Denoting v n (E x ) by J(x) and introducing ( the projection mapping ϕ : E R m by ϕ(x, y) = x we have v m+n ϕ 1 () ) = J for all boxes Rm. According to 7d13 we are interested in more general, nonlinear mappings ϕ. This leads ultimately to the curvilinear iterated integral treated in Analysis 4. Here are some preliminaries, to be used in the proof of Theorem 9a1. Consider a mapping ϕ : R m R n such that the inverse image ϕ 1 () of every box is a bounded set. (An example: ϕ : R 2 R, ϕ(x, y) = x 2 + y 2.) It leads to a pair of box functions F F (in dimension n), (9f1) F () = v (ϕ 1 ( )), F () = v (ϕ 1 ()), generally not additive but rather superadditive and subadditive: for every partition P of a box, F () C P F (C), F () C P F (C), which follows from (6g5), (6g6) and the fact that ϕ 1 (C 1) ϕ 1 (C 2) = ϕ 1 (C 1 C 2) = when C 1 C 2 =.

11 Tel Aviv University, 2014/15 Analysis-III,IV 153 If F () = F () then ϕ 1 () is Jordan measurable, and ϕ 1 ( ) is of volume zero; if this happens for all then the box function F () = v(ϕ 1 ()) is additive. A useful sufficient condition is given below in terms of functions J = (F ), J = (F ) ; that is, (9f2) v (ϕ 1 (i )) J (x) = inf lim, J v (ϕ 1 ( i )) (x) = sup lim ( i ) i i v( i ) ( i ) i i v( i ) where ( i ) i runs over all sequences of boxes such that i x 0 (see (9e2)) and lim i (... ) exists. 9f3 Proposition. If J, J are locally integrable and locally equivalent then F () = F () = J = J for every box. In this case 1 (9f4) v ( ϕ 1 () ) = where J is any function equivalent to J, J. 9f5 Exercise. Prove existence of J and calculate it for ϕ : R 2 R defined by (a) ϕ(x, y) = x 2 + y 2 ; (b) ϕ(x, y) = x 2 + y 2 ; (c) ϕ(x, y) = x + y, taking for granted that the area of a disk is πr 2. 9f6 Exercise. Prove existence of J and calculate it for ϕ : R 3 R 2 defined by ϕ(x, y, z) = ( x 2 + y 2, z ), taking for granted Prop. 9b9. 9f7 Exercise. For every partition P of a box, Prove it. 2 min C P We generalize 8c9, 8c11. F (C) v(c) F () v() F () v() max F (C) C P v(c). 1 Can this happen when m < n? If you are intrigued, try the inverse to the mapping of 6c5. 2 Hint: recall the proof of 8c7. J

12 Tel Aviv University, 2014/15 Analysis-III,IV 154 9f8 Exercise. Prove it. 1 v() inf J (x) F () F () v() sup J (x). x x 9f9 Exercise. Prove it. 2 3 J F () F () J. Prop. 9f3 follows immediately. 9f10 Remark. Similar statements hold for a mapping defined on a subset of R m (rather than the whole R m ). If ϕ : A R n for a given A R m then ϕ 1 () A for every, but nothing changes in (9f1), (9f2) and Prop. 9f3. In particular, if ϕ(a) is bounded, then A must be Jordan measurable; otherwise J, J cannot be integrable and equivalent. 9f11 Example (mind the boundary). Points of R n \ϕ(a) are irrelevant, but points of ϕ(a) \ ϕ(a) cannot be ignored, even if they are a set of volume zero. It can happen that v (A) > J. ϕ(a) Recall 8b9: G = (s 1, t 1 ) (s 2, t 2 )... is dense in (0, 1), k (t k s k ) = a < 1; v (G) = a, v (G) = 1. We take u k = i=k (t i s i ), then u 1 = a, u k 0, and u k u k+1 = t k s k. We define ϕ : G (0, a) by ϕ(x) = u k+1 + x s k for x (s k, t k ). For every box such that (0, a) the set ϕ 1 () is Jordan measurable, and v ( ϕ 1 () ) = v() (think, why). Thus, J = J = 1 on (0, a). Nevertheless, ϕ 1( (0, a) ) = G fails to be Jordan measurable, and v ( ϕ 1 ((0, a)) ) = 1 > a = v ( (0, a) ). Note that J (0) =, even though 0 / ϕ(g). 9f12 Remark. If J, J are integrable and equivalent on a given closed box (and not necessarily on every box) then v(ϕ 1 (C)) = J for every box C C. 1 Hint: similarly to the proof of 8c7; use 9f7. 2 Similar to 8c11. 3 Curiously, the left-hand and the right-hand sides differ four times:, ; lim inf, lim sup; v, v ; i, i.

13 Tel Aviv University, 2014/15 Analysis-III,IV 155 9f13 Exercise. Calculate J for the projection mapping ϕ(x, y) = x (a) from the disk A = {(x, y) : x 2 + y 2 1} R 2 to R; (b) from the annulus A = {(x, y) : 1 x 2 + y 2 4} R 2 to R. Is J (locally) integrable? 9f14 Exercise. Calculate J for the mapping ϕ(x) = sin x from the interval [0, 10π] R to R. Is J (locally) integrable? 9g Change of variable in general 9g1 Proposition. If ϕ : R m R n is such that 1 J, J are locally integrable and locally equivalent then for every integrable f : R n R the function f ϕ : R m R is integrable and f ϕ = fj. R m R n Proof. First, the claim holds when f = 1l is the indicator of a box, since fj = J (9f4) = v(ϕ 1 ()) = 1l ϕ 1 () = f ϕ. R n R m R m Second, by linearity in f the claim holds whenever f is a step function (on some box, and 0 outside). Third, given f integrable on a box (and 0 outside), we consider arbitrary step functions g, h on such that g f h. We have g ϕ f ϕ h ϕ and g ϕ = gj, h ϕ = hj, thus, R m R m gj f ϕ f ϕ hj ; R m R m gj fj hj. We take M such that J( ) M on and get hj gj = (h g)j M (h g) ; thus, integrability of f implies integrability of f ϕ and the needed equality for the integrals. 1 We still assume that the inverse image of a box is bounded.

14 Tel Aviv University, 2014/15 Analysis-III,IV 156 9g2 Corollary. If ϕ : R m R n is such that J, J are locally integrable and equivalent then: (a) for every Jordan measurable set E R n the set ϕ 1 (E) R m is Jordan measurable; (b) for every integrable f : E R the function f ϕ is integrable on ϕ 1 (E), and f ϕ = fj. ϕ 1 (E) Proof. (a) apply 9g1 to f = 1l E ; (b) apply 9g1 to f1l E. 9g3 Remark. If ϕ : A R n is such that J, J are integrable and equivalent on a given closed box (and not necessarily on every box) then for every integrable f : R the function f ϕ is integrable on the Jordan measurable set ϕ 1 (), and f ϕ = fj. Also, 9g2 holds for E. ϕ 1 () 9g4 Exercise. (a) Prove that f( x x 2 +y y ) 2 dxdy = 2π f(r)r dr [0,1] for every integrable f : [0, 1] R; (b) calculate +y 2 )/2 x 2 +y 2 1 e (x2 dxdy. (Could you do it by iterated integrals?) 9h Change of variable for a diffeomorphism 9h1 Proposition. Let U, V R n be open sets and ϕ : V U a diffeomorphism, then 1 2 J (x) = J (x) = det(dψ) x E for all x U; here ψ = ϕ 1 : U V. In the next lemma, 3 given a norm on R n, we denote = {x : x < 1}, = {x : x 1}, r + x 0 = {x : x x 0 < r}, r + x 0 = {x : x x 0 r} ; these are Jordan measurable by 9c3. 1 For J, J see (9f2). 2 det Dψ is called the Jacobian of ψ and often denoted by J ψ. 3 We may restrict ourselves to norms such that is a box, but this does not simplify the proof of the lemma.

15 Tel Aviv University, 2014/15 Analysis-III,IV 157 9h2 Lemma. Let be a norm on R n, A R n, f : A R n a mapping, ε (0, 1), and (1 ε) x 1 x 2 f(x 1 ) f(x 2 ) (1 + ε) x 1 x 2 for all x 1, x 2 A. Then v ( f(r + x) ) (1 ε) n v(r + x), v ( f(r + x) ) (1 + ε) n v(r + x) for all x and r such that r + x A and (1 ε)r + f(x) f(a). Proof. The second inequality follows from the inclusion f(r + x) (1 + ε)r + f(x) ; proof of this inclusion: for arbitrary x 1 r + x, y 1 = f(x 1 ) is defined, and belongs to (1 + ε)r + f(x) since y 1 f(x) (1 + ε) x 1 x (1 + ε)r. The first inequality follows from the inclusion f(r + x) (1 ε)r + f(x) ; proof of this inclusion: for arbitrary y 1 (1 ε)r +f(x) there exists x 1 A such that y 1 = f(x 1 ); and x 1 x 1 f(x 1 ε 1) f(x) = 1 y 1 ε 1 f(x) < 1 (1 ε)r = r, thus x 1 ε 1 r + x and y 1 = f(x 1 ) f(r + x). Proof of Prop. 9h1. Let x 0 U. Denote T = (Dψ) x0. y Theorem 9d1, v(t (E)) = det T v(e) for every Jordan measurable E R n. Note that ϕ 1 (E) = ψ(e). It is sufficient to prove that v (ψ( i )) v(t ( i )) 1, v (ψ( i )) v(t ( i )) 1 whenever i x 0. Similarly to the proof of 3c4 we downgrade the two copies of R n into a pair of affine spaces, and then upgrade them back to R n getting x 0 = 0, ψ(x 0 ) = 0, and T = id. (In contrast to the proof of 9d1, we do not need Euclidean metric in these affine spaces, since we work with the ratio of volumes of two sets ψ( i ), T ( i ) in the same space.) Similarly to the proof of Prop. 3c3 (and 4c7), for every ε > 0 there exists a neighborhood U ε of 0 such that (y 1 y 2 ) (x 1 x 2 ) ε x 1 x 2

16 Tel Aviv University, 2014/15 Analysis-III,IV 158 whenever x 1, x 2 U ε and y 1 = ψ(x 1 ), y 2 = ψ(x 2 ). The set V ε = ψ(u ε ) is a neighborhood of 0 (since ψ is open, recall 3b6). Given i x 0, we take M such that i α( i ) M/ n; using Lemma 9e3 we get points x i 0, numbers c i 0+ and norms i such that i = {x : x x i i c i } = c i i + x i and 1 M i. For all x 1, x 2 U ε, denoting y 1 = ψ(x 1 ) and y 2 = ψ(x 2 ), we have (y 1 y 2 ) (x 1 x 2 ) i (y 1 y 2 ) (x 1 x 2 ) ε x 1 x 2 Mε x 1 x 2, therefore (assuming ε < 1/M), (1 Mε) x 1 x 2 i y 1 y 2 i (1 + Mε) x 1 x 2 i. Claim: For all i large enough, c i i + x i U ε and c i i + y i V ε, where y i = f(x i ). Proof of the claim. First, all x such that x x i i c i satisfy x x x i + x i M x x i i + x i Mc i + x i 0 as i. Second, all y such that y y i i c i satisfy y y y i + y i M y y i i + y i < Mc i + y i 0 as i. Applying Lemma 9h2 to = i, A = U ε, f = ψ, Mε, x = x i and r = c i we get, for all i large enough, v ( ψ( i ) ) (1 Mε) n v( i ), v ( ψ( i ) ) (1 + Mε) n v( i ), that is, (1 Mε) n v (ψ( i )) v(t ( i )) v (ψ( i )) v(t ( i )) (1 + Mε)n. According to 9f11 we should examine J on the boundary of U. ut this is hard. A diffeomorphism V U need not extend to a homeomorphism V U. Here are two simple counterexamples: 1 (a) V = {x R 2 : 0 < x < 1}, U = {x R 2 : 1 < x < 2}, ϕ(x) = x + x ; x (b) V = U = {x R 2 : x < 1}, ϕ(r cos θ, r sin θ) = ( r cos(θ r ), r sin(θ r )). Fortunately, we have another way: approximation from inside (since, in contrast to 9f11, we assume that U, V are Jordan measurable). 1 Far not the worst case.

17 Tel Aviv University, 2014/15 Analysis-III,IV 159 Proof of Theorem 9a1. We prove the equivalent formulation given by 9a2: ϕ : V U a diffeomorphism, f : U R bounded on U, and f det Dψ bounded on U, where ψ = ϕ 1 : U V. We have to prove that f ϕ = V f det Dψ, provided that one integrand is integrable. U y 9h1, J and J are integrable and equivalent (moreover, continuous and equal) on every box such that U. y 9g3, ψ() V is Jordan measurable, and f ϕ = fj ψ() whenever f is integrable on. In particular, taking f = 1l E, we get v ( ψ(e) ) = J for all Jordan measurable E. Thus, E if E is of volume zero, then ψ(e) is also of volume zero. Similarly to the proof of 8e6 we take a box R n such that U, and for arbitrary partition P of we consider K P = C. C P,C U Taking into account that the boundaries C do not matter we get ψ(k P ) f ϕ = K P fj whenever f is integrable on K P. And if E K P is of volume zero, then ψ(e) is of volume zero. We take a sequence of partitions P 1, P 2,... of such that each P i+1 is a refinement of P i, and C P i diam(c) ε i, ε i 0. Claim: the corresponding sets K i = K Pi satisfy Ki U. Proof of the claim: clearly, K i. Given x U, we take i such that ε i < dist(x, R n \ U), and C P i such that x C, then C U (since diam(c) ε i ), thus x C K i ; and moreover, x Ki, since dist(y, R n \ U) > ε i for all y near x. If f has a compact support inside U and f J is integrable, then f is integrable (since the continuous function 1/J is bounded on the support of f), therefore f ϕ is integrable 1 and f ϕ = fj (since the support of V U f is contained in some Ki ). The same holds for the inverse diffeomorphism; if an integrable g has a compact support inside V, then g ψ is integrable. Taking g = f ϕ we see that integrability of f ϕ implies integrability of f (and fj). Thus, (f ϕ is integrable ) (fj is integrable ), and in this case f ϕ = fj 1 Alternatively you may prove the integrability via 8f1. V U

18 Tel Aviv University, 2014/15 Analysis-III,IV 160 whenever f has a compact support inside U. We have to get rid of this assumption. ( The relation Ki U implies ψ(ki ) V. y 8e9, v (Ki ) v (U) and v ψ(k i ) ) v (V ). Taking into account Jordan measurability of these sets we get v(u \ Ki ) 0, v(v \ ψ(ki )) 0 as i. Using boundedness of f (and therefore, f ϕ) and fj, we get integral convergence (recall Sect. 6e): (f ϕ) 1l ψ(k i ) f ϕ, fj 1l K i fj as i. It remains to use 6e3(b). Index aspect ratio, 151 centroid, 146 linear isometry, 147 Pappus theorem, 147 polar coordinates, 144 spherical coordinates, 145 i x 0, 151 F, F, 152 J, J, 153