Math 509: Advanced Analysis

Homework 7
Lawrence Tyler Rush
<me@tylerlogic.com>
April 6, 2015
http://coursework.tylerlogic.com/courses/upenn/math509/homework07

1 Fully prove steps 1 through 4 of the Change of Variables Theorem


Change of Variables Theorem: Let A Rn be an open set and g : A Rn a one-to-one, continuously differentiable map such that detg(x)0 for all x A. If f : g(A) R is a Riemann integrable function, then
∫       ∫
                    ′
 g(A)f =  A(f ∘ g)|detg |

Proof: The proof begins with several reductions which allow us to assume that f 1, that A is a small open set about a point a, and that g(a) is the identity matrix. Then the argument is completed by induction on n with the use of Fubini’s Theorem.

(a) Step 1


Suppose there is an open cover V of A such that for each U V and any integrable f, we have
∫       ∫
                    ′
 g(U)f =  U(f ∘ g)|detg |

Then the theorem is true for all A.

Proof.
The collection of all g(U) is an open cover of g(A). Let Φ be a partition of unity subordinate to this cover. For any Riemann integrable f : g(A) R, if φ = 0 outside of g(U), then, since g is one-to-one we have that (φf) g = 0 outside of U. Hence φf is integrable and the equation

∫        ∫              ′
    φf =   ((φf)∘ g)|detg |
 g(U)      U

can be written as

∫        ∫
    φf =   ((φf)∘ g)|detg′|
 g(A)      A

Summing over all φ Φ yields

φΦ g(A)φf = φΦ A((φf) g)|detg′|
g(A)(      )
  ∑
(    φ )
  φ∈Φf = A(            )
  ∑
(    ((φf) ∘g))
 φ∈ Φ|detg′|
g(A)f = A(( ( (     )  )    ))
(( ( ( ∑  φ) f) ∘ g))
      φ∈ Φ|detg′|
g(A)f = A(f g)|detg′|
as desired.

(b) Step 2


It suffices to prove the theorem for f = 1.

Proof.
If the theorem holds for f = 1 then it holds for f = constant. Let V be a rectangle in g(A) and P a partition of V . For each subrectangle S of P, let fS be the constant function mS(f). Then

L(f,P) = SPmS(f)v(S)
= SP intSfS
= SP g-1(intS)(fS g)|detg′|
SP g-1(intS)(f g)|detg′|
= g-1(V )(f g)|detg′|
Since V f = LUBP(f,P), this proves that
∫      ∫               ′
   f ≤       (f ∘ g)|detg |
  V     g-1(V)

Likewise, letting fS = MS(f), we get the opposite inequality, and so that conclude that

∫      ∫
   f =       (f ∘ g)|detg′|
  V     g-1(V)

Then as in Step 1, it follows that

∫       ∫
    f =   (f ∘ g)|detg′|
 g(A)     A

(c) Step 3


If the theorem is true for g : A Rn and for h : B Rn where g(A) B, then it is also true for h g : A Rn.

Proof.
To ease the proof slightly, define X = g(A) and f= (f h)|deth′|. Then we have

hg(A)f = h(g(A))f
= h(X)f
= X(f h)|deth′|
= Xf
= g(A)f
= A(f′∘ g)|detg′|
= A(((f ∘h)|det h′|)∘g)|detg′|
= A((f h) g)     ′
(|deth |∘g)|detg′|
= A(f (h g))     ′
(|deth |∘g)|detg′|
= A(f (h g))|det(h g)′|
as desired.

(d) Step 4


The theorem is true if g is a linear tranformation.

Proof.
By steps 1 and 2, it suffices to show for any open rectangle U that

∫       ∫       ′
     1 =   |detg|
 g(U)    U

Note that for a linear transformation g, we have g= g. Then this is just the fact from linear algebra that a linear tranformation g : Rn Rn multiplies volumes by |detg|.

2 Fully prove the Fundamental Theorem


Let A be a closed rectangle in Rn and f : A R a bounded function. Let
B = {x ∈ A : f is not continuous at x}

Then f is Riemann integrable on A if and only if B has measure zero.

Proof.
Suppose first that B has measure zero.

Let ε > 0. Define Bε = {x A : o(f,x) ε}. Now Bε B, hence Bε has measure zero. By problem 13 of our previous Chapter 2, the set Bε is closed. Since Bε is also bounded, it is compact, and so has content zero. Hence there is a finite collection U1,,Un of closed rectangles, whose interiors cover Bε, with total volume less than ε.

Now let P be a partition of the original rectangle A which “refines” the collection of rectangles Ui in the following sense. Each rectangle S P is in one of the following two groups:

  1. Group 1 (G1): S Ui for some i
  2. Group 2 (G2): otherwise; i.e. S is disjoint from Bε

Since the function f is, by hypothesis, bounded on A, choose M so that |f(x)| < M for all x A. Then

MS (f)- mS (f ) < 2M

for all S P. Thus since

                  ∑
U (f,P )- L(f,P) =    [MS (f)- mS (f)]vol(S)
                  S∈P

we can divide the above difference into two parts, the first corresponding to G1 and the other to G2. We have the following for the first part

 ∑  [MS (f)- mS (f)]vol(S) < 2M ∑  vol(Ui) < 2M ε
S∈G1                           i
(2.1)

As for the second part, since each point x S G2 has o(f,x) < ε, then any S G2 can be further partitioned into rectangles Sso that

∑   [MS ′(f) - mS′(f)]vol(S ′) < ∑   εvol(S′) < ε ∑ vol(S′) < εvol(S)
S′⊂S                         S′⊂S           S′⊂S

Thus replacing the partitions in G2 with these refined partitions implies the following bound

 ∑                          ∑              ∑
     [M ′S(f)- m ′S(f)]vol(S′) <     εvol(S ) = ε   vol(S) < εvol(A )
S′∈G2                       S∈G2          S∈G2
(2.2)

Putting together the partial sums from G1 and G2 of equations 2.1 and 2.2 yields

                 ∑
U(f,P)- L (f,P ) =   [MS (f )- mS(f)]vol(S) < 2M  ε+ εvol(A )
                 S∈P

The value on the right-hand side can be made arbitrarily small by appropriate choice of ε and so we conclude that f is Riemann integrable.

Conversely, suppose that f is Riemann integrable. We must show that the set B has measure zero. Since B = B1 B12 B13 ⋅⋅⋅, it is enough to show that each B1∕n has measure zero.

Since B1∕n is compact, that is the same as having content zero. Since f is Riemann integrable, then given any ε > 0 we can find a partition P of A such that

                  ε-
U(f,P )- L(f,P) < n

Let G be the subfamily of P consisting of rectangles which meet B1∕n. Then the rectangles S in G cover B1∕n. Expand slightly each of these rectangles S to a rectangles S, so that the interiors of the Snow cover B1∕n. Then each of the rectangles Scontains in its interior a point x B1∕n where the oscillation o(f,x) 1∕n. It follows from this that

MS ′(f)- mS ′(f) ≥ 1∕n

Hence

(1∕n) S′∈Gvol(S) S′∈G[MS(f) - mS(f)]vol(S) S′∈P[MS(f) - mS(f)]vol(S) < ε∕n
and therefore S′∈Gvol(S) < ε. Since the rectangles Sin Gcover B1∕n, and since ε > 0 is arbitrarily small, this shows the B1∕n has content zero, completing the proof.