Recall the change of variables formula. An assumption was made that the transformation was C^{1}. This
made the argument easy to push through with the use of the Besicovitch covering theorem. However, with
the Radon Nikodym theorem and the fundamental theorem of calculus, it is easy to give a much shorter
argument which actually gives a better result because it does not require the derivative to be continuous.
Recall that U ⊆ ℝ^{n}, was an open set on which h was differentiable and one to one with det
(Dh (x ))
≠0.
Then for x ∈ U, we had an inequality which came from the definition of the derivative and the Brouwer
fixed point theorem.
|detDh (x)|(1 − ε)n mn (B (x,rx))
≤ m (h (B(x,r )))
n x n
≤ |detDh (x)|(1 + ε) mn (B (x,rx))18octe1s (23.13)
this for all r_{x} sufficiently small. Now consider the following measure μ
(E)
≡ m_{n}
(h (E ))
. By Proposition
21.5.1 this is well defined. It is indeed a measure and μ ≪ m_{n} the details in Problem 15 on Page 1647.
Therefore, by Corollary 23.2.3, there is a nonnegative measurable g which is in L_{loc}^{1}
n
(ℝ )
such
that
∫
μ (E ) ≡ mn (h(E )) = gdmn
E
It is clearly a Radon measure. In particular,
∫
mn (h (B(x,r))) = B(x,r)gdmn
The idea is to identify g. Let x be a Lebesgue point of g, for all r small enough,
∫ ∫
X (y)f (y)dm = (X (x )(f ∘h)(x))|detDh (x)|dm
h(U) h(A) n U A n
Thus
∫ ∫
f (y)dmn = (f ∘h )(x) |detDh (x)|dmn
h(A) A
This gives a slight generalization.
Corollary 23.3.2Let A be any measurable subset of U and let f be a nonnegative Lebesgue measurablefunction. Let h : U → ℝ^{n}be one to one and differentiable with detDh
(x)
≠0. Then
∫ ∫
f (y)dmn = (f ∘ h)(x )|detDh (x)|dmn
h(A) A
As to the generalization when h is only assumed to be one to one and differentiable, this also follows as
before. The proof of Sard’s lemma in Theorem 21.5.6 did not require any continuity of the
derivative. Thus as before, h
(U0)
= 0 where U_{0}≡
{x : det Dh (x) = 0}
, Borel measurable because
detDh is Borel measurable due to the fact that h is continuous and the entries of the matrix
of Dh are each Borel measurable because they are limits of difference quotients which are
continuous functions. Letting U_{+}≡
∫ ∫
f (y)dm = f (y)dm
h(U) n h(U+) n
∫
= (f ∘h) (x)|detDh (x)|dmn
∫U+
= (f ∘h )(x)|detDh (x)|dmn
U
Then one can also obtain the result of Corollary 23.3.2 in the same way. This leads to the following version
of the change of variables formula.
Theorem 23.3.3Let A be any measurable subset of U, an open set in ℝ^{n}and let f be anonnegative Lebesgue measurable function. Let h : U → ℝ^{n}be one to one and differentiable.Then
∫ ∫
f (y)dm = (f ∘ h)(x )|detDh (x)|dm
h(A) n A n
You can also tweak this a little more to get a slightly more general result. You could assume, for
example, that h is continuous on the open set U and differentiable and one to one on some F ⊆ U where F
is measurable and m_{n}
(h(U ∖F ))
= 0 rather than assuming that h is differentiable on all of the
open set U. This would end up working out also, but the above is pretty close and is easier to
remember.