If f is nonnegative, then int Omega f(T omega)mu(d omega)=int_{Omega′}f(omega′)mu T^{-1}(d omega′). A function f (not necessarily nonnegative) is integrable with respect to mu T^{-1} if and only if fT is integrable with respect to mu

Hugo Stokes 2022-10-26 Answered
Theorem 16.13. If f is nonnegative, then
Ω f ( T ω ) μ ( d ω ) = Ω f ( ω ) μ T 1 ( d ω ) .
A function f (not necessarily nonnegative) is integrable with respect to μ T 1 if and only if fT is integrable with respect to μ, in which case (16.17) and
T 1 A f ( T ω ) μ ( d ω ) = A f ( ω ) μ T 1 ( d ω )
hold. For nonnegative f, (16.18) always holds.
(2.47) Theorem. Suppose Ω is an open set in R n and G : Ω R n is a C 1 diffeomorphism. (a) If f is a Lebesgue measurable function on G ( Ω ), then f G is Lebesgue measurable on Ω. If f 0 or f L 1 ( G ( Ω ) , m ), then
G ( Ω ) f ( x ) d x = Ω f G ( x ) | det D x G | d x
(b) If E Ω and E L n , then G ( E ) L n and m ( G ( E ) ) = E | det D x G | d x.
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Answers (2)

Szulikto
Answered 2022-10-27 Author has 22 answers
Step 1
We will see that, contrary to what was stated by the OP, none of these Theorems may be seen as generalizing the other!
To put it in perspective let me try to state both results in the most similar way possible. In order not to get lost in details, I will totally ignore questions about measurability and convergence of integrals and instead focus on their essential aspects.
In the first result, we are:
1. given a measure space ( X , μ ),
2. a map T : X Y,
3. where Y is another set, and
4. a function f : Y R
5. We then introduce a new measure on Y, the RANGE of T, namely the push-forward measure of μ by T, which I will denote by μ T (referred to as μ T 1 by the OP),
6. and then the conclusion says that
X f T d μ = Y f d μ T .
Step 2
The second result is similar in the sense that we are now:
1. given an open subset X R n (called Ω by the OP), which incidentally comes with a canonical measure, namely Lebesgue's measure,
2. a map T : X Y (called G by the OP),
3. where Y = T ( X ) R n is another set, which also comes with the canonical Lebesgue measure, and finally
4. a function f : Y R
The crucial difference is that here we do not mess with the measure on Y at all, sticking with Lebesgue measure, which I will call μ, all the time. Instead, we use our ingredients to:
5. introduce a new measure on X, the DOMAIN of T, namely μ T = | det D x T | d x.
6. and then the conclusion says that
X f T d μ T = Y f d μ .
Summarizing, in the first result, the transformation T is used to create a new measure μ T on Y, while in the second result, T is used to go the other way, namely introducing a new measure μ T on X.
In the second context it may be shown that the push forward measure of μ T under T is μ, but this is just a red herring, since it comes only as an afterthought.
It is therefore instructive to consider in which ways can a map T, as above, be used to give rise to new measures. Of course we can use T in a push-forward construction as in the first result above. Now, when all spaces come with preloaded measures, as is the case of Riemannian manifolds, a differentiable T leads to its Jacobian, which can be used as a density function against the default measure to create a new measure on its domain.
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ebendasqc
Answered 2022-10-28 Author has 3 answers
Step 1
First of all, let us consider a context in which we have two measure spaces ( X , μ ) and ( Y , ν ), and a measurable map T : X Y.
From the point of view of the change of variables in integrals, the ideal situation is that in which the identity
( ) Y f ( y ) d ν ( y ) = X f ( T ( x ) ) d μ ( x ) ,
holds for all nonnegative measurable functions f : Y R
The crux of the matter is thus how to find measures μ and ν providing for (∗), including situations in which one of these is given in advance, and we have to find the other one.
In that direction, perhaps the main observation to be made is that (∗) holds if and only if
( ) ν = T ( μ ) .
This can be easily proven by pluging in functions of the form f = 1 E , namely the characteristic function of a measurable subset E Y.
Thus, when we are facing the task of evaluating a concrete integral, we might try to identify our problem with either the right or left-hand-side of (∗), in the hope that the other side will be easier to compute.
In case we identify our concrete integral with the right-hand-side of (∗), in which case we will have already identified f, T, and μ, we will then be scrambling to find a measure ν satisfying (∗), but this is now easy since the only choice is ν = T ( μ ).
Step 2
On the other hand, if we identify our concrete integral with the left-hand-side of (∗), we will have identified f and ν, and then we will be required to provide T, as well as to solve equation (∗∗) for μ.
The content of the change of variables Theorem of Differential Calculus is precisely taylored for the above task, and it says that, when X and Y are open subsets of R n , T : X Y is a surjective differentiable map, and ν is Lebesgue's measure on Y, one possible choice for μ is
μ ( d x ) = | det D x T | d x ,
where dx denotes Lebesgue's measure on X.
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