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A Short Course on the Lebesgue Integral and
Measure Theory
- August 5, Steve Cheng
- 1 Motivation for the Lebesgue integral Contents
- 2 Basic measure theory
- 3 Measurable functions
- 4 Definition of the Lebesgue Integral
- 5 Convergence theorems
- 6 Some Results of Integration Theory
- 7 Lp spaces
- 8 Construction of Lebesgue Measure
- 9 Lebesgue Measure in Rn
- 10 Riemann integrability implies Lebesgue integrability
- 11 Product measures and Fubini’s Theorem
- 12 Change of variables in Rn
- 13 Vector-valued integrals
- 14 C∞ 0 functions are dense in Lp(Rn)
- 15 Other examples of measures
- 16 Egorov’s Theorem
- 17 Exercises
18 Bibliography 52
Preface
This article develops the basics of the Lebesgue integral and measure theory. In terms of content, it adds nothing new to any of the existing textbooks on the subject. But our approach here will be to avoid unduly abstractness and absolute generality, instead focusing on producing proofs of useful results as quickly as possible. Much of the material here comes from lecture notes from a short real analysis course I had taken, and the rest are well-known results whose proofs I had worked out myself with hints from various sources. I typed this up mainly for my own benefit, but I hope it will be interesting for anyone curious about the Lebesgue integral (or higher mathematics in general). I will be providing proofs of every theorem. If you are bored reading them, you are invited to do your own proofs. The bibliography outlines the background you need to understand this article.
Copyright matters
Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation; with no Invariant Sections, with no Front-Cover Texts, and with no Back-Cover Texts.
1 Motivation for the Lebesgue integral
If you have followed the rigorous definition of the Riemann integral in R or Rn, you may be wondering why do we need to study yet another integral. After all, why should we even care to integrate nasty functions like:
D(x) =
1 , x ∈ Q 0 , x ∈ R \ Q
Rephrased in another way, D(x) is actually the indicator function^1 of the set S = {x ∈ Q} ⊂ R ,
and we want to find its “length”. Continuing to rephrase this question, sup- pose we are taking many real-valued measurements x of a particular physical phenomenon. What is the probability, say, that x is rational? If we assume
(^1) For any set S, this is the function χS defined by χS (x) = 1 if x ∈ S and χS (x) = 0 if x /∈ S. Math people call this the “characteristic function”, while probability people call it the “indicator function” instead.
integral. (For example, why should
−∞
−∞ be the same as^
R^2? Note that in the Riemann theory, the iterated integral and the area integral are proven to be equal only for bounded sets of integration.) You will probably be able to find other sorts of limitations with the Riemann integral.
2 Basic measure theory
The setting of abstract integration is measure theory, which tells us what the areas or volumes of various sets are. Essentially we are given some function μ of sets which returns the area or volume — formally called the measure — of the given set. i.e. We assume at the beginning that such a function μ has already been defined for us. The abstract approach of the Lebesgue integral has the obvious advantage that the theory can be applied to many other measures besides volume in Rn. We begin with the axioms of measure theory.
Definition 2.1. Let X be any non-empty set. A sigma algebra^2 of subsets of X is a family A of subsets of X, with the properties:
- A is non-empty.
- If E ∈ A, then X \ E ∈ A.
- If {En}n∈N is a sequence of sets in A, then their union is in A. That is, A is closed under countable unions.
The pair (X, A) is called a measurable space, and the sets in A are called the measurable sets. Notice that the axioms always imply that X ∈ A. Also, by De Morgan’s laws, A is closed under countable intersections as well as countable union. Needless to say, we cannot insist that A is closed under arbitrary unions or intersections, as that would force A = 2X^ if A contains all the singleton sets. That would be uninteresting. On the other hand, we want closure under countable set operations, rather than just finite ones, as we will want to take countable limits.
Example 2.1. Let X be any (non-empty) set. Then A = 2X^ is a sigma algebra.
Example 2.2. Let X be any (non-empty) set. Then A = {X, ∅} is a sigma algebra.
To get non-trivial sigma algebras to work with we need the following, a very unconstructive(!) construction: If we have a family of sigma algebras on X, then the intersection of all the sigma algebras from this family is also a sigma algebra on X. If all of the sigma
(^2) I do not know why it has such a ridiculous name, other than the fact that it is often denoted by the Greek letter.
algebras from the family contains some fixed G ⊆ 2 X^ , then the intersection of all the sigma algebras from the family, of course, is a sigma algebra containing G. Now if we are given G, and we take all the sigma algebras on X that contain G, and intersect all of them, we get the smallest sigma algebra that contains G.
Definition 2.2. The smallest sigma algebra containing any given G ⊆ 2 X^ , as constructed above, is denoted 〈G〉, and is also called the sigma algebra generated by G.
The following is an often-used sigma algebra.
Definition 2.3. If X is a topological space, we can construct the sigma algebra 〈T 〉, where T is the set of all open sets. This is called the Borel sigma algebra and is denoted B(X). When topological spaces are involved, we will always take the sigma algebra to be the Borel sigma algebra unless stated otherwise.
B(X), being generated by the open sets, then contains all open sets, all closed sets, and countable unions and intersections of open sets and closed sets. It seems unlikely, however, that every set in B(X) is expressible as a countable union and/or intersection of open sets and closed sets, although it is tempting to think that. By the way, Theorem 9.3 shows the Borel sigma algebra is generally not all of 2X^.
Sigma algebras are the domain on which measures are defined.
Definition 2.4. Let (X, A) be a measurable space. A positive measure on this space is a function μ : A → [0, ∞] such that
- μ(∅) = 0
- Countable additivity: For any sequence of mutually disjoint sets En ∈ A,
μ
n=
En
∑^ ∞
n=
μ(En).
The set (X, A, μ) will be called a measure space. Whenever convenient we will abbreviate this expression, as in “let X be a measure space”, etc. Also, in this article, when we say “measure”, we will be dealing with positive measures only. (There are also theories about signed measures and complex measures.)
Example 2.3. Let X be an arbitrary set, and A be a sigma algebra on X. Define μ : A → [0, ∞] as
μ(A) =
|A| , if A is a finite set ∞ , if A is an infinite set.
This is called the counting measure. We will be able to model the infinite series
n=1 an^ in Lebesgue integration theory by using X = N and the counting measure, since integrals are essentially sums of the integrand values weighted by areas or measures.
- Monotonicity: If E, F ∈ A, and E ⊆ F , then μ(E) ≤ μ(F ).
- If E ⊆ F has finite measure (μ(E) < ∞), then μ(F \ E) = μ(F ) − μ(E).
- If A or B has finite measure, then μ(A ∪ B) = μ(A) + μ(B) − μ(A ∩ B).
Proof. The first fact is obvious. For the second fact, we have
μ(F ) = μ((F \ E) ] E) = μ(F \ E) + μ(E) ,
and μ(F \ E) ≥ 0. For the third fact, just subtract μ(E) from both sides. (The funny union symbol means that the union is disjoint.) Of course the fact that A is a sigma algebra is used throughout to know that the new sets also belong to A.) For the fourth fact, we decompose each of A, B, and A ∪ B into disjoint parts, to obtain the following:
μ(A) = μ(A ∩ Bc) + μ(A ∩ B). μ(B) = μ(B ∩ Ac) + μ(B ∩ A). μ(A ∪ B) = μ(A ∩ Bc) + μ(Ac^ ∩ B) + μ(A ∩ B).
Adding the first two equations and then substituting in the third one,
μ(A) + μ(B) = μ(A ∩ Bc) + μ(A ∩ B) + μ(B ∩ Ac) + μ(B ∩ A) = μ(A ∪ B) + μ(A ∩ B).
Since one of A or B has finite measure, so does A ∩ B ⊆ A, B, by the second fact, so we may subtract μ(A ∩ B) from both sides. Of course if one of A or B has infinite measure, the resulting equation says nothing interesting.
The preceding theorem, as well as the next ones, are quite intuitive and you should have no trouble remembering them.
Theorem 2.2. Let (X, A, μ) be measure space, and let E 1 ⊆ E 2 ⊆ E 3 ⊆ · · · be subsets in A with union E. (The sets En are said to increase to E, and henceforth we will write {En} ↗ E for this.) Then
μ(E) = μ
n=
En
= lim n→∞ μ(En).
Proof. The sets Ek and E can be written as the disjoint unions
Ek = E 1 ∪ (E 2 \ E 1 ) ∪ (E 3 \ E 2 ) ∪ · · · ∪ (Ek \ Ek− 1 ) E = E 1 ∪ (E 2 \ E 1 ) ∪ (E 3 \ E 2 ) ∪ · · · ,
(and set E 0 = ∅), so that
μ(E) =
∑^ ∞
k=
μ(Ek \ Ek− 1 ) = lim n→∞
∑^ n
k=
μ(Ek \ Ek− 1 ) = lim n→∞ μ(En).
Theorem 2.3. For any En ∈ A,
μ
n=
En
∑^ ∞
n=
μ(En).
Proof.
μ
n=
En
= μ
n=
En \ (E 1 ∪ E 2 ∪ · · · ∪ En− 1 )
∑^ ∞
n=
μ
En \ (E 1 ∪ E 2 ∪ · · · ∪ En− 1 )
∑^ ∞
n=
μ(En).
Theorem 2.4. Let {En} ↘ E (that is, En are decreasing and their intersection is E), and μ(E 1 ) < ∞. Then
lim n→∞ μ(En) = μ(E) = μ
n=
En
Proof. We have {E 1 \ En} ↗ (E 1 \ E). So
μ(E 1 \ E) = μ
n=
(E 1 \ En)
= lim n→∞ μ(E 1 \ En) ,
μ(E 1 ) − μ(E) = lim n→∞ [μ(E 1 ) − μ(En)] = μ(E 1 ) − lim n→∞ μ(En) ,
and cancel μ(E 1 ) on both sides.
3 Measurable functions
To do integration theory, we of course need functions to integrate. You should not expect that arbitrary functions can be integrated, but only the “measurable” ones. The following definition is not difficult to motivate.
Definition 3.1. Let (X, A) and (Y, B) be measurable spaces. A map f : X → Y is measurable if
for all B ∈ B, the set f −^1 (B) = [f ∈ B] is in A.
Example 3.1. A constant map is always measurable, for f −^1 (B) is either ∅ or X.
Theorem 3.1. The composition of two measurable functions is measurable.
Proof. Immediate from the definition.
Theorem 3.6. Let fn be a sequence of measurable R-valued functions. Then the functions
sup n
fn, inf n fn, max n fn, min n fn, lim sup n
fn, lim inf n fn
(the limits are pointwise) are all measurable.
Proof. If g(x) = supn fn(x), then [g > c] =
n[fn^ > c], and we apply Theorem 3.4. Similarly, if g(x) = infn fn(x), then [g < c] =
n[fn^ < c].^ The rest can be expressed as in terms of supremums and infimums (over a countable set), so they are measurable also.
Not surprisingly, we will need to do arithmetic in integration theory, so we better know that
Theorem 3.7. If f, g : X → R are measurable, then so are f + g, f g, and f /g.
Proof. Consider the countable union
[f + g < c] =
r∈Q
[f < c − r] ∩ [g < r].
The set equality is justified as follows: Clearly f (x) < c−r and g(x) < r together imply f (x) + g(x) < c. Conversely, if we set g(x) = t, then f (x) < c − t, and we can increase t slightly to a rational number r such that f (x) < c − r, and g(x) < t < r. This shows that f + g is measurable (by Theorem 3.4). Since [−f < c] = [f > −c], we see that −f is measurable. Therefore the functions
f +(x) = max{+f (x), 0 } (positive part of f ) f −(x) = max{−f (x), 0 } (negative part of f )
are measurable (from Theorem 3.6 and Example 3.1). Since f = f +^ − f −, f is measurable if f +^ and f −^ are measurable separately also. Since
f g = (f +^ − f −)(g+^ − g−) = f +g+^ − f +g−^ − f −g+^ + f −g−^ ,
to prove that f g is measurable, it suffices to assume that f and g are both non-negative. Then just as with the sum,
[f g < c] =
r∈Q
[f < c/r] ∩ [g < r].
Finally, for 1/g,
[1/g < c] =
[1/c < g, cg > 0] ∪ [1/c > g, cg > 0] , c 6 = 0 [g < 0] , c = 0.
Remark 3.8. You probably have already noticed there may be difficulty in defin- ing what the arithmetic operations mean when the operands are infinite (or when dividing by zero). The usual way to deal with these problems is to simply redefine the functions whenever they are infinite to be some fixed value. In particular, if the function g is obtained by changing the original measurable function f on a measurable set A to be a constant c, we have:
[g ∈ B] =
[g ∈ B] ∩ A
[g ∈ B] ∩ Ac
[g ∈ B] ∩ A =
A , c ∈ B ∅ , c /∈ B [g ∈ B] ∩ Ac^ = [f ∈ B] ∩ Ac^ ,
so the resultant function g is also measurable. Very conveniently, any sets like A = [f = +∞], [f = 0] are automatically measurable. Thus the gaps in the previous proof with respect to infinite values can be repaired with this device.
As a final note, one intermediate result from the proof is quite useful and should be formally recognized:
Theorem 3.9. An R-valued function f is measurable if and only if f +^ and f − are measurable. Moreover, if f is measurable, so is |f | = f +^ + f −.
Remark 3.10. Of course the converse to the second statement is not true. You may construct a counterexample to convince yourself of this fact.
4 Definition of the Lebesgue Integral
The idea behind Riemann integration is to try to measure the sums of area of the rectangles “below a graph” of a function and then take some sort of limit. The Lebesgue integral uses a similar approach: we perform integration on the “simple” functions first:
Definition 4.1. A function is simple if its range is a finite set.
An R-valued simple function ϕ always has a representation
ϕ =
∑^ n
k=
ak χEk ,
where ak are the distinct values of ϕ, and Ek = ϕ−^1 ({ak}). Conversely, any expression of the above form, where ak need not be distinct, and Ek is not neces- sarily ϕ−^1 ({ak}), also defines a simple function. For the purposes of integration, however, we will require that Ek be measurable, and that they partition X. It should be mentioned that χS is measurable if and only if S is.
Intuitively, the simple functions in Sf are supposed to approximate f as close as we like, and we find the integral of f by computing the integrals of these approximations. But logically we need to know that these approximations really do exist. This is the essence of the following theorem.
Theorem 4.2 (Approximation Theorem). Let f : X → [0, ∞] be measur- able. Then there exists a sequence of non-negative functions {ϕn} ↗ f , meaning ϕn are increasing pointwise and converging pointwise to f. Moreover, if f is bounded, it becomes possible for the ϕn to converge to f uniformly.
Proof. We prove the second statement first. Let N be any integer > sup f , and set
ϕn =
N ∑ 2 n
k=
k − 1 2 n^
χEn,k , En,k = f −^1
([
k − 1 2 n^
k 2 n
We have 0 ≤ f − ϕn < 2 −n^ uniformly. The detailed verification is left to the reader. The construction for the first statement is very similar. Set
ϕn =
∑^ n^2 n
k=
k − 1 2 n^
χEn,k + χFn , Fn = [f ≥ n].
I leave it to you to check that 0 ≤ f (x) − ϕn(x) < 2 −n^ whenever n > f (x), and ϕn(x) = n whenever f (x) = ∞.
In case you were worrying about whether this new definition of the integral agrees with the old one in the case of the non-negative simple functions, well, it does. Use monotonicity to prove this.
Definition 4.4. If f is not necessarily non-negative, we define ∫
X
f dμ =
X
f +^ dμ −
X
f −^ dμ ,
provided that the two integrals on the right are not both ∞.
Of course we will want to integrate over subsets of X also. This can be accomplished in two ways. Let A be a measurable subset of X. Either we simply consider integrating over the measure space restricted to subsets of A, or we define (^) ∫
A
f dμ =
X
f χA dμ.
If ϕ is non-negative simple, a simple working out of the two definitions of the integral over A shows that they are equivalent. To prove this for the case of arbitrary measurable functions, we will need the tools of the next section.
Let us note some other basic properties of our integral:
Let f, g be non-negative. Since
Scf = c · Sf = {cϕ : ϕ ∈ Sf } , 0 ≤ c < ∞.
we have (we freely omit the “dμ” and/or the integration limit “X” when they are implied by the context) (^) ∫
cf = c
f.
This rule about constant multiplication also holds for f and c not necessarily non-negative, as you can easily check, but proving linearity requires the tools of the next section. Moreover, if 0 ≤ f ≤ g, then Sf ⊆ Sg , and therefore
f ≤
g. In particular, if A ⊆ B, then f χA ≤ f χB , so ∫
A
f ≤
B
f.
Unsurprisingly
f ≤
g also holds if f, g are not necessarily non-negative, and that is proven by considering the positive and negative parts of f, g separately. Then since −|f | ≤ f ≤ |f |, we also obtain
|f | ≤
f ≤
|f |, i.e.
f
|f |.
(This last inequality is sometimes called the “generalized triangle inequality”, as integrals can be viewed as an advanced form of summing.)
Next, we make one more definition related to integrals.
Definition 4.5. A (μ-)measurable set is said to have (μ-)measurable zero if μ(E) = 0. Typical examples of a measure-zero set are the singleton points in Rn, and lines and curves in Rn, n ≥ 2. By countable additivity, any countable set in Rn has measure zero also. A particular property is said to hold almost everywhere if the set of points for which the property fails to hold is a set of measure zero. For example, “a function vanishes almost everywhere”. Clearly, if you integrate anything on a set of measure zero, you get zero.
Assuming that linearity of the integral has been proved, we can demonstrate the following intuitive result.
Theorem 4.3. A measurable function f : X → [0, ∞] vanishes almost every- where if and only if
X f^ = 0.
Proof. Let A = [f = 0], and μ(Ac) = 0. Then
∫
X
f =
X
f · (χA + χAc ) =
X
f χA +
X
f χAc =
A
f +
Ac
f = 0 + 0.
Using this theorem, we are now in the position to prove linearity of the Lebesgue integral for non-simple functions. Given any two non-negative mea- surable functions f, g, by the approximation theorem (Theorem 4.2), we know that are non-negative simple functions {ϕn} ↗ f , and {ψn} ↗ g. Then {ϕn + ψn} ↗ f + g, and so ∫ f + g = lim n→∞
ϕn + ψn = lim n→∞
ϕn +
ψn =
f +
g.
(The second equality follows because we already know the integral is linear for simple functions. For the first and third equality we apply the Monotone Convergence Theorem.) And if f, g not necessarily non-negative, then ∫ f + g =
(f +^ − f −) + (g+^ − g−)
f +^ + g+^ − (f −^ + g−)
f +^ + g+^ −
f −^ + g−
f +^ +
g+^ − (
f −^ +
g−) =
f +
g.
(Only at the fourth equality we apply what we had just proved for non-negative functions. The third and last equality are just by the definition of the integral.) Here is another application of the Monotone Convergence Theorem.
Theorem 5.2 (Beppo Levi). Let fn : X → [0, ∞] be measurable. Then
∫ (^) ∑∞
n=
fn =
∑^ ∞
n=
fn.
Proof. Let gN =
∑N
n=1 fn, and^ g^ =^
n=1 fn.^ The Monotone Convergence Theorem applies to gN , and:
∫ g =
lim N →∞
gN = lim N →∞
gN = lim N →∞
∑^ N
n=
fn =
∑^ ∞
n=
fn.
There are many more applications like this. We postpone those for now, since you will probably be even more amazed by the next convergence theorem. We first need a lemma.
Lemma 5.3 (Fatou’s Lemma). Let fn : X → [0, ∞] be measurable. Then ∫ lim inf n→∞ fn ≤ lim inf n→∞
fn.
Proof. Set gn = infk≥n fk, so that gn ≤ fn, and {gn} ↗ lim infn fn. Then ∫ lim inf n fn =
lim n gn = lim n
gn = lim inf n
gn ≤ lim inf n
fn.
Remark 5.4. By adding and subtracting a constant, the hypotheses may be weakened to allow functions that are bounded below by any fixed number, not just non-negative functions. (This lower bound condition cannot be dropped.) The same considerations apply to the Monotone Convergence Theorem.
The following definition is used to formulate a crucial hypothesis of the theorem that is about to follow.
Definition 5.1. A function f : X → R is called integrable if it is measurable and
X |f^ |^ <^ ∞. It is immediate that f is integrable if and only if f +^ and f −^ are both integrable. It is also helpful to know, that
|f | < ∞ must imply |f | < ∞ almost everywhere.
Theorem 5.5 (Dominated Convergence Theorem). Let (X, μ) be a mea- sure space. Let fn : X → R be a sequence of measurable functions converging pointwise to f. Moreover, suppose that there is an integrable function g such that |fn| ≤ g, for all n. Then fn and f are also integrable, and
n^ lim→∞
X
|fn − f | dμ = 0.
Proof. Obviously fn and f are integrable. Also, 2g − |fn − f | is measurable and non-negative. By Fatou’s lemma, ∫ lim inf n (2g − |fn − f |) ≤ lim inf n
(2g − |fn − f |).
Since fn converges to f , the left-hand quantity is just
2 g. The right-hand quantity is:
lim inf n
2 g −
|fn − f |
2 g + lim inf n
|fn − f |
2 g − lim sup n
|fn − f |.
Since
2 g is finite, it may be cancelled from both sides. Then we obtain
lim sup n
|fn − f | ≤ 0 , i.e. lim n→∞
|fn − f | = 0.
Remark 5.6. It obviously suffices to only require that fn converge to f pointwise almost everywhere, or that |fn| is bounded above by g almost everywhere. (Of course, if fn only converges to f almost everywhere, then the theorem would not automatically say that f is measurable.)
Then ν is a measure on (X, A), and for any measurable function f on X, ∫
X
f dν =
X
f g dμ ,
often written as dν = g dμ.
Proof. We prove ν is a measure. ν(∅) = 0 is trivial. For countable additivity, let {En} be measurable with union E, so that χE =
n=1 χEn , and
ν(E) =
E
g dμ =
X
gχE dμ
X
∑^ ∞
n=
gχEn dμ =
∑^ ∞
n=
X
gχEn dμ =
∑^ ∞
n=
ν(En).
Next, if f = χE for some E ∈ A, then ∫
X
f dν =
X
χE dν = ν(E) =
X
χE g dμ =
X
f g dμ.
By linearity, we see that
f dν =
f g dμ whenever f is non-negative simple. For general non-negative f , we use a sequence of simple approximations {ϕn} ↗ f , so {ϕng} ↗ f g. Then by the monotone convergence, ∫
X
f dν = lim n→∞
X
ϕn dν = lim n→∞
X
ϕng dμ =
X
lim n→∞ ϕng dμ =
X
f g dμ.
Finally, for f not necessarily non-negative, we apply the above to its positive and negative parts, and use linearity.
The procedure of proving some fact about integrals by first reducing to the case of simple functions and non-negative functions is used quite often. (It will get quite monotonous if we had to detail the procedure every time we use it, so we won’t anymore if the circumstances permit.) Also, we should note that if f is only measurable but not integrable, then the integrals of f +^ or f −^ might be infinite. If both are infinite, the integral of f is not defined, although the equation of the theorem might still be interpreted as saying that the left-hand and right-hand sides are undefined at the same time. For this reason, and for the sake of the clarity of our exposition, we will not bother to modify the hypotheses of the theorem to state that f must be integrable. Problem cases like this also occur for some of the other theorems we present, and there I will also not make too much of a fuss about these problems, trusting that you understand what happens when certain integrals are undefined.
Theorem 6.3 (Change of variables). Let X, Y be measure spaces, and g : X → Y , f : Y → R be measurable. Then ∫
X
(f ◦ g) dμ =
Y
f dν ,
where ν(B) = μ(g−^1 (B)) is a measure defined for all measurable B ⊆ Y.
Proof. First suppose f = χB. Let A = g−^1 (B) ⊆ X. Then f ◦ g = χA, and we have ∫
Y
f dν =
Y
χB dν = ν(B) = μ(g−^1 (B)) = μ(A) =
X
(f ◦ g) dμ.
Since both sides of the equation are linear in f , the equation holds whenever f is simple. Applying the “standard procedure” mentioned above, the equation is then proved for all measurable f.
Remark 6.4. The change of variables theorem can also be applied “in reverse”. Suppose we want to compute
Y f dν, where^ ν^ is already given to us. Further assume that g is bijective and its inverse is measurable. Then we can define μ(A) = ν(g(A)), and it follows that
Y f dν^ =^
X (f^ ◦^ g)^ dμ. Our theorem (especially when stated in the reverse form) is clearly related to the usual “change of variables” theorem in calculus. If g : X → Y is a bijec- tion between open subsets of Rn, and both it and its inverse are continuously differentiable (i.e. g is a diffeomorphism), and ν = λ is the Lebesgue measure in Rn, then (as we shall prove rigorously in Lemma 12.1),
μ(A) = λ(g(A)) =
A
|det Dg| dλ.
Appealing to Theorems 6.2 and 6.3, we obtain:
Theorem 6.5 (Differential change of variables in Rn). Let g : X → Y be a diffeomorphism of open sets in Rn. If A ⊆ X is measurable, and f : Y → R is measurable, then ∫
g(A)
f dλ =
A
(f ◦ g) dμ =
A
(f ◦ g) · |det Dg| dλ.
The next two theorems are the Lebesgue versions of well-known results about the Riemann integral.
Theorem 6.6 (First Fundamental Theorem of Calculus). Let I ⊆ R be an interval, and f : I → R be integrable (with Lebesgue measure in R). Then the function
F (x) =
∫ (^) x
a
f (t) dt
is continuous. Furthermore, if f is continuous at x, then F ′(x) = f (x).
Proof. To prove continuity, we compute:
F (x + h) − F (x) =
∫ (^) x+h
x
f (t) dt =
I
f (t) · χx,x+h dt.
