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Lecture Notes in Real Analysis
Lewis Bowen∗
University of Texas at Austin
December 8, 2014
Contents
1 Outer measure and measurable sets 3
2 Measures and measurable sets 4
3 Cantor sets and the Cantor-Lebesgue function 5
4 Measurable functions 5
5 Borel functions (tangential and optional) 7
6 Semi-continuity (tangential) 8
7 Littlewood’s 3 principles 8 7.1 An aside...................................... 11
8 Convergence in measure 11
9 Integration for bounded functions 12
10 Integration for nonnegative functions 15
11 Integrable functions 16
12 Convergence Theorems 17
13 Riemannian integration 20
∗supported in part by NSF grant DMS-0968762, NSF CAREER Award DMS-0954606 and BSF grant 2008274
- 14 Differentiation
- 15 Differentiation of monotone functions
- 15.1 Functions of Bounded Variation
- 16 Absolutely continuous functions
- 17 Convex functions
- 18 Lp spaces
- 19 Hilbert spaces
- 20 Signed measures
- 21 Radon-Nikodym Theorem
- 22 The Dual of Lp
- 23 Outer measure
- 24 Carath´eodory’s Extension Theorem
- 25 Product measures
- 26 Fubini’s Theorem
- 27 Convolution
- 27.1 Norm inequalities
- 27.2 Smoothness
- 27.3 Approximate identities
- 28 The space of measures
- 28.1 Regularity
- 28.2 Some examples
- 29 Fourier series
- 29.1 Definition and convolution
- 29.2 L
- 29.3 Derivatives and absolute convergence
- 29.4 Localization
2 Measures and measurable sets
Exercise 10. If E ⊂ R is measurable then for every > 0 there exists a closed set F ⊂ E with m∗(E \ F ) < .
Proof. Because E is meas., its complement Ec^ is also meas. So if > 0 then there exists an open set O ⊃ Ec^ with m∗(O \ Ec) < . Now Oc^ is a closed set, Oc^ ⊂ E and E \ Oc^ = E ∩ O = O \ Ec. So m∗(E \ Oc) < .
Exercise 11. If E 1 , E 2 ,... ⊂ R are measurable and pairwise disjoint then
m∗(∪iEi) =
i
m∗(Ei).
Definition 6. Let X be a set and C a σ-algebra on X. We say that (X, C) is a measurable space (or Borel space). Also let μ : C → [0, ∞] be a function satisfying: if E 1 , E 2 ,... ∈ C are pairwise disjoint then
μ(∪iEi) =
i
μ(Ei).
Then μ is a measure on (X, C) and (X, C, μ) is a measure space. Often we omit C from the notation and just say “μ is a measure on X”.
Observation 4. m∗^ is a measure on (R, M) where M denotes the collection of measurable subsets. From now on, we let m denote the restriction of m∗^ to M. This is called Lebesgue measure on R.
Exercise 12. Let (X, C, μ) be a measure space. Suppose E 1 ⊂ E 2 ⊂ · · · ∈ C and F 1 ⊃ F 2 ⊃ · · · ∈ C. Then lim i
μ(Ei) = μ(∪iEi).
If μ(F 1 ) < ∞ then limi μ(Fi) = μ(∩iFi).
Definition 7. A countable intersection of open subsets is called a set of type Gδ. A countable union of closed subsets is called a set of type Fσ.
Example 1. The irrational numbers are a dense Gδ subset of the real line.
Exercise 13. Let E ⊂ R. Prove that the following are equivalent.
- E is measurable.
- for every > 0 there exists an open set O ⊃ E with m∗(O \ E) < ;
- there exists a set G ⊃ E of type Gδ such that m∗(G \ E) = 0;
- for every > 0 there exists a closed set F ⊃ E with m∗(E \ F ) < ;
- there exists a set F ⊃ E of type Fσ such that m∗(E \ F ) = 0;
- there exists a Borel set B such that m(E M B) = 0 where E M B = (E \ B) ∪ (B \ E) is the symmetric difference of B and E;
- for every set A ⊂ R, m∗(A) = m∗(A ∩ E) + m∗(A \ E).
3 Cantor sets and the Cantor-Lebesgue function
...
4 Measurable functions
Definition 8. Let X, Y be topological spaces. We will assume X is endowed with a sigma- algebra so that we can meaningfully discuss measurable subsets of X. A function f : X → Y is
- continuous if for every open O ⊂ Y , f −^1 (O) is open;
- measurable if for every open O ⊂ Y , f −^1 (O) is measurable.
Observation 5. Every continuous function is measurable.
We will concern ourselves with measurable functions into the extended reals R∪{−∞, +∞}.
Exercise 14. Let f : X → R ∪ {±∞} be a function. TFAE
- f is measurable;
- for every a ∈ R, f −^1 (a, +∞] is measurable;
- for every a ∈ R, f −^1 [a, +∞] is measurable;
- for every Borel subset B ⊂ R ∪ {±∞}, f −^1 (B) is measurable.
Proof. Clearly (1) ⇒ (2). So assume (2). By taking complements, we see f −^1 [−∞, a] is measurable ∀a. Since f −^1 ([−∞, a)) = ∪r∈Q,r<af −^1 [−∞, r]
it follows that f −^1 ([−∞, a)) is measurable. By taking complements again we see that f −^1 [a, +∞] is measurable. So (2) ⇒ (3). Now assume (3). We will prove (1). Since every open subset is a countable union of open intervals, it suffices to show that f −^1 (I) is measurable whenever I is an open interval. By taking complements we see that this is true if I = [−∞, a) for some a. Because f −^1 (a, +∞] = ∪r∈Q,r>af −^1 [r, +∞], it is also true if I = f −^1 (a, +∞] for some a. So it’s true whenever I is an infinite interval. If I is finite then I = (a, b) for some a, b ∈ R in which case f −^1 (I) = f −^1 (a, +∞] ∩ f −^1 [−∞, b). So it’s true in this case too.
Definition 9. Let f, g : X → R be functions. We say that f = g almost everywhere (a.e.) if {x ∈ X : f (x) 6 = g(x)} has measure zero.
Exercise 17. Suppose f = g a.e. If f is measurable then so is g.
Proof. Let Z = {x ∈ X : f (x) 6 = g(x)} and let O ⊂ R ∪ {±∞} be open. Then g−^1 (O) \ Z = f −^1 (O) \ Z. So there exists a subset Z′^ ⊂ Z such that g−^1 (O) = Z′^ ∪ (f −^1 (O) \ Z). Because Z, Z′^ have measure zero, they are both measurable. Thus g−^1 (O) is measurable. Since O is arbitrary, this proves g is measurable.
Definition 10. The characteristic function or indicator function of a set E ⊂ X is the function χE : X → R given by χE (x) = 1 if x ∈ E and χE (x) = 0 otherwise. A simple function is a finite linear combination of characteristic functions of measurable subsets.
Exercise 18. Let f : X → R be a function. TFAE
- f is simple
- f is measurable and the range of f is finite
- there exist disjoint measurable sets E 1 ,... , Ek and real numbers r 1 ,... , rk such that
f (x) =
i
riχEi.
Later on, we will use simple functions to develop integration theory. It will be useful to have the following approximation result:
Exercise 19. Let f be a measurable function on X. Then there exist simple functions {fi}i such that f = limi fi pointwise. Moreover, if f ≥ 0 then we can choose fi so that 0 ≤ f 1 ≤ f 2 ≤ · · · ≤ f.
Proof. Define fn by
f (x) =
−n f (x) < −n k 2 −n^ k 2 −n^ ≤ f (x) < (k + 1)2−n n + 1 f (x) ≥ n + 1
5 Borel functions (tangential and optional)
Definition 11. Let X, Y be topological spaces. A function f : X → Y is Borel if for every open O ⊂ Y , f −^1 (O) is Borel.
Observation 6. Every continuous function is Borel and every Borel function is measurable (as long as Borel sets are measurable which is usually the case).
Exercise 20. A function f : X → Y is Borel if and only if for every Borel set B ⊂ Y , f −^1 (Y ) is Borel. Hence compositions of Borel functions are Borel.
Exercise 21. There is a Borel function f : [0, 1] → [0, 1] and a measurable set X ⊂ [0, 1] such that f −^1 (X) is not measurable.
Exercise 22. There are measurable functions f, g : R → R whose composition is not mea- surable.
6 Semi-continuity (tangential)
We won’t use this much but it is good to have in your vocabulary:
Definition 12. f : X → R∪{±∞} is upper semi-continuous if f −^1 [−∞, a) is open for every a ∈ R (where X is a topological space). f is lower semi-continuous if f −^1 (a, +∞] is open for every a ∈ R.
Exercise 23. Let X be a compact metric space. Then f : X → R ∪ {±∞} is upper semi- continuous if and only if there exist continuous functions f 1 , f 2 ,... such that f = infi fi. Similarly, f is lower semi-continuous if and only if there exist continuous functions f 1 , f 2 ,... such that f = supi fi.
7 Littlewood’s 3 principles
Littlewood’s three principles are:
- Every subset of the real line of finite measure is nearly a finite union of intervals.
- Every measurable function is nearly continuous.
- Every convergent sequence of functions is nearly uniformly convergent.
Let us make this rigorous:
Exercise 24 (First principle). Suppose E ⊂ R is measurable and has finite measure. Prove: for every > 0 there exists a finite union of open intervals O such that m(O M E) < .
Proof. There exists an open set O ⊃ E with m(O \ E) < . Since O is a countable union of intervals, this means there is a finite union of intervals O′^ ⊂ O with m(O \O′) < . Therefore m(O′^ M E) < 2 .
We will prove:
Theorem 7.1 (Second principle: Lusin’s Theorem). Suppose f : R → R measurable func- tion. For every > 0 there exists a continuous function g such that
m({x ∈ R : f (x) 6 = g(x)}) < .
We now start with the proof of Lusin’s theorem.
Exercise 27. If f : E → R is simple and > 0 then there exists a closed set F ⊂ E such that m(E \ F ) < and f F is continuous.
Proof. Because f is simple f =
∑n i=1 ciχEi^ for some coefficients^ ci^ and some pairwise disjoint measurable sets Ei. Let Fi ⊂ Ei be closed sets with m(Ei \ Fi) < /n. Let F = ∪Fi. Note m(E \ F ) < and f restricted to F is continuous.
Exercise 28. If f : R → R is measurable and > 0 then there exists a closed set F ⊂ R such that m(R \ F ) < and f F is continuous.
Proof. Let us first assume that m(E) < ∞. Let {fk} be a sequence of simple functions that converge pointwise to f. For each k, let Fk ⊂ E be a closed set with m(R \ Fk) < / 2 k^ such that fk Fk is continuous. By Egorov’s Theorem there exists B ⊂ R such that m(B) < and {fk} converges uniformly to f on R \ B. Wlog, we may require B to be open so that F 0 := R \ B is closed. (if B is not open we may replace it with an open set O ⊃ B such that m(O) < ). Because a uniformly convergent sequence of continuous functions is continuous, f F∞ := F 0 ∩
k Fk^ is continuous. Note
m(E \ F∞) ≤
∑^ ∞
k=
m(E \ Fk) < 2 .
So this handles the case m(E) < ∞. To obtain the general case, we find finite measure sets E 0 ⊂ E 1 ⊂ E 2 ⊂ · · · ⊂ R such that R = ∪Ei. Then for each i there exists a closed set Fi ⊂ Ei with m(Ei \ Fi) < / 2 i^ such that f Fi is continuous. Note that m(R \ ∪iFi) < . So there exists finite I such that m(R\∪Ii=0Fi) < . Observe that f is continuous on ∪Ii=0Fi.
Exercise 29. Prove Tietze’s Extension Theorem: if F ⊂ R is any closed set and f : F → R any continuous function then there exists a continuous function g : R → R whose restriction to F equals f.
Proof. There exist pairwise disjoint open intervals O 1 , O 2 ,... such that the complement of F is the union ∪iOi. Write Oi = (ai, bi). Now define g by g(x) = f (x) if x ∈ F and
g(x) =
bi − x bi − ai
f (ai) +
x − ai bi − ai
f (bi)
if x ∈ (ai, bi).
Remark 1. Tietze’s Extension Theorem holds in much greater generality: you can allow F to be a closed subset of any normal topological space. (Normal means that every two disjoint closed subsets have disjoint open neighborhoods. For example, metric spaces are normal).
Lusin’s Theorem is an immediate consequence of the previous two exercises. Lusin’s Theorem can be generalized a great deal. To explain we need:
Definition 13. A measure m on a topological space X is regular if for every measurable set E ⊂ X, every > 0 there exists an open set O ⊃ E and a compact set F ⊂ E such that
m(O \ E) < , m(E \ F ) < .
For example, Lebesgue measure is regular.
Regular measures are common:
Theorem 7.3. Every Borel measure on a locally compact secound countable space is regular.
(We won’t prove this here).
Exercise 30 (Lusin’s Theorem). Suppose X is a normal Hausdorff topological space and m is a σ-finite regular measure on X. Then for any measurable function f : X → R and > 0 there exists a continuous g : X → R such that
m({x ∈ X : f (x) 6 = g(x)}) < .
7.1 An aside
It can be shown that a measurable function f : [a, b] → R is Riemann-integrable if and only if the set of all x ∈ [a, b] such that f is discontinuous at x has measure zero. Example: Let f = χQ. Claim: f is discontinuous everywhere. Proof: For any x ∈ R and any δ > 0 we have that (x − δ, x + δ) ∩ Q 6 = ∅ and (x − δ, x + δ) ∩ Qc^6 = ∅. So there y, z with |x − y| < δ such that y ∈ Q and z /∈ Q. Then either |f (x) − f (y)| = 1 or |f (x) − f (z)| = 1. In any case, f is discontinuous at x.
8 Convergence in measure
Definition 14. Let f 1 , f 2 ,... , f∞ be measurable functions all defined on some measurable subset E ⊂ R. We say
- {fn} converges pointwise to f∞ if f∞(x) = limn fn(x) for every x ∈ E;
- {fn} converges pointwise a.e. to f∞ if f∞(x) = limn fn(x) for a.e. x ∈ E;
- {fn} converges in measure to f∞ if for every > 0,
lim n m({x ∈ X : |fn(x) − f∞(x)| > }) = 0.
Exercise 31. Suppose E has finite measure and fn → f∞ on E pointwise a.e. Show that fn → f∞ in measure.
Proof. This follows from Egorov’s Theorem. Note: it is necessary that E have finite measure.
We observe ∫
f + g dm =
c∈R
cm({x ∈ X : f (x) + g(x) = c})
c∈R
c
b∈R
m({x ∈ X : f (x) = b, g(x) = c − b})
b∈R
c∈R
(b + c)m({x ∈ X : f (x) = b, g(x) = c})
b∈R
c∈R
bm({x ∈ X : f (x) = b, g(x) = c}) +
b∈R
c∈R
cm({x ∈ X : f (x) = b, g(x) = c})
b∈R
bm({x ∈ X : f (x) = b}) +
c∈R
cm({x ∈ X : g(x) = c})
f dm +
g dm.
Exercise 35. Suppose E ⊂ X has finite measure and f : E → R is a bounded measurable function.
sup φ≤f
φ dm = inf f ≤ψ
ψ dm
where the sup and inf are over simple functions from E to R.
Proof. It’s easy to see that ≤ must occur. Indeed, if φ ≤ f ≤ ψ are as above and
φ =
∑^ n
i=
ciχEi , ψ =
∑^ m
i=
diχFi
then after refining the partition {E 1 ,... , En}, {F 1 ,... , Fm} we may assume that they are equal. That is, we may assume that
φ =
∑^ k
i=
ciχGi , ψ =
∑^ k
i=
diχGi
for some measuable sets G 1 ,... , Gk. Since ci ≤ di, the inequality ≤ follows. Because f is bounded there is some M > 0 such that |f (x)| ≤ M for all x. For n > 0 and k ∈ Z, |k| ≤ 2 n^ let
En,k = {x ∈ X : k 2 −n^ ≤ f (x)/M < (k + 1)2−n}.
Let φn(x) = M k 2 −n^ and ψn(x) = M (k + 1)2−n^ on En,k. These are both simple functions and by design |φn − ψn| ≤ M 2 −n. So ∣ ∣ ∣ ∣
φn dm −
ψndm
∣ ≤^ M^2
−n ∑ k
m(En,k) = M 2 −nm(E).
Since this tends to 0 as n → ∞, it proves the exercise.
Definition 16. With f, X as above we define ∫
E
f dm = sup φ≤f
φ dm = inf f ≤ψ
ψ dm.
If Y ⊂ X is measurable then (^) ∫
Y
f dm =
f χY dm.
If [a, b] ⊂ X ⊂ R is an interval and m is Lebesgue measure then
∫
[a,b]
f dm =
∫ (^) b
a
f (x) dx.
If X = E or if E is understood from the context then we usually drop the subscript and simply write
f dm.
Exercise 36. Suppose E ⊂ X has finite measure and f, g : E → R are bounded measurable functions and a, b ∈ R. Then
- (linearity)
E af^ +^ bg dm^ =^ a^
f dm + b
g dm,
- (monotonicity) if f ≤ g a.e. then
f dm ≤
g dm. So f = g a.e. then
∫ f dm^ = g dm.
- If A ≤ f ≤ B then Am(E) ≤
f dm ≤ Bm(E),
- (finitely additive) if∫ A 1 , A 2 ,... , An ⊂ X are disjoint, measurable and ∪iAi = E then f dm =
i
Ai f dm.
Proof. Suppose that a, b > 0. If φ 1 ≤ f ≤ ψ 1 and φ 2 ≤ g ≤ ψ 2 are simple functions then
aφ 1 + bφ 2 ≤ af + bg ≤ aψ 1 + bψ 2
are simple functions and ∫
E
aφ 1 + bφ 2 dm = a
φ 1 dm + b
φ 2 dm
and a similar formula holds with ψi in place of φi. Using the definition of the integral, this implies (1).
∫ To prove (2) we simply notice that if^ φ^ ≤^ f^ is a simple function since^ φ^ ≤^ g^ we also have φ dm ≤
g dm by definition. So taking the sup over all φ ≤ f we obtain (2). Item (3) follows from item (4) by letting f (or g) be a constant. To see item (4), observe that f =
i f χAi. So (4) follows from (1).
Exercise 38. Let f : R → R be a nonnegative function and R = {(x, y) : 0 ≤ y ≤ f (x)} be the region under the graph of f. Then
∫ f dm = m(R).
Proof. By linearity, it suffices to prove this in the special case in which f has finite support. This is clear if f is the characteristic function of an interval. Therefore, it is also true if f is a finite linear combination of such characteristic functions. Since any measurable set with finite measure is nearly a finite union of intervals (Littlewood’s first principle), it is also true if f = χE where E is measurable. Therefore it is true if f is simple. Using Egorov’s Theorem and approximation of f by simple functions, we obtain the result for arbitrary f.
11 Integrable functions
Definition 18. A measurable function f : X → R is integrable if ∫ |f | dm < ∞.
Because |f | is a nonnegative measurable function,
|f | dm is well-defined. In this case, let
f +^ = max(f, 0)
f −^ = max(−f, 0).
Note: f +, f −^ are measurable functions, f = f +^ − f −^ and |f | = f +^ + f −. Define
∫ f dm =
f +^ dm −
f −^ dm.
Exercise 39. Suppose f, g : X → R are integrable and c ∈ R is a constant. Then
- cf is integrable and
cf dm = c
f dm,
- f + g is integrable and
f + g dm =
f dm +
g dm,
- if f ≤ g a.e. then
f dm ≤
g dm. In particular if f = g a.e. then
f dm =
g dm.
- if X = A ∪ B where A, B are disjoint measurable sets, then ∫ f dm =
A
f dm +
B
f dm.
Proof. Suppose that c > 0. Then (cf )+^ = cf +^ and (cf )−^ = cf −. This implies (1) when c > 0. The case c ≤ 0 is similar. To prove (2), suppose f = f 1 − f 2 where f 1 , f 2 are nonnegative measurable functions. Then f = f +^ − f −^ = f 1 − f 2
implies f +^ + f 2 = f −^ + f 1. By the previous section on integrals of nonnegative functions, ∫ f +^ dm +
f 2 dm =
f +f 2 dm =
f −^ + f 1 dm =
f −^ dm +
f 1 dm.
Therefore, (^) ∫
f dm =
f +^ dm −
f −^ dm =
f 1 dm −
f 2 dm.
Now let us apply this to the situation at hand: ∫ f + g dm =
f +^ − f −^ + g+^ − g−^ dm
(f +^ + g+) − (f −^ + g−) dm
(f +^ + g+) dm −
(f −^ + g−) dm
f +^ dm +
g+^ dm −
f −^ dm +
g−^ dm
f dm +
g dm.
This proves (2). (3) follows from (2) since
g − f dm =
g dm −
f dm.
(4) follows from (2) since f = χAf + χB f.
12 Convergence Theorems
There are four convergence theorems which state that if fn converges in some sense to a function f then
fn dm →
f dm (or at least we have an inequality if not an equality).
Exercise 40 (Bounded Convergence Theorem). Suppose m(X) < ∞. Let {fn} be a uni- formly bounded sequence of measurable functions on X that converge pointwise a.e. to f. (uniformly bounded means there is a number M > 0 such that |fn(x)| ≤ M for a.e. x and every n). Then (^) ∫
fn dm →
f dm
as n → ∞.
Exercise 44. If X 1 , X 2 ,... are pairwise disjoint measurable sets whose union is X and f ≥ 0 is measurable then (^) ∫
f dm =
∑^ ∞
i=
Xi
f dm.
(therefore if μ(E) =
E f dm^ then^ μ^ is a measure on^ X).^ More generally, if^ {fn}^ are nonnegative functions and f =
n fn^ then ∫ f dm =
n
fn dm.
Exercise 45 (Lebesgue’s Dominated Convergence Theorem). Let {fn} be a sequence of mea- surable functions on X that converge pointwise a.e. to f. Suppose there is an integrable function g such that 0 ≤ fn ≤ g for all n. Then ∫ fn dm →
f dm
Proof. Proof #1:One way to do this is to realize that g dm is a measure. That is: we define a new measure μ on X by
μ(E) =
E
g dm.
We’ve already proven that this really defines a measure. Moreover, if k is any bounded measurable function then (^) ∫
k dμ =
kg dm.
(To see this, note that it’s true for characteristic functions and therefore true for simple functions and therefore true for arbitrary bounded measurable functions). After replacing X with the support of g if necessary we may assume that g > 0 a.e. By the Bounded Convergence Theorem, ∫ fn/g dμ →
f /g dμ.
(This is because μ(X) =
g dm < ∞ and 0 ≤ fn/g ≤ 1 is bounded). Since
k dμ =
kg dm for any bounded measurable function k, this implies the exercise. Proof #2: By Fatou’s Lemma,
lim inf n
fn dm ≥
f dm.
On the other hand, g − fn ≥ 0 converges pointwise a.e. to g − f. So Fatou’s Lemma implies
lim inf n
g − fn dm ≥
g − f dm
In other words,
lim sup n
fn ≤
f dm.
Exercise 46. The hypotheses of Lebesgue’s Dominated Convergence Theorem can be weak- ened by replacing 0 ≤ fn ≤ g with h ≤ fn ≤ g where h, g are integrable functions.
Proof. Observe that 0 ≤ fn −h ≤ g −h. By the previous exercise,
fn −h dm →
f −h dm. Cancelling
−h dm from both sides proves the result.
Definition 19. If (X, m) is a measure space and p > 0, we let Lp(X, m) be the set of all equivalence classes classes of measurable functions f : X → C such that ∫ |f |p^ dm < ∞.
Here: two functions f, g are equivalent if they agree a.e. By abuse of notation, we may write f ∈ L^1 (X, m) to mean that f is integrable (for example). Later on, we will show that if p ≥ 1 and f, g ∈ Lp(X, m) then
‖f − g‖p =
|f − g|p^ dm
) 1 /p
is a metric on Lp(X, m). We will study the topological aspects of these spaces later. Also, we let L∞(X, m) denote the set of equivalence classes of bounded measurable func- tions. We write ‖f ‖∞ = sup{a ≥ 0 : m(f −^1 [a, ∞]) > 0 }.
(This is called the essential supremum of f ). This also gives a metric on L∞(X, m): the distance between f and g is ‖f − g‖∞.
13 Riemannian integration
Let f be a function on an interval [a, b] ⊂ R. Let Σ = {x 0 , x 1 ,... , xn} ⊂ [a, b] with x 0 = a, xn = b. Define
IΣ+ (f ) =
∑^ n
i=
(sup f [xi− 1 , xi])|xi − xi− 1 |
and
I Σ− (f ) =
∑^ n
i=
(inf f [xi− 1 , xi])|xi − xi− 1 |.
Let diam(Σ) = maxi |xi − xi− 1 |. We define the Riemannian integral of f (if it exists) by
R
f dx = lim diam(Σ)→ 0
I+Σ (f ) = lim diam(Σ)→ 0
I Σ− (f ).
We say f is Riemann-integrable if its Riemann integral exists and is finite. We will prove:
