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Math 3333 - Intermediate Analysis - David Blecher
Notes on integration (contd)
The following condition is very useful:
Theorem. (Riemann condition) A bounded function f : [a, b] → R is integrable if and only if ∀ > 0, ∃ a partition P of [a, b] such that U (f, P ) − L(f, P ) < .
Proof: By definition, f is integrable if and only if U (f ) = L(f ). (⇐) Suppose that ∀ > 0, ∃ a partition P of [a, b] such that U (f, P ) − L(f, P ) < . Using this, and the definition of L(f ) and U (f ),
U (f ) ≤ U (f, P ) < L(f, P ) + ≤ L(f ) + .
Since this is true for every > 0, we have U (f ) ≤ L(f ), by Theorem 11.7. But L(f ) ≤ U (f ) by Theorem 29.6, so L(f ) = U (f ).
(⇒) If U (f ) = L(f ), and if > 0 is given, choose (by definition of L(f ) and U (f )), partitions Q and R such that L(f, Q) > L(f ) − 2 , and U (f, R) < U (f ) + 2 . Let P = Q ∪ R, the refinement of both Q and R obtained by taking their union. Then by the last equations, and Theorem 29.4, we have U (f, P ) ≤ U (f, R) < U (f ) +
= L(f ) +
< L(f, Q) +
≤ L(f, P ) + .
Looking at the left and right side of the last line, we see that U (f, P ) − L(f, P ) < , which is what was required.
Definition. A function f : D → R is called uniformly continuous if ∀ > 0, ∃ δ > 0 such that |f (x) − f (y)| < whenever x, y ∈ D, and |x − y| < δ.
Can you spot the difference between the definitions of f being continuous, and f being uniformly continuous? The only difference is that in uniformly continuity, the δ does not depend on the points in D.
Theorem. If D is compact, and f : D → R is continuous, then f is uniformly continuous.
Proof: Suppose that f is not uniformly continuous. Thus ∃ > 0 such that ∀δ > 0, ∃ x and y in D such that |x − y| < δ but |f (x) − f (y)| ≥ . Let us take δ = (^1) n , for n ∈ N. Thus ∃ xn and yn in D such that |xn − yn| < (^1) n , but |f (xn) − f (yn)| ≥ . Since D is compact, it is bounded, so by Theorem 19.7 the sequence (xn) has a convergent subsequence, xn 1 , xn 2 , xn 3 , · · · say. Suppose that this subsequence converges to a point x say. Since D is closed, by the Bonus problem we must have x ∈ D. Similarly, the subsequence yn 1 , yn 2 , yn 3 , · · · has a convergent subsequence ym 1 , ym 2 , ym 3 , · · · say, converging to a point y ∈ D. By Theorem 19.4, the further subsequence xm 1 , xm 2 , xm 3 , · · · converges to x too. Now |xmk − ymk | ≤ (^) m^1 k → 0 as k → ∞. On the other hand, by Fact 9 on sequences, xmk − ymk → x − y, and so by the squeezing (Fact 5 on sequences) |x − y| = 0. Thus x = y. Now f (xmk ) → f (x) and f (ymk ) → f (y) by Theorem 21.2(b), so by an argument similar to the above, |f (xmk ) − f (ymk )| → |f (x) − f (y)| = 0. On the other hand, |f (xmk ) − f (ymk )| ≥ , so limk |f (xmk ) − f (ymk )| ≥ (by Fact 4 on sequences). This is a contradiction. 1
Theorem. If f : [a, b] → R is continuous, then f is integrable.
Proof: By the previous theorem, f is uniformly continuous. Thus given > 0, there is a number δ > 0 such that |f (x) − f (y)| < (^) b−a whenever x, y ∈ [a, b] and |x − y| < δ. Choose a partition P = {x 0 , x 1 , · · · , xn} of [a, b] such that ∆xk = xk − xk− 1 < δ for every k = 1, 2 , · · · , n. Consider the interval [xk− 1 , xk]. By the Min-max theorem, f has a maximum value Mk and a minimum value mk on this interval; so there are numbers s and t in [xk− 1 , xk] with f (s) = Mk, f (t) = mk. Since |s − t| ≤ ∆xk < δ, we conclude that
Mk − mk = |f (s) − f (t)| <
b − a
Now
U (f, P ) − L(f, P ) =
∑^ n
k=
∆xk Mk −
∑^ n
k=
∆xk mk =
∑^ n
k=
∆xk (Mk − mk),
and so
U (f, P ) − L(f, P ) <
∑^ n
k=
∆xk
b − a
= (b − a)
b − a
Thus f satisfies the ‘Riemann condition’ Theorem above, and so f is integrable.
Example. Calculate
0 x
(^2) dx, using upper or lower sums.
Solution. You probably saw this example worked in Calculus 1. If you did not, please read 29. in the Text for more details if you need them. Here we take the partition Pn = { 0 , (^1) n , (^) n^2 , · · · , n− n 1 , 1 }.
So ∆xk = (^1) n for all k = 1, 2 , · · · , n. If you draw a picture, it is clear that the numbers Mk = ( kn )^2 and mk = (k− n 1 )^2. Thus
U (f, Pn) =
∑^ n
k=
Mk ∆xk =
∑^ n
k=
k n
)^2
n
n^3
∑^ n
k=
k^2.
We know (see section on Mathematical induction), that the last sum equals n(n+1)(2 6 n+1). Thus
U (f, Pn) =
n^3
n(n + 1)(2n + 1) 6
n
n
as n → ∞. A similar argument shows that
L(f, Pn) =
n
n
as n → ∞. It is easy to see from these, and using also Fact 4 for sequences, that U (f ) ≤ 13 and
L(f ) ≥ 13. But L(f ) ≤ U (f ), so L(f ) = U (f ) = 13. Hence
0 x
(^2) dx = 1
More properties of the Riemann integral.
Many of the properties of the Riemann integral follow from its definition, and from general properties of ‘inf’s’ and ‘sup’s’, which are applied to the mk and Mk. Lets illustrate this idea, by first proving the following properties of infima and suprema, and then applying these properties to prove a well-known fact about the integral.
Finally, if K is any negative number, then combining the facts we have proved, we get: ∫ (^) b
a
Kf dx =
∫ (^) b
a
(−(−K)f ) dx = −
∫ (^) b
a
(−K)f dx = −(−K)
∫ (^) b
a
f dx = K
∫ (^) b
a
f dx.
We already proved earlier that: Fact I2: If f : [a, b] → R is integrable, and if m ≤ f (x) ≤ M for all x ∈ [a, b], then
m(b − a) ≤
∫ (^) b
a
f dx ≤ M (b − a).
From this it is easy to deduce that if f : [a, b] → R is integrable, and if f (x) ≥ 0 for all x ∈ [a, b],
then
∫ (^) b a f dx^ ≥^ 0. Indeed, simply take^ m^ = 0 in Fact I2. Fact I3:
∫ (^) b a K dx^ =^ K(b^ −^ a), if^ K^ is a constant. Proof: Apply Fact I2, with m = M = K, and f (x) = K: we get K(b − a) ≤
∫ (^) b a K dx^ ≤^ K(b^ −^ a), which gives what we want.
Fact I4: If f and g are integrable on [a, b], and if f (x) ≤ g(x) for all x ∈ [a, b], then
∫ (^) b
∫ (^) b a^ f dx^ ≤ a g dx. Proof: If we take a partition P of [a, b], and if mk are the infimums used in the definition of L(f, P ), and if m′ k are the infimums used in the definition of L(g, P ), then
mk = inf{f (x) : x ∈ [xk− 1 , xk]} ≤ inf{g(x) : x ∈ [xk− 1 , xk]} = m′ k.
Thus
L(f, P ) =
∑^ n
k=
mk ∆xk ≤
∑^ n
k=
m′ k ∆xk = L(g, P ) ≤ L(g) =
∫ (^) b
a
g dx.
Taking the supremum over all partitions P we deduce that L(f ) ≤
∫ (^) b a g dx, which is what we need since L(f ) =
∫ (^) b a f dx.^ Fact I5:
∫ (^) b a (f^ +^ g)^ dx^ =^
∫ (^) b a f dx^ +^
∫ (^) b a g dx, if^ f^ and^ g^ are integrable on [a, b]. Proof: This uses infima and suprema in the kinds of ways we have done in the last few Facts. See textbook for details.
Fact I6: If f is integrable on [a, b], then so is |f (x)|, and |
∫ (^) b a f dx| ≤^
∫ (^) b a |f^ |^ dx. Proof: To see that |f | is integrable on [a, b], we show that it satisfies Riemann’s condition. Let > 0 be given. Since f satisfies Riemann’s condition, let P = {x 0 , x 1 , · · · , xn} be a partition of [a, b] with
U (f, P) − L(f, P) =
∑^ n
k=
Mk∆xk −
∑^ n
k=
mk∆xk =
∑^ n
k=
(Mk − mk)∆xk < .
If s, t ∈ [xk− 1 , xk] then |f (s)|−|f (t)| ≤ |f (s)−f (t)| ≤ Mk −mk. Thus |f (s)| ≤ Mk −mk +|f (t)|, and so sup{|f (t)| : xk− 1 ≤ t ≤ xk} ≤ Mk −mk +|f (t)|. Hence sup{|f (t)| : xk− 1 ≤ t ≤ xk}−(Mk −mk) ≤ |f (t)|, and so sup{|f (t)| : xk− 1 ≤ t ≤ xk} − (Mk − mk) ≤ inf{|f (t)| : xk− 1 ≤ t ≤ xk}. This gives
sup{|f (t)| : xk− 1 ≤ t ≤ xk} − inf{|f (t)| : xk− 1 ≤ t ≤ xk} ≤ Mk − mk.
Multiplying this last equation by ∆xk = xk − xk− 1 , and adding over all k, we have that
∑^ n
k=
∆xk sup{|f (t)| : xk− 1 ≤ t ≤ xk} − ∆xk inf{|f (t)| : xk− 1 ≤ t ≤ xk} ≤
∑^ n
k=
(Mk − mk)∆xk.
But this says precisely that
U (|f |, P) − L(|f |, P) ≤ U (f, P) − L(f, P) < .
So |f | satisfies Riemann’s condition. We have −|f (x)| ≤ f (x) ≤ |f (x)|. By Fact I4, and I1, we have that
−
∫ (^) b
a
|f | dx =
∫ (^) b
a
(−|f |) dx ≤
∫ (^) b
a
f dx ≤
∫ (^) b
a
|f | dx.
That is, |
∫ (^) b a f dx| ≤^
∫ (^) b a |f^ |^ dx.^ Fact I7:
∫ (^) b a f dx^ =^
∫ (^) c a f dx^ +^
∫ (^) b c f dx^ if^ a^ ≤^ c^ ≤^ b, and if^ f^ is integrable on [a, c] and on [c, b]. Proof: We will first show that f satisfies Riemann’s condition on [a, b]. So suppose that > 0 is given. Since f is integrable on [a, c] and on [c, b] it satisfies Riemann’s condition on those intervals. So there are partitions P 1 of [a, c] and P 2 of [c, b] such that
U (f, P 1 ) − L(f, P 1 ) < / 2 , U (f, P 2 ) − L(f, P 2 ) < / 2.
Let P = P 1 ∪ P 2 , a partition of [a, b]. It is easy to check that L(f, P) = L(f, P 1 ) + L(f, P 2 ) (draw a picture of the rectangles under the curve). Similarly, U (f, P) = U (f, P 1 ) + U (f, P 2 ). Thus
U (f, P) − L(f, P) = U (f, P 1 ) − L(f, P 1 ) + U (f, P 2 ) − L(f, P 2 ) < /2 + /2 = .
Thus f satisfies Riemann’s condition on [a, b] and so is integrable on [a, b]. Also ∫ (^) b
a
f dx ≤ U (f, P) = U (f, P 1 ) + U (f, P 2 ) ≤
∫ (^) c
a
f dx + +
∫ (^) b
c
f dx + .
Since was arbitrary, we have
∫ (^) b a f dx^ ≤^
∫ (^) c a f dx^ +^
∫ (^) b c f dx. Similarly, ∫ (^) b
a
f dx ≥ L(f, P) = L(f, P 1 ) + L(f, P 2 ) ≥
∫ (^) c
a
f dx +
∫ (^) b
c
f dx − 2 ,
and so
∫ (^) b a f dx^ ≥^
∫ (^) c a f dx^ +^
∫ (^) b c f dx. Thus^
∫ (^) b a f dx^ =^
∫ (^) c a f dx^ +^
∫ (^) b c f dx.^ Fact I8: If f is monotone on [a, b], then f is integrable on [a, b]. Proof: This one is relatively easy to prove, but since it is not so important we refer to the text for the proof.
∫ Fact I9:^ (The first fundamental theorem of Calculus)^ If^ f^ is integrable on [a, b], define^ F^ (x) = x a f dt, for^ x^ ∈^ [a, b]. If^ f^ is continuous at a point^ c^ ∈^ (a, b), then^ F^
′(c) = f (c).
