= t. Here I will write dt for dF. The generalized Riemann
integral does something very significant which is far superior to what can be achieved with
other integrals. It can always integrate derivatives. Suppose f is defined on an interval,
[a,b]
and that f^{′}
(x)
exists for all x ∈
[a,b]
, taking the derivative from the right or left at the
endpoints. What about the formula
∫
b ′
a f (t)dt = f (b)− f (a)? (12.14)
(12.14)
Can one take the integral of f^{′}? If f^{′} is continuous there is no problem of course. However,
sometimes the derivative may exist and yet not be continuous. Here is a simple
example.
Example 12.2.1Let
{ x2sin (12) if x ∈ (0,1]
f (x) = 0 if x =x0 .
You can verify that f has a derivative on
[0,1]
but that this derivative is not continuous.
The fact that derivatives are generalized Riemann integrable depends on the following
simple lemma called the straddle lemma by McLeod [28].
Lemma 12.2.2Suppose f :
[a,b]
→ ℝ is differentiable. Then there exist δ
(x )
> 0 suchthat if u ≤ x ≤ v and u,v ∈
[x− δ(x),x + δ(x)]
, then
|f (v) − f (u)− f′(x)(v − u)| < ε|v − u|.
Proof: Consider the following picture.
PICT
From the definition of the derivative, there exists δ
(x)
> 0 such that if
|v− x|
,
|x − u|
< δ
(x)
, then
′ ε
|f (u) − f (x)− f (x )(u − x)| < 2 |u− x|
and
|f′(x)(v− x)− f (v) + f (x)| < ε|v− x|
2
Now add these and use the triangle inequality along with the above picture to write
|f ′(x)(v− u) − (f (v) − f (u))| < ε |v − u |.
This proves the lemma.
The next proposition says 12.14 makes sense for the generalized Riemann integral.
Proposition 12.2.3Suppose f :
[a,b]
→ ℝ is differentiable. Then f^{′}∈ R^{∗}
[a,b]
and
∫
f (b)− f (a) = bf′dx
a
where the integrator function is F
(x)
= x.
Proof:Let ε > 0 be given and let δ
(x)
be such that the conclusion of the above lemma
holds for ε replaced with ε∕
(b − a)
. Then let P =
{(I ,t)}
i i
_{i=1}^{n} be δ fine. Then using
the triangle inequality and the result of the above lemma with Δx_{i} = x_{i}− x_{i−1},
| | | |
|| ∑n ′ || ||∑n ′ ||
||f (b)− f (a)− f (ti)Δxi || = || f (xi)− f (xi−1)− f (ti)Δxi ||
i=1 i=n1
≤ ∑ ε∕(b− a)Δx = ε.
i=1 i
This proves the proposition.
With this proposition there is a very simple statement of the integration by parts formula
which follows immediately.
Corollary 12.2.4Suppose f,g are differentiable on
[a,b]
. Then f^{′}g ∈ R^{∗}
[a,b]
if andonly if g^{′}f ∈ R^{∗}
[a,b]
and in this case,
∫ ∫
fg|b− bfg′dx = bf′gdx
a a a
The following example, is very significant. It exposes an unpleasant property of the
generalized Riemann integral. You can’t multiply two generalized Riemann integrable
functions together and expect to get one which is generalized Riemann integrable. Also, just
because f is generalized Riemann integrable, you cannot conclude
|f|
is. This is very
different than the case of the Riemann integrable. It is unpleasant from the point of
view of pushing symbols. The reason for this unpleasantness is that there are so
many functions which can be integrated by the generalized Riemann integral. When
you say a function is generalized Riemann integrable, you do not say much about
it.
Example 12.2.5Consider the function
{ ( )
f (x) = x2sin 1x2- if x ⁄= 0
0 if x = 0
Then f^{′}
(x)
exists for all x ∈ ℝ and equals
{ ( ) ( )
f′(x) = 2x sin 1x2- − 2x cos x12 if x ⁄= 0
0 if x = 0
Then f^{′}is generalized Riemann integrable on
[0,1]
because it is a derivative. Now let ψ
(x)
denote the sign of f
(x)
. Thus
(
{ 1 if f (x) > 0
ψ (x) ≡ ( − 1 if f (x) < 0
0 if f (x) = 0
Then ψ is a bounded function and you can argue it is Riemann integrable on