THE ADDITIVITY OF THE INTEGRAL
FOKKO VAN DE BULT
In this note we prove the following Theorem.
Theorem 1.
Let
f
and
g
be integrable functions on the interval
[
a,b
]
. Then
f
+
g
is also integrable on
[
a,b
]
and we have
Z
b
a
(
f
+
g
)(
x
)
dx
=
Z
b
a
f
(
x
)
dx
+
Z
b
a
g
(
x
)
dx.
Proof.
The proof given here is almost the same as the one described in the book
on pages 85 and 86, and probably written down better there.
Suppose
f
and
g
are integrable on [
a,b
]. Choose an arbitrary
n
∈
N
. Then
we see that there exists a step function
s
n
≤
f
such that
R
b
a
s
n
(
x
)
dx
≥
I
(
f
)

1
4
n
=
R
b
a
f
(
x
)
dx

1
4
n
, as otherwise I
(
f
)

1
4
n
would be an upperbound to
S
=
{
R
b
a
s
(
x
)
dx

s
≤
f,s
a step function
}
less than I
(
f
), in violation of the deﬁnition
I
(
f
) = sup(
S
). Likewise we ﬁnd
t
n
, ˜
s
n
and
˜
t
n
with
R
b
a
t
n
(
x
)
dx
≤
R
b
a
f
(
x
)
dx
+
1
4
n
,
R
b
a
˜
s
n
(
x
)
dx
≥
R
b
a
g
(
x
)
dx

1
4
n
and
R
b
a
˜
t
n
(
x
)
dx
≤
R
b
a
g
(
x
)
dx
+
1
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 Spring '10
 323
 Logic, Binary relation, dx, Step function

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