Next: C. Prerequisites
Up: Math 111 Calculus I
Previous: A. Hints and Answers
  Index
B. Proofs of Some Area Theorems
B.1
Theorem (Addition Theorem.) For any bounded sets
and
in
 |
(B.2) |
and consequently
Proof: We have
and
Hence by the additivity of area
 |
(B.3) |
and
 |
(B.4) |
If we solve equation (B.4) for
and use this result
in
equation (B.3) we get the desired result.
Proof: The proof is by induction. If
, then (B.6) says
, which is true. Suppose now that
is a generic element of
, and that (B.6)
is true when
.
Let
be bounded sets in
. Then
Hence (B.6) is true when
, and by induction
the formula holds for all
B.7
Theorem (Monotonicity of Area.)
Let
be bounded sets such that
. Then
.
Proof: If
then
, and in this case
equation (B.4)
becomes
Since
, it follows that
.
Proof: The proof is by induction on
. For
, equation (B.9)
says
that
, and this is true. Now suppose
is a family of mutually almost-disjoint sets. Then
and this is a finite union of zero-area sets, and hence is a zero-area set.
Hence,
by the addition theorem,
i.e.,
The theorem now follows from the induction principle.
Next: Prerequisites
Up: Math 111 Calculus I
Previous: A. Hints and Answers
  Index
Ray Mayer
2007-09-07