Assume
is integrable. We want to show that if
satisfies
for all but a finite number of points in
, then
is integrable as well and
.
Proof:
Assume there exists a point
in
where
and
differ.
Since
is integrable, by definition, for any given
, there exists a partition
of
such that the upper sum
.
Let's consider the partition
which includes the point
. Without loss of generality, assume
is the right endpoint of a subinterval
in
.
Since
for all but a finite number of points, we can select a subinterval
such that
for all
in
except
.
Now, we have:
![\sf\:\sup_{x \in [x_i, x_(i+1)]} g(x) - \inf_{x \in [x_i, x_(i+1)]} g(x) = \sup_{x \in [x_i, x_(i+1)]} f(x) - \inf_{x \in [x_i, x_(i+1)]} f(x) \\](https://img.qammunity.org/2024/formulas/mathematics/high-school/k6gcw8p6lxhxnosfah6o7z19u4ryike26s.png)
Using this, we can rewrite the upper sum and lower sum as:
![\sf\:U(g, P) = U(f, P) + \sup_{x \in [x_i, x_(i+1)]} g(x) - \inf_{x \in [x_i, x_(i+1)]} g(x) \\](https://img.qammunity.org/2024/formulas/mathematics/high-school/ek6qopfvtgrjwkbu6oxaxi4tl2gjj2cqp2.png)
![\sf\:L(g, P) = L(f, P) + \sup_{x \in [x_i, x_(i+1)]} g(x) - \inf_{x \in [x_i, x_(i+1)]} g(x) \\](https://img.qammunity.org/2024/formulas/mathematics/high-school/trijvxmy5gheza3r5ylgw760po10si17qw.png)
Since
is integrable, we have
. Therefore,
as well.
This holds for any partition
of
. Hence,
is integrable and
.