7.1. Riemann Integral
Corollary 7.1.17: Riemann Integral of almost Continuous Function
Proof:
We can prove this easily by applying Lebesgue's Theorem, noting that any set with at most countably many points has measure zero.
To prove this is directly, however, is not easy; we will start with a simpler version of this theorem: if f is continuous and bounded over the interval [a, b] except at one point xk, then f is Riemann integrable over [a, b].
We know that f is bounded by some number M
over the interval [a, b].
Take any
> 0 and choose a
partition P that includes the point xk
such that
| P | </ 12M
Then in particular
|xk+1 - xk-1| </ 6M
We also know that f is
uniformly continuous over [a, xk-1] as well as
uniformly continuous over [xk+1, b]. Therefore, for
our chosen there exists
- a
' such that |f(x) - f(y)| < 1/3
/ (b - a) for all x, y inside [a, xk-1] with |x - y| <
'
- a
'' such that |f(x) - f(y)| < 1/3
/ (b - a) for all x, y inside [xk+1, b] with |x - y| <
''
Now refine the partition P by adding points on the left side of
xk-1 so that the mesh on that side is less than
', and by adding points on the
right side of xk+1 so that the mesh there is
less than
''. For simplicity,
call that new partition again P. Then we have:
| U(f,P) - L(f,P) |![]()
|cj - dj| (xj - xj-1) =
![]()
For the first term we have:
|c1 - d1| (x1 - x0) + ... + |ck-1 - dk-1| (xk-1 - xk-2)
< 1/3/(b-a) (xk-1 - x0) < 1/3
/(b-a) (b - a) = 1/3
![]()
because of uniform continuity to the left of xk and our choice of the partition. The third term can be estimated similarly:
|ck+2 - dk+2| (xk+2 - xk+1) + ... + |cn - dn| (xn - xn-1)
< 1/3/(b-a) (xn - xk+1) < 1/3
/(b-a) (b - a) = 1/3
![]()
Since f is bounded by M we know that |cj - dj| < 2M for all j so that the middle term can be estimated by:
|ck - dk| (xk - xk-1) + |ck+1 - dk+1| (xk+1 - xk)
< 2M (xk+1 - xk-1) < 2M/ 6M = 1/3
![]()
Taking everything together we have:
|U(f,P) - L(f,P)| < 1/3+ 1/3
+ 1/3
=
![]()
Therefore, by Riemann's Lemma, the function f is Riemann integrable.