3-25 Listed Main properties of the definite integral that we will prove:
Included: Monotonicity. Linearity. Additivity. Bounded Constraint. Continuity of Integral Function for integrable functions. Continuous Functions are integrable. Fundamental Theorem (Derivative form) for Continuous Functions. FTofC (Evaluation form) for Continuous Functions. Mean Value Theorem for Integrals for Continuous Functions.

3-26. Additivity. Continuity of Integral Function for Integrable Functions. Continuous Functions are Integrable.

Excerpt from Spivak, Calculus, Ch 13
Theorem [Additivity]










Proof: [Converse proof omitted]

































Theorem: Suppose `f ` is integrable on `[a,b]` and there exist m and M so that for all `x in [a,b]` `m \le f(x) \le M`. Then `m(b-a) \le \int_a^b f \le M(b-a).

Proof: Consider any partition `P` of `[a,b]`, then `m(b-a) \le L(f,P)` and `U(f,P) \le M(b-a)`.
Now since `f` is integrable,
`m(b-a) \le L(f,P) \le lub {L(f,P)} =  \int_a^b f = glb{U(f,P)} \le U(f,P) \le M(b-a)`.  EOP

Excerpted from Spivak, Calculus, Chapter 13:

Theorem: If `f ` is integrable on `[a,b]` and  `F(x) = \int_a^x f ` for `x in [a,b] then `f` is continuous on `[a,b]`.

Proof:  Since `f` is bounded on the interval `[a,b]`, we have a number `M >0` so that for all `x in [a,b]`,  `-M \le f(x) \le M`.
Now consider `F` on the interval `[c, c+h]`, of length `h>0`.
Then  `F(c+h) = \int_a^{c+h} f ` and using additivity we have `F(c+h) - F(c) = \int_c^{c+h} f `.
By the previous result we have

Excerpted from Spivak, Calculus, Chapter 14:

Theorem: If `f ` is continuous on `[a,b]` then `f` is integrable on `[a,b]`.

Proof [From Spivak, Calculus, Ch 14]

3-29 Discussion of Improper Integral for Discontinuities on bounded intervals.
This discussion defines the integral for a function with a single point of discontinuity `c in [a,b]` by `\int_a^b f = lim_{c*->c^-}\int_a^{c*}f + lim_{c*->c^+}\int_{c*}^b f `.
Example: Find `\int_{-1}^1 1/{sqrt{|x|}}`.  Work left for reader... [done in class.]

Sets of Measure Zero introduced. [revised slightly from class]

We say a set S of real numbers has measure zero  if and only if given any positive number ε, there is a family {I_n} of closed intervals such that S is contained in the union of the I_n and the total length of the I_n is less than ε.

Countable Sets have measure zero.
Suppose `S = {a_1,a_2,a_3, ... }` then `S` has measure zero.

Proof: Given  `epsilon >0`, consider the family of intervals, `[a_1,a_1+1/3 epsilon],  [a_2,a_2+1/9epsilon], [a_3,a_3+1/27epsilon],  ... ` then the sum of these lengths is `epsilon(1/3+1/9+1/27+...) = 1/2 epsilon < epsilon.` EOP.


Continuity, integrability, and measure zero sets related.
Consider the function `f` where `f(x) = 0` if `x` is an irrational number and `f(x) =1/q` when `x = p/q` where the gcd of `p` and `q` is `1`. Then `f` is discontinuous only at the rational numbers and `\int_0^1 f` = 0. Discuss how the discontinuity values are nullified by using the fact that the rational numbers have measure zero.


The Cantor Set: has measure 0 but is uncountable.
Discussion: Consider the sequence of unions of closed intervals:
`C_0=[0,1]; C_1 = [0,1/3] cup [2/3,1]; C_2=[0,1/9] cup [2/9,1/3] cup [2/3, 7/9] cup [8/9,1];...`
The intersection of the family of these sets is called the Cantor Set. It has measure zero and is one of the "monsters" of the real numbers since it is not countably infinite!
See  Cantor set - Wikipedia, the free encyclopedia