3-21 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.
3-22.
Recall the definition of the Darboux Integral for `f` a bounded function on `[a,b]`.
Suppose `P =\{ a=x_0<x_1<...<x_{n-1}<x_n=b\}; I_k = [x_{k-1},x_k]; Delta x_k=x_k-x_{k-1}` for `k=1,2,...,n`.
`M_k = lub\{f(x): x in I_k\}; m_k= glb\{f(x): x in I_k\}`
` U(P)=\sum_{k=1}^n M_k cdot Delta x_k ;L(P)=\sum_{k=1}^n m_k cdot Delta x_k ;` and
`int_U f= glb\{U(P)\}; \int_L f= glb\{L(P)\}`.
Then `f` is (Darboux) integrable over `[a,b]` if `\int _U f =\int_L f `, which is then denoted `\int_a^b f`.
Note: We have shown that `f` is integrable if and only if for any `epsilon > 0` there is a partition `P` where `U(P)-L(P) < epsilon`.
Also If `f` is Darboux integrable then `f` is Riemann integrable. [For any partition, `L(P) \le R(P,C) \le U(P)` ]
Linearity and Monotonicity. [Proof omitted here.]
i. If `f` and `g` are integrable over `[a,b]` and `alpha in R` then `alpha f + g` is integrable over `[a,b]` and
`\int_a^b alpha f + g = alpha \int_a^b f + \int_a^b g`.
ii.a If `f(x) \ge 0 ` for all `x in [a,b]` and `f` is integrable then `\int_a^b f \ge 0`.
ii.b If `f(x) \ge g(x) ` for all `x in [a,b]` and `f` and `g` are integrable then `\int_a^b f \ge \int_a^b g`.
Additivity. Continuity of Integral Function for Integrable Functions. Continuous Functions are Integrable.
Excerpt from Spivak, Calculus, Ch 13
Theorem [Additivity]
Proof: [Converse proof omitted]
EOP
Theorem:[Boundedness] 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
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.
3-22.
Recall the definition of the Darboux Integral for `f` a bounded function on `[a,b]`.
Suppose `P =\{ a=x_0<x_1<...<x_{n-1}<x_n=b\}; I_k = [x_{k-1},x_k]; Delta x_k=x_k-x_{k-1}` for `k=1,2,...,n`.
`M_k = lub\{f(x): x in I_k\}; m_k= glb\{f(x): x in I_k\}`
` U(P)=\sum_{k=1}^n M_k cdot Delta x_k ;L(P)=\sum_{k=1}^n m_k cdot Delta x_k ;` and
`int_U f= glb\{U(P)\}; \int_L f= glb\{L(P)\}`.
Then `f` is (Darboux) integrable over `[a,b]` if `\int _U f =\int_L f `, which is then denoted `\int_a^b f`.
Note: We have shown that `f` is integrable if and only if for any `epsilon > 0` there is a partition `P` where `U(P)-L(P) < epsilon`.
Also If `f` is Darboux integrable then `f` is Riemann integrable. [For any partition, `L(P) \le R(P,C) \le U(P)` ]
Linearity and Monotonicity. [Proof omitted here.]
i. If `f` and `g` are integrable over `[a,b]` and `alpha in R` then `alpha f + g` is integrable over `[a,b]` and
`\int_a^b alpha f + g = alpha \int_a^b f + \int_a^b g`.
ii.a If `f(x) \ge 0 ` for all `x in [a,b]` and `f` is integrable then `\int_a^b f \ge 0`.
ii.b If `f(x) \ge g(x) ` for all `x in [a,b]` and `f` and `g` are integrable then `\int_a^b f \ge \int_a^b g`.
Additivity. Continuity of Integral Function for Integrable Functions. Continuous Functions are Integrable.
Excerpt from Spivak, Calculus, Ch 13
Theorem [Additivity]
Proof: [Converse proof omitted]
EOP
Theorem:[Boundedness] 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 bounded and 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
`-Mh \le \int_c^{c+h}f
\le Mh` or
(1) `-Mh \le F(c+h) - F(c) \le Mh`
Now consider `F` on the interval `[c+h, c]`, where `h`<0, and length `-h>0`.
Then using the additive property we have `F(c) = \int _a^c f = \int _a^{c+h} f + \int_{c+h}^c f` and so
`F(c+h)-F(c) = -\int_{c+h}^c f.`
Using boundedness we have then that `-M(-h) \le \int_{c+h}^c f \le M(-h)` or `M(-h) \ge -\int_{c+h}^c f \ge M(h)`.
Thus we have when `h <0` that `Mh \le F(c+h) - F(c) \le M(-h)`. In either case we see that as `h to 0` the squeeze theorem can apply , showing that `F(c+h)-F(c) to 0` or `lim_{x to c} f(x) = f(c)`, and thus `F` is continuous at `x=c` for all `c in [a,b]`. EOP
(1) `-Mh \le F(c+h) - F(c) \le Mh`
Now consider `F` on the interval `[c+h, c]`, where `h`<0, and length `-h>0`.
Then using the additive property we have `F(c) = \int _a^c f = \int _a^{c+h} f + \int_{c+h}^c f` and so
`F(c+h)-F(c) = -\int_{c+h}^c f.`
Using boundedness we have then that `-M(-h) \le \int_{c+h}^c f \le M(-h)` or `M(-h) \ge -\int_{c+h}^c f \ge M(h)`.
Thus we have when `h <0` that `Mh \le F(c+h) - F(c) \le M(-h)`. In either case we see that as `h to 0` the squeeze theorem can apply , showing that `F(c+h)-F(c) to 0` or `lim_{x to c} f(x) = f(c)`, and thus `F` is continuous at `x=c` for all `c in [a,b]`. EOP
Theorem: If `f ` is continuous on
`[a,b]` then `f` is integrable on `[a,b]`.
Proof [From Spivak,
Calculus, Ch 14]
Mean Value Theorem For Integrals:
If `f` is continuous on `[a,b]`, then there is a number `c in (a,b)` where `f(c) * (b-a) = \int_a^b f`.
Proof: [Outline]. Let `M = max\{f(x): x in [a.b]\}` and `m = min\{f(x): x in [a.b]\}` then we have `m(b-a) \le \int_a^b f \le M(b-a)`. Consider `g(x) = x* (b-a)` for `x in [m,M]`. Then `g` is a continuous function on `[m,M]` and by the IVT, there is some `L in [m,M]` where `g(L) = L(b-a) = \int_a^b f`. [ IVT]. But then by the IVT applied to `f`, there is some `c in [a,b]` where `L = f(c)`. Hence `f(c)*(b-a) = \int_a^b f`. EOP.
Note: MVT for Integrals can also be used to prove `F'(x)=f(x)` in the FTofC Theorem 1.
3-25 Discussion of Improper Integral for Discontinuities on bounded intervals.
Practivity 4* from Week 8-9:
Suppose `f: [0,4] to R` and `f(x) = 5` for `x in [0,2)` and `f(x) = 7` for `x in [2,4].
Use the definition of the definite integral [ Riemann or Darboux- your choice] to show that `\int_0^4 f = 24`.
Note: `f` is not continuous at `x=2`. This is a "step" discontinuity.
Solution: Suppose `epsilon >0`. Let `M` be a natural number so that `1/{M} lt epsilon / 2`. Use a partition `P` with `Delta x = 1/{M}` and
`x_k = k Delta x` for `k = 0, 1, 2, ...,4M`. In particular `x_{2M} = {2M}/M = 2` and so we can find `U(P)` and `L(P)` as follows:
`U(P) = [5 Delta x + .... 5 Delta x + 7Delta x] + [7 Delta x + .... +7 Delta x]= (2M-1)5 Delta x + 7 Delta x + ( 2M )* 7 Delta x`
` = 2M* 5 Delta x + 2 Delta x + 14=10 + 2 Delta x + 14 = 24 + 2Delta x.`
`L(P) = [5 Delta x + .... 5 Delta x + 5Delta x] + [7 Delta x + .... +7 Delta x] = 2M* 5Delta x + 2M* 7 Delta x = 10 + 14 = 24.`
Therefore with this partition, `U(P) - L(P) = 2 Delta x < 2 epsilon / 2 = epsilon`. So by the definition of the Darboux integal, `f` is Darboux integrable, and since `L(P) = 24` for all `P` where `Delta x = 1/n` for any natural number `n >0` , the lower integral, ` L\int_0^2 f = 24`, so `\int_0^4 f = 24`.
As in beginning calculus, we can define the integral for a bounded 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 `. This is actually unnecessary as we can show that in this circumstance, `f` is Darboux integrable, and that the integral is the same as the number found by the limit definition.
Example: Find `\int_{-1}^1 1/{sqrt{|x|}}`. This function is unbounded, but by letting `f(x)= 1/{sqrt{|x|}}` when `x \ne 0` and `f(x) = 0 ` it can be shown that this function is Darboux integrable, though to find the integral the limit process is most convenient. Work left for reader...
Insight: The key to having these integrals exist for functions with discontinuities is the ability to isolate the discontinuity and bound the effect of the discontinuity on the upper and lower sums. One way to see this for a given `epsilon > 0` have a partition where each point of an isolated step or removable discontinuity is included in a partition with an interval of size `Delta x_k` where `(U_k - L_k)* Delta x_k` is sufficiently small so that `U(P) - L(P) < epsilon`. [Check how this strategy was applied in the practivity.]
Sets of Measure Zero :
Definition: The length of a closed and bounded interval `[a,b]` is denoted `l([a,b])` and is equal to `b-a`.
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 = [a_n,b_n]: n = 0,1,2, ... \}`
of closed (and bounded) intervals such that S
is contained in the union of the In
and the total length of the In `= \sum_0^{oo) l(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.
Application: The set of rational numbers between `0` and `1` has measure 0.
Proof: The rational numbers are countably infinite. Use the sequence `0,1,1/2,1/3,2/3,1/4,3/4,1/5,2/5,3/5,4/5, 1/6,5/6,1/7, ...`
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.
Notice how the strategy for dealing with discontinuities can be implemented here so the non-zero values are nullified by using the fact that the rational numbers have measure zero.
Theorem: If `f : [a,b] to R` is a bounded function that is continuous except on a set `S` of measure zero, then `f` is Darboux integrable.`
Proof: Omitted.
The Cantor Set: This is a set that 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!
Let `C =` the cantor set `= cap \{C_j: j∈N\}`.
If `f(x)=1` for `x in C` and `f(x)=0` for `x in [0,1]-C`, then `f ` is Darboux integrable and `\int_0^1 f = 0` .
Proof: If `P_n=\{0,1/{3^n},2/{3^n},...,1\}` then `U(P_n)=(2/3)^n` and `L(P_n)=0`.
See Cantor set - Wikipedia, the free encyclopedia