10. Integral

Partitions

A partition $P$ of $[a, b]$ is a finite set of points from $[a, b]$ that includes both $a$ and $b$ . The notational convention is to always list the point of a partition $P = {x_{0}, x_{2}, \dots, x_{n}}$

a = x_{0} < x_{1} < x_{2} < \dots < x_{n} = b

for each $[x_{k - 1}, x_{k}]$ of $P$ , let

m_{k} = inf {f (x) : x \in [x_{k - 1}, x_{k}]} and M_{k} = sup {f (x) : x \in [x_{k - 1}, x_{k}]}

Lower Sum

L (f, P) = \sum_{k = 1}^{n} m_{k} (x_{k} - x_{k - 1})

Upper Sum

U (f, P) = \sum_{k = 1}^{n} M_{k} (x_{k} - x_{k - 1})

Clearly $L (f, P) \leq U (f, P)$

Further Concepts

Δ x_{i} = x_{i} - x_{i - 1}

$f : [a, b] \to R$ is bounded, let $m, M \in R$ so that for all $x \in [a, b]$ have $m \leq f (x) \leq M$ . Then for every partition $P$ of $[a, b]$

m (b - a) \leq L (P, f) \leq U (P, f) \leq M (b - a)

Lower Darboux Integral := $\underset{―}{\int_{a}^{b}} f = inf {U (P, f) : P is a partition of [a, b]}$
Upper Darboux Integral := $\overset{―}{\int_{a}^{b}} = sup {L (P, f) : P is a partition of [a, b]}$

A partition $Q$ is a refinement of a partition $P$ if $P \subseteq Q$
$Q = \tilde{P}$

If $P \subseteq \tilde{P}$ then $L (f, P) \leq L (f, \tilde{P})$ and $U (f, P) \geq U (f, \tilde{P})$

Proof

Adding a singe point $z$ to some subinterval $[x_{k - 1}, x_{k}]$ of $P$

\begin{aligned} m_{k} (x_{k} - x_{k - 1}) & = m_{k} (x_{k} - z) + m_{k} (z - x_{k - 1}) \\ \leq m_{k}^{'} (x_{k} - z) + m_{k}^{″} (z - x_{k - 1}) \end{aligned}

Use induction

If $P_{1}$ and $P_{2}$ are any two partitions of $[a, b]$ , then

L (f, P_{1}) \leq U (f, P_{2})

Proof

Let $Q = P_{1} \cup P_{2}$

L (f, P_{1}) \leq L (f, Q) \leq U (f, Q) < U (f, P_{2})

Upper and Lower Integrals

Let $P$ be the collection of all possible partitions of the interval $[a, b]$

Upper Integral

U (f) = inf {U (f, P) : P \in P}

Lower Integral

L (f) = sup {L (f, P) : P \in P}

Thm Bounded Functions and Upper and Lower Integrals

For any bounded function $f$ on $[a, b]$ it is always the case that

U (f) \geq L (f)

Proof

If $U (f) < L (f, P)$ then since $U (f) = inf {U (f, P) : P \in P}$ there exists a $P_{1}$ such that $U (f, P_{1}) < L (f, P)$ which is not possible by our common refinement theorem. Same if $U (f) < L (f)$ .

Riemann Integrability

A bounded function $f$ on the interval $[a, b]$ is Riemann-integrable if $U (f) = L (f)$

Alt Definition

$f$ is Riemann integrable iff for all $ε > 0$ there exists $P_{ε}$ of $[a, b]$ st

U (P_{ε}, f) - L (P_{ε}, f) < ε

Thm

$f : [a, b] \to R$ is bounded let $m, M \in R$ s.t for all $x \in [a, b]$ $m \leq f \leq M$ then

m (b - a) \leq \underset{―}{\int_{a}^{b}} f \leq \overset{―}{\int_{a}^{b}} f \leq M (b - a)

Proof

we know $m (b - a) \leq L (P, f) \leq U (P, f) \leq M (b - a)$

Have:

{\begin{cases} m (b - a) \leq L (P, f) ⟹ m (b - a) \leq {\int_{―}}_{a}^{b} f \\ M (b - a) \geq U (P, f) ⟹ M (b - a) \geq {\int^{¯}}_{a}^{b} f \end{cases}

Pasted image 20250320100218.webp

Thm if integrable then bounded

If $f : [a, b] \to R$ is Riemann integrable and there exists $m, M$ with $m \leq f \leq m$ then:

m (b - a) \leq \int_{a}^{b} f \leq M (b - a)

Example

$f : [0, 2] \to R$

f (x) = {\begin{cases} 1 & x < 1 \\ \frac{1}{2} & x = 1 \\ 0 & x > 1 \end{cases}

Claim: $\int_{0}^{2} f = 1$

Let $1 > ε > 0$ , let $P = {0, 1 - ε, 1 + ε, 2}$

Then,

\begin{aligned} m_{1} & = inf {f (x) : x \in [0, 1 - ε]} \\ m_{2} & = inf {f (x) : x \in [1 - ε, 1 + ε]} \\ m_{3} & = inf {f (x) : x \in [1 + ε, 2]} \end{aligned}

\begin{aligned} M_{1} & = sup {f (x) : x \in [0, 1 - ε]} \\ M_{2} & = sup {f (x) : x \in [1 - ε, 1 + ε]} \\ M_{3} & = sup {f (x) : x \in [1 + ε, 2]} \end{aligned}

Δ x_{1} = 1 - ε, Δ x_{2} = 2 ε, Δ x_{3} = 1 - ε

\begin{array}{r} L (P, f) = \sum_{}^{} m_{i} Δ x_{i} = (1 - ε) \\ U (P, f) = \sum_{}^{} M_{i} Δ x_{i} = 1 - ε \end{array}

\overset{―}{\int_{0}^{2}} f - \underset{―}{\int_{0}^{2}} \leq U (P, f) - L (P, f) = 1 - ε - (1 - ε) = 2 ε

1 - ε = L (P, f) \leq \int_{0}^{2} f \leq U (P, f) = 1 + ε ⟹ \int_{0}^{2} f = 1

Example

Not integrable

Drichlet $f : [0, 1] \to R$

f (x) = {\begin{cases} 1 & x \in Q \\ 0 & x \in R ∖ Q \end{cases}

Fix a partition:

\begin{aligned} m_{i} & = inf {f} = 0 \\ M_{i} & = sup {f} = 1 \end{aligned}

Thus $L (P, f) = \sum_{}^{} m_{i} Δ x_{i} = 0$ and $U (P, f) = \sum_{}^{} M_{i} Δ x_{i} = \sum_{}^{} Δ x_{i} = 1$

\underset{―}{\int_{0}^{1}} f = 0 but \overset{―}{\int_{0}^{1} f} = 1

Thm Summing Integrals

Let $a < c < b$ then $f : [a, b] \to R$ is R. integrable iff $f$ is integrable on $[a, c]$ and $[c, b]$ then $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$

Proof

for all $ε > 0$ there exists a $P$ so that $U (P, f) - L (P, f) < ε$
$\tilde{P}$ be a refinement.
Let:

\begin{array}{r} P_{1} = \tilde{P} \cap [a, c] \\ P_{2} = \tilde{P} \cap [c, b] \end{array}

U (P_{1}, f) - L (P_{1}, f) < ε and L (P_{2}, f) - L (P_{2}, f) < ε

$⟸$ if $f \in R [a, c]$ and $f \in R [c, b] \exists P_{1} and P_{2}$ of $[a, c]$ and $[c, b]$

with

\begin{aligned} U (P_{1}, f) - L (P_{1}, f) & < ε / 2 \\ U (P_{2}, f) - L (P_{2}, f) & < \frac{ε}{2} \end{aligned}

Take $P = P_{1} \cup P_{2} ⟹ U (P, f) - L (P, f) < ε$

\begin{aligned} \int_{a}^{b} f & \leq U (P, f) < ε + L (P, f) \\ = ε + L (P_{1}, f) + L (P_{2}, f) \\ \leq \int_{a}^{c} f + \int_{c}^{b} f \end{aligned}

\begin{array}{r} \int_{a}^{c} f + \int_{c}^{b} f \leq U (P_{1}, f) + U (P_{2}, f) \\ < ε + L (P_{1}, f) + L (P_{2}, f) \\ = ε + L (P, f) \leq ε + \int_{a}^{b} f \end{array}

Other Properties

$f, g : [a, b] \to R$ is bounded

Then

\overset{―}{\int_{a}^{b}} (f + g) \leq \overset{―}{\int_{a}^{b}} f + \overset{―}{\int_{a}^{b}} g

\underset{―}{\int_{a}^{b}} f \geq \underset{―}{\int_{a}^{b}} f + \underset{―}{\int_{a}^{b}} g

If $f \leq g$
$⟹ \overset{―}{\int_{a}^{b}} f \leq \overset{―}{\int_{a}^{b}} g$
$⟹ \underset{―}{\int_{a}^{b}} f \leq \underset{―}{\int_{a}^{b}} g$

If $f, g$ are integrable on $[a, b]$ then:

i) $f + g$ is integrable with $\int_{a}^{b} (f + g) = \int_{a}^{b} f + \int_{a}^{b} g$
ii) For $k \in R$ the function $k \cdot f$ is integrable with $\int_{a}^{b} k \cdot f = k \int_{a}^{b} f$
iii) If $m \leq f (x) < M$ on $[a, b]$ then $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$
iv) If $f (x) < g (x)$ on $[a, b]$ then $\int_{a}^{b} f \leq \int_{a}^{b} g$
v) The function $| f |$ is integrable and $| \int_{a}^{b} f | \leq \int_{a}^{b} | f |$

Thm Continuity and Integral

If $f : [a, b] \to R$ is continuous then $f \in R [a, b]$

Proof

If $f : [a, b] \to R$ is continuous then $f \in R [a, b]$

Since it in continuous on a compact set then $f$ is uniformly continuous so let $ε > 0$ there exists a $δ > 0$ so that $| x - y | < δ ⟹ | f (x) - f (y) | < \frac{ε}{b - a}$

Consider $P = {x_{0}, x_{1}, \dots, x_{n}}$ with $Δ x_{i} < δ \forall i$
If we partitioned uniformly , Take $n st \frac{b - a}{n} < δ$

x_{i} = \frac{i}{n} (b - a) + a

For all $x, y \in [x_{i}, x_{i + 1}]$ $| x - y | \leq Δ x_{i} < δ$ so $| f (x) - f (y) | < \frac{ε}{b - a}$
In general since $f$ is continuous on $[x_{i - 1}, x_{i}]$ so $f$ attains a max and min on $[x_{i - 1}, x_{i}]$
Let $x :$ max and $y :$ min, $f (x) = M_{i}, f (y) = m_{i} ⟹ M_{i} - m_{i} \leq f (x) - f (y) \leq \frac{ε}{b - a}$

\begin{aligned} \overset{―}{\int_{a}^{b}} f - \underset{―}{\int_{a}^{b}} f & \leq U (P, f) - L (P, f) \\ = \sum_{}^{} M_{i} Δ x_{i} - \sum_{}^{} m_{i} Δ x_{i} \\ = \sum_{}^{} (M_{i} - m_{i}) Δ x_{i} \\ < \frac{ε}{b - a} \sum_{}^{} Δ x_{i} = ε \end{aligned}

Since $ε$ is arbitrary $\overset{―}{\int_{a}^{b}} f = \underset{―}{\int_{a}^{b}} f ⟹ f \in R [a, b]$

Thm Integrand Sequences

$f : [a, b] \to R$ is bounded, ${a_{n}}, {b_{n}}$ with $a < a_{n} < b_{n} < b$ and $a_{n} \to a$ and $b_{n} \to b$
If $f$ is Riemann integrable on $[a_{n}, b_{n}]$ then $f$ is Riemann integrable on $[a, b]$ and

\int_{a}^{b} f = lim \int_{a_{n}}^{b_{n}} f

Thm Discontinuity and Integration

Let $f : [a, b] \to R$ is bounded with finitely many discontinuities then $f$ is Riemann integrable

Thm Integrability and Interior Points

If $f : [a, b] \to R$ is bounded and $f$ is integrable on $[c, b]$ for all $c \in (a, b)$ then $f$ is integrable on $[a, b] .$
useless

Integrability of Thomae

f (x) = {\begin{cases} \frac{1}{q} & if x = \frac{p}{q} \\ 1 & if x = 0 \\ 0 & otherwise \end{cases}

Thomae is continuous on $Q$

Choose $N \in N$ s.t $\frac{1}{n + 1} < ε / 2$ let

B_{N} = {1, \frac{1}{2}, \frac{1}{3}, \frac{2}{3}, \dots, \frac{1}{N}, \dots, \frac{N - 1}{N}}

If $x \notin B_{N}$ then $f (x) < \frac{1}{N + 1} < \frac{ε}{2}$
let $m = | B_{N} |$ and $P = {x_{0}, \dots, x_{n}}$ where $n > m$ and
$Δ x_{i} < \frac{ε}{4 m}$

Then there are at most $2 m$ sub-intervals of $[0, 1]$ s.t $[x_{i - 1}, x_{i}] \cap B_{N} \neq 0$

Take $M_{i} = sup_{x \in [x_{i - 1}, x_{i}]} f (x)$
Then

\begin{aligned} U (P, f) & = \sum_{}^{} M_{i} Δ x_{i} \\ = \underset{[x_{i - 1}, x_{i}] \cap B_{N} \neq \emptyset}{\sum_{}^{}} M_{i} Δ x_{i} + \sum_{[x_{i - 1}, x_{i} \] \cap B_{N} = \emptyset}^{} M_{i} Δ x_{i} \\ \leq \frac{1}{N + 1} (1 - 0) + 1 \cdot 2 \cdot (\frac{ε}{4 m}) \\ < \frac{ε}{2} + \frac{ε}{2} = ε \end{aligned}

(U - L) < ε

integrable

Integrable Limit Theorem

Assume that $f_{n} ⇉ f$ on $[a, b]$ and that each $f_{n}$ is integrable. Then $f$ is integrable and

lim_{n \to \infty} \int_{a}^{b} f_{n} = \int_{a}^{b} f

Proof

| \int_{a}^{b} f_{n} - \int_{a}^{b} f | = | \int_{a}^{b} (f_{n} - f) | \leq \int_{a}^{b} | f_{n} - f |

Take $ε > 0$ , since $f_{n} ⇉ f$ $\exists N$ such that

| f_{n} - f (x) | < ε / (b - a) for all n \geq N and x \in [a, b]

and for $n \geq N$

\begin{array}{r} | \int_{a}^{b} f_{n} - \int_{a}^{b} f | \leq \int_{a}^{b} \frac{ε}{b - a} = ε \end{array}

Fundamental Theorem of Calculus

1)

If $f : [a, b] \to R$ is integrable and $F : [a, b] \to R$ satisfies $F^{'} (x) = f (x)$ for all $x \in [a, b]$ then

\int_{a}^{b} f = F (b) - F (a)

2)

Let $g : [a, b] \to R$ be integrable and for $x \in [a, b]$ define

G (x) = \int_{a}^{x} g

Then $G$ is continuous on $[a, b]$ . If $g$ is continuous at some point $c \in [a, b]$ then $G$ is differentiable at $c$ and $G^{'} (c) = g (c)$

Measure Zero

$\forall ε > 0 \exists$ a countable collection of ${(a_{k}, b_{k})}$ open intervals such that:

A \subseteq ⋃_{k = 1}^{\infty} (a_{k}, b_{k}) and \sum_{}^{} (b_{k} - a_{k}) < ε

So "Almost everywhere" (except for a measure zero set)

Lebesgue’s Theorem

Let f be a bounded function deﬁnedon the interval [a, b].Then,f is Riemann-integrable if and only if the set ofpoints where f is not continuous has measure zero.

Other Integrals

Reimann Stieltjes

f : [a, b] \to R, g : R \to R

\int_{a}^{b} f (x) d g (x)

\sum_{}^{} f (c_{i}) | g (x_{i}) - g (x_{i - 1}) |

Henstock K Integral

Choose $δ : [a, b] \to R$ so that $\forall i$ , $[x_{i - 1}, x_{i}] \subset [t_{i} - δ_{i}, t_{i} + δ_{i}]$ where $t_{i} \in Δ x_{i}$ , want:

| \int f - \sum_{P}^{} f | < ε

Now more general so that Drichlet can be integrated

10. Integral

Partitions

Refinement

Thm Refinement Inequalities

Proof

Thm Common Refinement and Upper and Lower Sums

Proof

Upper and Lower Integrals

Thm Bounded Functions and Upper and Lower Integrals

Proof

Riemann Integrability

Alt Definition

Thm

Proof

Thm if integrable then bounded

Example

Example

Thm Summing Integrals

Proof

Other Properties

Thm Continuity and Integral

Proof

Thm Integrand Sequences

Thm Discontinuity and Integration

Thm Integrability and Interior Points

Integrability of Thomae

Integrable Limit Theorem

Proof

Fundamental Theorem of Calculus

1)

2)

Measure Zero

Lebesgue’s Theorem

Other Integrals

Reimann Stieltjes

Henstock K Integral