riemann integration

Partition

A partition of

[a, b]

is a finite tuple sorted in ascending order

P = (x_{1}, x_{2}, \dots x_{n + 1})

such that

a = x_{1}

and

b = x_{n + 1}

i-th Section of a Partition

Suppose that

P = (x_{1}, \dots x_{n + 1})

is a partition, then for each

i \in [1, \dots, n]

we define

s_{i} : = [x_{i}, x_{i + 1}]

Supremum of a Bounded Function on a Section

Suppose that

f : [a, b] \to R

and that

P = (x_{1}, \dots, x_{n + 1})

is a partition of

[a, b]

then for each

i \in [1, \dots, n]

we have

M_{i} (f, P) = sup ({f (x) : x \in [x_{i}, x_{i + 1}]})

i-th Delta of a Partition

Suppose that

P = (x_{1}, \dots, x_{n + 1})

is a partition, then for each

i \in [1, \dots n]

we define:

Δ_{i} = x_{i + 1} - x_{i}

Note that if a partition has $k$ elements then there will be $k - 1$ deltas.

A Delta Sum Telescopes

Suppose that

P = (x_{1}, \dots, x_{n + 1})

is a partition of

[a, b]

then

\sum_{i = 1}^{n} Δ_{i} = b - a

Recall that

\sum_{i = 1}^{n} Δ_{i} = \sum_{i = 1}^{n} (x_{i + 1} - x_{i}) \stackrel 1 = x_{n + 1} - x_{1} = b - a

where we've used fact 1.

Mesh of a Partition

Suppose that

P

is a partition of

[a, b]

such that

| P | = n + 1

for some

n \in N_{0}

then

mesh (P) = max ({Δ_{i} : i \in [1, \dots, n]})

Upper Sum of a Bounded Function over a Partition

Suppose that

f : [a, b] \to R

is bounded, and that

P

is a partition of

[a, b]

then we define

U (f, P) : = \sum_{j = 1}^{n} M_{j} (f, P) Δ_{j}

Lower Sum of a Bounded Function over a Partition

Suppose that

f : [a, b] \to R

is bounded, and that

P

is a partition of

[a, b]

then we define

L (f, P) : = \sum_{j = 1}^{n} m_{j} (f, P) Δ_{j}

Riemann Integrable

Suppose that

f : [a, b] \to R

is bounded, then we say that it is Riemann integrable if

{sup}_{P} (L (f, P)) = {inf}_{Q} (U (f, Q))

In that case we write

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

as their common value

Riemann Integrable Characterizations

Let

f (x)

be bounded on

[a, b]

then the following are equivalent

$f$ is Riemann integrable
For each $ϵ \in R^{+}$ there is a partition $P$ such that

U (f, P) - L (f, P) < ϵ

For every $ϵ \in R^{+}$ there is some $δ \in R^{+}$ such that for every partition $Q$ such that $mesh (Q) < δ$ wherein

U (f, Q) - L (f, Q) < ϵ

There exists an $I \in R$ such that for every $ϵ \in R^{+}$ there is a $δ \in R^{+}$ such that for every partition $Q$ with $mesh (Q) < δ$ and every evaluation sequence $X$ of $Q$ we have

| I (f, Q, X) - I | < ϵ

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

Note that bullet point 2 is a the most tractable for explicit functions, as one can construct a partition that works for a given function.

Jump Discontinuity

Suppose

f : [a, b] \to R

, and

c \in (a, b)

then if

{lim}_{x \to c^{+}} f (x) \neq {lim}_{x \to c^{-}} f (x)

then we say that

f

has a jump discontinuity at

c

Function Difference Set is the Same as its Negative

Suppose that

f : A \to R

and define

F = {f (x) - f (y) : x, y \in A}

, then

F = - F

Note that

F = {f (x) - f (y) : x, y \in A} = {- (f (y) - f (x)) : y, x \in A} = - F

as needed.

Supremum of the Function Difference Set is the Same with Absolute Values

Let

f : A \to R

and define

F = {f (x) - f (y) : x, y \in A}

then

sup (F) = sup (| F |)

Recall that if

F = - F

then we know

sup (F) = sup (| F |)

Supremum of the Sum of Two Functions is Less than or Equal to the Sum of the Supremums

Suppose that

f, g : A \to R

then

sup ({f (x) + g (x) : x \in A}) \leq sup ({f (x) : x \in A}) + sup ({g (x) : x \in A})

Let $α = sup ({f (x) : x \in A}), β = sup ({g (x) : x \in A})$ and $γ = sup ({f (x) + g (x) : x \in A})$ .

We know that for any $x \in A$ we have that $f (x) \leq α$ and $g (x) \leq β$ therefore we have that $f (x) + g (x) \leq α + β$ , this proves that $α + β$ is an upper bound to the set ${f (x) + g (x) : x \in A}$ therefore we have $sup ({f (x) + g (x) : x \in A}) \leq sup ({f (x) : x \in A}) + sup ({f (x) : x \in A})$ As $γ$ is the least upper bound.

Supremum of the Sum of two Functions is the Same as the Sum of the Supremums if the Variables are Independent

Suppose that

f, g : A \to R

then

sup ({f (x) + g (y) : x, y \in A}) = sup ({f (x) : x \in A}) + sup ({g (x) : x \in A})

Let $α = sup ({f (x) : x \in A}), β = sup ({g (x) : x \in A})$ and $γ = sup ({f (x) + g (y) : x, y \in A})$ .

For any $x \in A$ we have that $f (x) \leq α$ and for any $y \in A$ we have $g (x) \leq β$ thus we have $f (x) + g (y) \leq α + β$ so that $α + β$ is an upper bound of the set $γ$ is supping over there fore we have that $γ \leq α + β$

... [TODO] not sure how to prove the other direction ...

Bound on the Upper minus Lower Sum for The Same Partition

Suppose that

f : [a, b] \to R

is bounded, and let

P = {x_{1}, \dots, x_{n + 1}}

be a partition of

[a, b]

then we have that

U (f, P) - L (f, P) \leq 2 ‖ f ‖_{\infty} (b - a)

U (f, P) - L (f, P) & = \sum_{i = 1}^{n} ({sup}_{x \in [x_{i}, x_{i + 1}]} f (x) - inf (x \in [x_{i}, x_{i + 1}] f (x))) Δ_{i} & \stackrel 1 = \sum_{i = 1}^{n} ({sup}_{x \in [x_{i}, x_{i + 1}]} f (x) + sup (x \in [x_{i}, x_{i + 1}] - f (x))) Δ_{i} & \stackrel 2 = \sum_{i = 1}^{n} ({sup}_{x, y \in [x_{i}, x_{i + 1}]} (f (x) - f (y))) Δ_{i} & \stackrel 3 = \sum_{i = 1}^{n} ({sup}_{x, y \in [x_{i}, x_{i + 1}]} (| f (x) - f (y) |)) Δ_{i} & \stackrel 4 \leq \sum_{i = 1}^{n} (2 {sup}_{x, y \in [x_{i}, x_{i + 1}]} (| f (x) |)) Δ_{i} & \stackrel 5 \leq \sum_{i = 1}^{n} (2 ‖ f ‖_{\infty}) Δ_{i} & = 2 ‖ f ‖_{\infty} \sum_{i = 1}^{n} Δ_{i} & \stackrel 6 = 2 ‖ f ‖_{\infty} (b - a)

Where we've used

fact 1
fact 2
fact 3
fact 4
fact 5: The supremum gets bigger on a superset
fact 6

Piecewise Continuous

A function

f : [a, b] \to R

is called piecewise continuous if for every

[c, d] \subseteq [a, b]

if it only has a finite number of discontinuities all of which are jump discontinuities

Riemann Integrable on Two Parts of the Interval Means Riemann Integrable on the Whole Interval

Suppose that

f : [a, b] \to R

and let

x \in [a, b]

then if

f

restricted to

[a, x]

and

f

restricted to

[x, b]

(which we denote as

g_{1}, g_{2}

) are both riemann integrable then so is

f

[a, b]

and then

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

Let

ϵ \in R^{+}

using bullet point 2 from the RMI characterization we know that there are partitions

P_{1}, P_{2}

[a, x], [x, b]

such that

(U (g_{1}, P_{1}) - L (g_{1}, P_{1})) + (U (g_{2}, P_{2}) - L (g_{2}, P_{2})) < ϵ

Then note that

U (g_{1}, P_{1}) + U (g_{2}, P_{2}) = U (f, P_{1} \cup P_{2})

which also holds for the lower sums, and therefore from the previous inequality we have

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

therefore by (2) of the RMI characterization

f

is Riemann integrable.

Every Piecewise Continuous Function is Riemann Integrable

Every piecewise continuous function is Riemann integrable

Since $f$ is piecewise continuous, then suppose its jump discontinuity points are given by $D = {d_{1}, d_{2}, \dots d_{n}}$ . We will show this by using strong induction on the number of discontinuities within $[a, b]$ .

Note that one can manage to prove that if $f$ is RMI over $[a, b]$ with $1 \leq i \leq k$ jump discontinuities, then it is RMI over $[a, b]$ if it has $k + 1$ discontinuities. This can be done because we can break $[a, b]$ into two intervals $I_{1}, I_{2}$ such that $I_{1} ⊔ I_{2} = [a, b]$ , where $I_{1}, I_{2}$ each have a number of discontinuities within the range $[1, \dots, k]$ so that $f ↾ I_{1}$ and $f ↾ I_{2}$ are both integrable on their respective intervals, so that $f$ is RMI on $[a, b]$ .

Therefore all that remains is the base case, which is that $f$ is RMI if there is exactly $1$ jump discontinuity of $f$ on the set $[a, b]$ , let $d$ be this discontinuity, since $f$ is continuous on $L : = [a, d - δ]$ and $R : = [d + δ, b]$ for any small $δ$ , then we know that $f$ is RMI when restricted to these intervals.

Let $ϵ \in R^{+}$ by (2) of the RMI characterization we obtain $P_{L}, P_{R}$ which are partitions of $L, R$ respectively (note that $P_{L} \cap P_{R} = \emptyset$ such that $U (f ↾ L, P_{L}) - L (f ↾ L, P_{L}) < \frac{ϵ}{3}$ and $U (f ↾ R, P_{R}) - L (f ↾ R, P_{R}) < \frac{ϵ}{3}$

Finally Consider $P_{L} \cup P_{R}$ which is a partition of $[a, b]$ which contains the section $[d - δ, d + δ]$ along with the left and right partitions. Focusing on this section we can see it will contribute $({sup}_{x \in [d - δ, d + δ]} f (x) - {sup}_{y \in [d - δ, d + δ]} f (y)) 2 δ$ to $U (f, P_{L} \cup P_{R}) - L (f, P_{L} \cup P_{R})$ , now:

U (f, P_{L} \cup P_{R}) - L (f, P_{L} \cup P_{R}) & = (U (f ↾ L, P_{L}) + ({sup}_{x \in [d - δ, d + δ]} f (x)) 2 δ + U (f ↾ R, P_{R})) - (L (f ↾ L, P_{L}) + ({inf}_{x \in [d - δ, d + δ]} f (x)) 2 δ + L (f ↾ R, P_{R})) & = (U (f ↾ L, P_{L}) - L (f ↾ L, P_{L})) + ({sup}_{x \in [d - δ, d + δ]} f (x) - {inf}_{x \in [d - δ, d + δ]} f (x)) 2 δ + (U (f ↾ R, P_{R}) - L (f ↾ L, P_{L})) & < \frac{2 ϵ}{3} + {sup}_{x, y \in [d - δ, d + δ]} (| f (x) - f (y) |) 2 δ & \leq \frac{2 ϵ}{3} + 2 {sup}_{x \in [d - δ, d + δ]} | f (x) | & \leq \frac{2 ϵ}{3} + 2 ‖ f ‖_{\infty} 2 δ & = \frac{2 ϵ}{3} + 4 ‖ f ‖_{\infty} δ

Thus a selection of

δ = \frac{ϵ}{4 ‖ f ‖_{\infty} 3}

shows that

U (f, P_{L} \cup P_{R}) - L (f, P_{L} \cup P_{R}) < ϵ

as needed.

🏗️ ΘρϵηΠατπ🚧 (under construction)

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