Riemann integral over rectangles · Selected prerequisite proofs

L10.1.2: Full Darboux bounds, with the used volume exercise now proved.

Proposition 10.1.2.

Suppose

R \subset \R^n

is a closed rectangle and

f \colon R \to \R

is a bounded function. Let

m, M \in \R

be such that for all

x \in R\text{,}

we have

m \leq f(x) \leq M\text{.}

Then for every partition

P

of

R\text{,}

\begin{equation*} m \, V(R) \leq L(P,f) \leq U(P,f) \leq M\, V(R) . \end{equation*}

Proof.

Let

P

be a partition of

R\text{.}

For all

i\text{,}

we have

m \leq m_i \leq M_i \leq M\text{.}

Also

\sum_{i=1}^N V(R_i) = V(R)\text{.}

Therefore,

\begin{gathered} m \, V(R) = m \left( \sum_{i=1}^N V(R_i) \right) = \sum_{i=1}^N m \, V(R_i) \leq \sum_{i=1}^N m_i \, V(R_i) \leq \\ \leq \sum_{i=1}^N M_i \, V(R_i) \leq \sum_{i=1}^N M \,V(R_i) = M \left( \sum_{i=1}^N V(R_i) \right) = M \,V(R) . \qedhere \end{gathered}

L10.1.5: Both refinement inequalities; upper half is P17.1.

Proposition 10.1.5.

Suppose

R \subset \R^n

is a closed rectangle,

P

is a partition of

R\text{,}

and

\widetilde{P}

is a refinement of

P\text{.}

If

f \colon R \to \R

is bounded, then

\begin{equation*} L(P,f) \leq L(\widetilde{P},f) \qquad \text{and} \qquad U(\widetilde{P},f) \leq U(P,f) . \end{equation*}

Proof.

We prove the first inequality, and the second follows similarly. Let

R_1,R_2,\ldots,R_N

be the subrectangles of

P

and

\widetilde{R}_1,\widetilde{R}_2,\ldots,\widetilde{R}_{\widetilde{N}}

be the subrectangles of

\widetilde{P}\text{.}

Let

I_k

be the set of all indices

j

such that

\widetilde{R}_j \subset R_k\text{.}

For example, in figures 10.1 and 10.2,

I_4 = \{ 6, 7, 8, 9 \}

as

R_4 = \widetilde{R}_6 \cup \widetilde{R}_7 \cup \widetilde{R}_8 \cup \widetilde{R}_9\text{.}

Then,

\begin{equation*} R_k = \bigcup_{j \in I_k} \widetilde{R}_j, \qquad V(R_k) = \sum_{j \in I_k} V(\widetilde{R}_j). \end{equation*}

Let

m_k \coloneqq \inf \bigl\{ f(x) : x \in R_k \bigr\}\text{,}

and

\widetilde{m}_j \coloneqq \inf \bigl\{ f(x) : x \in \widetilde{R}_j \bigr\}

as usual. If

j \in I_k\text{,}

then

m_k \leq \widetilde{m}_j\text{.}

Then

\begin{equation*} \begin{aligned} L(P,f) = \sum_{k=1}^N m_k V(R_k) & = \sum_{k=1}^N \sum_{j\in I_k} m_k V(\widetilde{R}_j) \\ & \leq \sum_{k=1}^N \sum_{j\in I_k} \widetilde{m}_j V(\widetilde{R}_j) = \sum_{j=1}^{\widetilde{N}} \widetilde{m}_j V(\widetilde{R}_j) = L(\widetilde{P},f) . \qedhere \end{aligned} \end{equation*}

L10.1.6: Complete common-refinement proof of lower <= upper integral and volume bounds.

Proposition 10.1.6.

Let

R \subset \R^n

be a closed rectangle and

f \colon R \to \R

a bounded function. Let

m, M \in \R

be such that for all

x \in R\text{,}

we have

m \leq f(x) \leq M\text{.}

Then

\begin{equation*} m \, V(R) \leq \underline{\int_R} f \leq \overline{\int_R} f \leq M \, V(R).\tag{10.1} \end{equation*}

Proof.

For every partition

P\text{,}

via Proposition 10.1.2,

\begin{equation*} m\,V(R) \leq L(P,f) \leq U(P,f) \leq M\,V(R). \end{equation*}

Taking the supremum of

L(P,f)

and the infimum of

U(P,f)

over all partitions

P\text{,}

we obtain the first and the last inequality in (10.1).

The key inequality in (10.1) is the middle one. Let

P=(P_1,P_2,\ldots,P_n)

and

Q=(Q_1,Q_2,\ldots,Q_n)

be partitions of

R\text{.}

Define

\widetilde{P} = ( \widetilde{P}_1,\widetilde{P}_2,\ldots,\widetilde{P}_n )

by letting

\widetilde{P}_k \coloneqq P_k \cup Q_k

for every

k\text{.}

Then

\widetilde{P}

is a partition of

R\text{,}

and

\widetilde{P}

is a refinement of

P

and also a refinement of

Q\text{.}

By Proposition 10.1.5,

L(P,f) \leq L(\widetilde{P},f)

and

U(\widetilde{P},f) \leq U(Q,f)\text{.}

Therefore,

\begin{equation*} L(P,f) \leq L(\widetilde{P},f) \leq U(\widetilde{P},f) \leq U(Q,f) . \end{equation*}

In other words, for two arbitrary partitions

P

and

Q\text{,}

we have

L(P,f) \leq U(Q,f)\text{.}

Via Proposition 1.2.7, we obtain

\begin{equation*} \sup \, \bigl\{ L(P,f) : P \text{ a partition of } R \bigr\} \leq \inf \, \bigl\{ U(P,f) : P \text{ a partition of } R \bigr\} . \end{equation*}

In other words,

\underline{\int_R} f \leq \overline{\int_R} f\text{.}

L10.1.12: Both directions of the small-gap criterion with supremum/infimum operations explicitly available.

Proposition 10.1.12.

Let

R \subset \R^n

be a closed rectangle and

f \colon R \to \R

a bounded function. Then

f \in \sR(R)

if and only if for every

\epsilon > 0\text{,}

there exists a partition

P

of

R

such that

\begin{equation*} U(P,f) - L(P,f) < \epsilon . \end{equation*}

Proof.

First, if

f

is integrable, then the supremum of

L(P,f)

and infimum of

U(Q,f)

over all partitions

P

and

Q

are equal and hence the infimum of

U(P,f)-L(Q,f)

is zero. Taking a common refinement

\widetilde{P}

of

P

and

Q

we find

U(\widetilde{P},f)-L(\widetilde{P},f) \leq U(P,f)-L(Q,f)\text{.}

Hence the infimum of

U(P,f)-L(P,f)

over all partitions

P

is zero, and so for every

\epsilon > 0\text{,}

there must be some partition

P

such that

U(P,f) - L(P,f) < \epsilon\text{.}

For the other direction, given an

\epsilon > 0

find

P

such that

U(P,f) - L(P,f) < \epsilon\text{.}

\begin{equation*} \overline{\int_R} f - \underline{\int_R} f \leq U(P,f) - L(P,f) < \epsilon . \end{equation*}

As

\overline{\int_R} f \geq \underline{\int_R} f

and the above holds for every

\epsilon > 0\text{,}

we conclude

\overline{\int_R} f = \underline{\int_R} f

and

f \in \sR(R)\text{.}

L10.1.13: Restriction to a subrectangle including the locally supplied degenerate cases.

Proposition 10.1.13.

Let

S \subset \R^n

be a closed rectangle. If

f \colon S \to \R

is integrable and

R \subset S

is a closed rectangle, then

f

is integrable on

R\text{.}

Proof.

Given

\epsilon > 0\text{,}

find a partition

P=(P_1,\ldots,P_n)

of

S

such that

U(P,f)-L(P,f) < \epsilon\text{.}

By making a refinement of

P

if necessary, assume that the endpoints of

R

are in

P\text{.}

That is, if

R = [a_1,b_1] \times [a_2,b_2] \times \cdots \times [a_n,b_n]\text{,}

then

a_i,b_i \in P_i\text{.}

Let

\widetilde{P} = (\widetilde{P}_1,\ldots,\widetilde{P}_n)

be the partition of

R

given by

\widetilde{P}_i = P_i \cap [a_i,b_i]\text{.}

Subrectangles of

\widetilde{P}

are subrectangles of

P\text{,}

that is,

R

is a union of subrectangles of

P\text{.}

Divide the subrectangles of

P

into two collections: Let

R_1,R_2\ldots,R_K

be the subrectangles of

P

that are also subrectangles of

\widetilde{P}

and let

R_{K+1},\ldots, R_N

be the rest. See Figure 10.3. Let

m_k

and

M_k

be the infimum and supremum of

f

on

R_k

as usual. Then,

\begin{equation*} \begin{split} \epsilon & > U(P,f)-L(P,f) = \sum_{k=1}^K (M_k-m_k) V(R_k) + \sum_{k=K+1}^N (M_k-m_k) V(R_k) \\ & \geq \sum_{k=1}^K (M_k-m_k) V(R_k) = U(\widetilde{P},f|_R)-L(\widetilde{P},f|_R) . \end{split} \end{equation*}

Therefore,

f|_R

is integrable.

L10.1.14: Full rectangle-diameter inequality.

Proposition 10.1.14.

If a rectangle

R \subset \R^n

has longest side at most

\alpha\text{,}

then for all

x,y \in R\text{,}

\begin{equation*} \snorm{x-y} \leq \sqrt{n} \, \alpha . \end{equation*}

Proof.

\begin{equation*} \begin{split} \snorm{x-y} & = \sqrt{ {(x_1-y_1)}^2 + {(x_2-y_2)}^2 + \cdots + {(x_n-y_n)}^2 } \\ & \leq \sqrt{ {(b_1-a_1)}^2 + {(b_2-a_2)}^2 + \cdots + {(b_n-a_n)}^2 } \\ & \leq \sqrt{ {\alpha}^2 + {\alpha}^2 + \cdots + {\alpha}^2 } = \sqrt{n} \, \alpha . \qedhere \end{split} \end{equation*}

L10.1.15: Complete continuous-integrability argument, with its fine-grid and zero-case omissions supplied.

Theorem 10.1.15.

Let

R \subset \R^n

be a closed rectangle. If

f \colon R \to \R

is continuous, then

f \in \sR(R)\text{.}

Proof.

The proof is analogous to the one-variable proof with some complications. The set

R

is a closed and bounded subset of

\R^n\text{,}

and hence compact. So

f

is uniformly continuous by Theorem 7.5.11. Let

\epsilon > 0

be given. Find a

\delta > 0

such that

\snorm{x-y} < \delta

implies

\babs{f(x)-f(y)} < \frac{\epsilon}{V(R)}\text{.}

Let

P

be a partition of

R\text{,}

where the longest side of every subrectangle is strictly less than

\frac{\delta}{\sqrt{n}}\text{.}

If

x, y \in R_k

for a subrectangle

R_k

of

P\text{,}

then, by the proposition,

\snorm{x-y} < \sqrt{n} \frac{\delta}{\sqrt{n}} = \delta\text{.}

Therefore,

\begin{equation*} f(x)-f(y) \leq \babs{f(x)-f(y)} < \frac{\epsilon}{V(R)} . \end{equation*}

As

f

is continuous on

R_k\text{,}

which is compact,

f

attains a maximum and a minimum on this subrectangle. Let

x

be a point where

f

attains the maximum and

y

be a point where

f

attains the minimum. Then

f(x) = M_k

and

f(y) = m_k

in the notation from the definition of the integral. Thus,

\begin{equation*} M_k-m_k = f(x)-f(y) < \frac{\epsilon}{V(R)} . \end{equation*}

And so

\begin{equation*} \begin{split} U(P,f) - L(P,f) & = \left( \sum_{k=1}^N M_k V(R_k) \right) - \left( \sum_{k=1}^N m_k V(R_k) \right) \\ & = \sum_{k=1}^N (M_k-m_k) V(R_k) \\ & < \frac{\epsilon}{V(R)} \sum_{k=1}^N V(R_k) = \epsilon. \end{split} \end{equation*}

Proposition 10.1.12 then says that

f \in \sR(R)\text{.}

L10.1.19: Independence of support-containing rectangle, with the exact zero-extension exercise and empty-support case supplied in P17.3.

Proposition 10.1.19.

Suppose

f \colon \R^n \to \R

is a continuous function with compact support. If

R

and

S

are closed rectangles such that

\operatorname{supp}(f) \subset R

and

\operatorname{supp}(f) \subset S\text{,}

then

\begin{equation*} \int_S f = \int_R f . \end{equation*}

Proof.

As

f

is continuous, it is automatically integrable on the rectangles

R\text{,}

S\text{,}

and

R \cap S\text{.}

Applying Exercise 10.1.7 twice,

\int_S f = \int_{S \cap R} f = \int_R f\text{.}

A diagram of a rectangle divided by two horizontal and two vertical lines into 9 smaller rectangles. The four vertices on the horizontal axis from left to right are labeled x sub 1,0, x sub 1,1, x sub 1,2, and x sub 1,3. On the vertical axis from bottom to top the vertices are labeled x sub 2,0, x sub 2,1, x sub 2,2, and x sub 2,3. The smaller rectangles are labeled in a snaking pattern from top left to bottom right R sub 1, R sub 2, and so on until R sub 9. — Figure 10.1. Example partition of a rectangle in $\R^2\text{.}$ The order of the subrectangles is not important.

A diagram of a divided rectangle as the previous one with extra marks. There are two more vertical cuts and one horizontal one marked in dashed line, now dividing the rectangle into 20 smaller rectangles labeled R tilde sub 1 through R tilde sub 20 in no particular order. The vertices on the horizontal and the vertical axis are marked similarly to before but now with x tilde sub 1,0 through x tilde sub 1,5 and x tilde sub 2,0 through x tilde sub 2,4 in order. On the old cuts both the old and new markings are shown, so for example x sub 1,2 has also the mark x tilde sub 1,3. — Figure 10.2. Example refinement of the partition from Figure 10.1. New “cuts” are marked in dashed lines. The exact order of the new subrectangles does not matter.

A rectangle subdivided into subrectangles in the same manner as before, this time with 2 horizontal and 3 vertical cuts into 12 smaller rectangles. A set of 4 smaller rectangles from the second and third row and second and third column is marked in dark shade and these 4 rectangles are labeled R sub 1, R sub 2, R sub 3, and R sub 4. — Figure 10.3. A partition of a large rectangle $S\text{,}$ that also gives a partition of a smaller rectangle (shaded and outlined) $R \subset S\text{.}$ The subrectangles $R_1,R_2,R_3,R_4$ are the subrectangles of $\widetilde{P} = \bigl( \{ x_{1,1}, x_{1,2} , x_{1,3} \} , \{ x_{2,1}, x_{2,2} , x_{2,3} \} \bigr)\text{.}$