Jordan measurable sets · Selected prerequisite proofs

L10.5.1: Jordan measurability iff null boundary, using an interior containing rectangle.

Proposition 10.5.1.

A bounded set

S \subset \R^n

is Jordan measurable if and only if the boundary

\partial S

is a measure zero set.

Proof.

Suppose

R

is a closed rectangle such that

S

is contained in the interior of

R\text{.}

If

x \in \partial S\text{,}

then for every

\delta > 0\text{,}

the sets

S \cap B(x,\delta)

(where

\chi_S

is 1) and the sets

(R \setminus S) \cap B(x,\delta)

(where

\chi_S

is 0) are both nonempty. So

\chi_S

is not continuous at

x\text{.}

If

x

is either in the interior of

S

or in the complement of the closure

\widebar{S}\text{,}

then

\chi_S

is either identically 1 or identically 0 in a whole neighborhood of

x

and hence

\chi_S

is continuous at

x\text{.}

Therefore, the set of discontinuities of

\chi_S

is precisely the boundary

\partial S\text{.}

The proposition follows.

L10.5.3: Full equality of Jordan volume and outer measure, with the finite-disjoint-rectangle and empty-family steps supplied.

Proposition 10.5.3.

If

S \subset \R^n

is Jordan measurable, then

V(S) = m^*(S)\text{.}

Proof.

Given

\epsilon > 0\text{,}

let

R

be a closed rectangle that contains

S\text{.}

Let

P

be a partition of

R

such that

\begin{equation*} U(P,\chi_S) \leq \left( \int_R \chi_S \right) + \epsilon = V(S) + \epsilon \qquad \text{and} \qquad L(P,\chi_S) \geq \left( \int_R \chi_S \right) - \epsilon = V(S)-\epsilon. \end{equation*}

Let

R_1,R_2,\ldots,R_k

be all the subrectangles of

P

such that

\chi_S

is not identically zero on each

R_j\text{.}

That is, there is some point

x \in R_j

such that

x \in S

(i.e.

\chi_S(x)=1

). Let

O_j

be an open rectangle such that

R_j \subset O_j

and

V(O_j) < V(R_j) + \nicefrac{\epsilon}{k}\text{.}

Notice that

S \subset \bigcup_j O_j\text{.}

Then

\begin{equation*} U(P,\chi_S) = \sum_{j=1}^k V(R_j) > \left(\sum_{j=1}^k V(O_j)\right) - \epsilon \geq m^*(S) - \epsilon . \end{equation*}

As

U(P,\chi_S) \leq V(S) + \epsilon\text{,}

then

m^*(S) - \epsilon \leq V(S) + \epsilon\text{,}

or in other words

m^*(S) \leq V(S)\text{.}

Let

R'_1,R'_2,\ldots,R'_\ell

be all the subrectangles of

P

such that

\chi_S

is identically one on each

R'_j\text{.}

In other words, these are the subrectangles contained in

S\text{.}

The interiors of the subrectangles

R'^\circ_j

are disjoint and

V(R'^\circ_j) = V(R'_j)\text{.}

Via Exercise 10.3.16,

\begin{equation*} m^*\Bigl(\bigcup_{j=1}^\ell R'^\circ_j\Bigr) = \sum_{j=1}^\ell V(R'^\circ_j) . \end{equation*}

Hence

\begin{equation*} m^*(S) \geq m^*\Bigl(\bigcup_{j=1}^\ell R'_j\Bigr) \geq m^*\Bigl(\bigcup_{j=1}^\ell R'^\circ_j\Bigr) = \sum_{j=1}^\ell V(R'^\circ_j) = \sum_{j=1}^\ell V(R'_j) = L(P,\chi_S) \geq V(S) - \epsilon . \end{equation*}

Therefore

m^*(S) \geq V(S)

as well.

L10.5.5: Every bounded continuous function on a bounded Jordan set is Riemann integrable; full proof.

Proposition 10.5.5.

If

S \subset \R^n

is a bounded Jordan measurable set and

f \colon S \to \R

is a bounded continuous function, then

f

is integrable on

S\text{.}

Proof.

Define the function

\widetilde{f}

as above for some closed rectangle

R

with

S \subset R\text{.}

If

x \in R \setminus \widebar{S}\text{,}

then

\widetilde{f}

is identically zero in a neighborhood of

x\text{.}

Similarly, if

x

is in the interior of

S\text{,}

then

\widetilde{f} = f

on a neighborhood of

x

and

f

is continuous at

x\text{.}

Therefore,

\widetilde{f}

is only ever possibly discontinuous at

\partial S\text{,}

which is a set of measure zero, and we are finished.

L10.5.9: Full compact Jordan image theorem, with the local inverse theorem already closed; source f|V is the map g|V.

Proposition 10.5.9.

Suppose

U \subset \R^n

is open and

S \subset U

is a compact Jordan measurable set. Suppose

g \colon U \to \R^n

is a one-to-one continuously differentiable mapping such that the Jacobian determinant

J_g

is never zero on

S\text{.}

Then

g(S)

is bounded and Jordan measurable.

Proof.

Let

T \coloneqq g(S)\text{.}

By Lemma 7.5.5, the set

T

is also compact and so closed and bounded. We claim

\partial T \subset g(\partial S)\text{.}

Suppose the claim is proved. As

S

is Jordan measurable,

\partial S

is measure zero. Then

g(\partial S)

is measure zero by Proposition 10.3.10. As

\partial T \subset g(\partial S)\text{,}

then

T

is Jordan measurable.

It is therefore left to prove the claim. As

T

is closed,

\partial T \subset T\text{.}

Suppose

y \in \partial T\text{.}

There must exist an

x \in S

such that

g(x) = y\text{,}

and by hypothesis

J_g(x) \neq 0\text{.}

We use the inverse function theorem (Theorem 8.5.1). We find a neighborhood

V \subset U

of

x

and an open set

W

such that the restriction

f|_V

is a one-to-one and onto function from

V

to

W

with a continuously differentiable inverse. In particular,

g(x) = y \in W\text{.}

As

y \in \partial T\text{,}

there exists a sequence

\{ y_k \}_{k=1}^\infty

in

W

with

\lim_{k\to\infty} y_k = y

and

y_k \notin T\text{.}

As

g|_V

is invertible and in particular has a continuous inverse, there exists a sequence

\{ x_k \}_{k=1}^\infty

in

V

such that

g(x_k) = y_k

and

\lim_{k\to\infty} x_k = x\text{.}

Since

y_k \notin T = g(S)\text{,}

clearly

x_k \notin S\text{.}

Since

x \in S\text{,}

we conclude that

x \in \partial S\text{.}

The claim is proved:

\partial T \subset g(\partial S)\text{.}