8.1: Elementary and Measurable Functions

From set functions, we now return to point functions

$$f : S \rightarrow\left(T, \rho^{\prime}\right)$$

whose domain $D_{f}$ consists of points of a set $S .$ The range space $T$ will mostly be $E,$ i.e., $E^{1}, E^{*}, C, E^{n},$ or another normed space. We assume $f(x)=0$ unless defined otherwise. (In a general metric space $T,$ we may take some fixed element $q$ for $0 .$ ) Thus $D_{f}$ is all of $S$, always.

We also adopt a convenient notation for sets:

$$"A(P)" \text { for } "\{x \in A | P(x)\}."$$

Thus

$$\begin{aligned} A(f \neq a) &=\{x \in A | f(x) \neq a\}, \\ A(f=g) &=\{x \in A | f(x)=g(x)\}, \\ A(f>g) &=\{x \in A | f(x)>g(x)\}, \text { etc. } \end{aligned}$$

Definition

A measurable space is a set $S \neq \emptyset$ together with a set ring $\mathcal{M}$ of subsets of $S,$ denoted $(S, \mathcal{M})$.

Henceforth, $(S, \mathcal{M})$ is fixed.

Definition

An M-partition of a set $A$ is a countable set family $\mathcal{P}=\left\{A_{i}\right\}$ such that

$$A=\bigcup_{i} A_{i}(d i s j o i n t),$$

with $A, A_{i} \in \mathcal{M}$.

We briefly say "the partition $A=\bigcup A_{i} .$"

An $\mathcal{M}$-partition $\mathcal{P}^{\prime}=\left\{B_{i k}\right\}$ is a refinement of $\mathcal{P}=\left\{A_{i}\right\}\left(\text { or } \mathcal{P}^{\prime} \text { refines }\right.$ $\mathcal{P},$ or $\mathcal{P}^{\prime}$ is finer than $\mathcal{P} )$ iff

$$(\forall i) \quad A_{i}=\bigcup_{k} B_{i k}$$

i.e., each $B_{i k}$ is contained in some $A_{i}$.

The intersection $\mathcal{P}^{\prime} \cap \mathcal{P}^{\prime \prime}$ of $\mathcal{P}^{\prime}=\left\{A_{i}\right\}$ and $\mathcal{P}^{\prime \prime}=\left\{B_{k}\right\}$ is understood to be the family of all sets of the form

$$A_{i} \cap B_{k}, \quad i, k=1,2, \dots$$

It is an $\mathcal{M}$ -partition that refines both $\mathcal{P}^{\prime}$ and $\mathcal{P}^{\prime \prime}$.

Definition

A map (function) $f : S \rightarrow T$ is elementary, or $\mathcal{M}$-elementary, on a set $A \in \mathcal{M}$ iff there is an M-partition $\mathcal{P}=\left\{A_{i}\right\}$ of $A$ such that $f$ is constant $\left(f=a_{i}\right)$ on each $A_{i} .$

If $\mathcal{P}=\left\{A_{1}, \ldots, A_{q}\right\}$ is finite, we say that $f$ is simple, or $\mathcal{M}$-simple, on $A .$

If the $A_{i}$ are intervals in $E^{n},$ we call $f$ a step function; it is a simple step function if $\mathcal{P}$ is finite.

The function values $a_{i}$ are elements of $T$ (possibly vectors). They may be infinite if $T=E^{*} .$ Any simple map is also elementary, of course.

Definition

A map $f : S \rightarrow\left(T, \rho^{\prime}\right)$ is said to be measurable (or $\mathcal{M}$ -measurable $)$ on a $\operatorname{set} A$ in $(S, \mathcal{M})$ iff

$$f=\lim _{m \rightarrow \infty} f_{m} \quad(\text { pointwise }) \text { on } A$$

for some sequence of functions $f_{m} : S \rightarrow T,$ all elementary on $A .$ (See Chapter 4, §12 for "pointwise.")

Note 1. This implies $A \in \mathcal{M},$ as follows from Definitions 2 and $3 .(\mathrm{Why} \text { ? })$

Corollary $\PageIndex{1}$

If $f : S \rightarrow\left(T, \rho^{\prime}\right)$ is elementary on $A,$ it is measurable on $A .$

Proof

Set $f_{m}=f, m=1,2, \ldots,$ in Definition $4 .$ Then clearly $f_{m} \rightarrow f$ on $A$. $square$

Corollary $\PageIndex{2}$

If $f$ is simple, elementary, or measurable on $A$ in $(S, \mathcal{M}),$ it has the same property on any subset $B \subseteq A$ with $B \in \mathcal{M}$.

Proof

Let $f$ be simple on $A ;$ so $f=a_{i}$ on $A_{i}, i=1,2, \ldots, n,$ for some finite $\mathcal{M}$ -partition, $A=\bigcup_{i=1}^{n} A_{i}$.

If $A \supseteq B \in \mathcal{M},$ then

$$\left\{B \cap A_{i}\right\}, \quad i=1,2, \ldots, n,$$

is a finite $\mathcal{M}$ -partition of $B(\text { why? }),$ and $f=a_{i}$ on $B \cap A_{i} ;$ so $f$ is simple on $B$.

For elementary maps, use countable partitions.

Now let $f$ be measurable on $A,$ i.e.,

$$f=\lim _{m \rightarrow \infty} f_{m}$$

for some elementary maps $f_{m}$ on $A .$ As shown above, the $f_{m}$ are elementary on $B,$ too, and $f_{m} \rightarrow f$ on $B ;$ so $f$ is measurable on $B . \quad \square$

Corollary $\PageIndex{3}$

If $f$ is elementary or measurable on each of the (countably many $)$ sets $A_{n}$ in $(S, \mathcal{M}),$ it has the same property on their union $A=\bigcup_{n} A_{n}$.

Proof

Let $f$ be elementary on each $A_{n}$ (so $A_{n} \in \mathcal{M}$ by Note 1$)$.

By Corollary 1 of Chapter 7, §1,

$$A=\bigcup A_{n}=\bigcup B_{n}$$

for some disjoint sets $B_{n} \subseteq A_{n}\left(B_{n} \in \mathcal{M}\right)$.

By Corollary $2, f$ is elementary on each $B_{n} ;$ i.e., constant on sets of some $\mathcal{M}$ -partition $\left\{B_{n i}\right\}$ of $B_{i}$.

All $B_{n i}$ combined (for all $n$ and all $i )$ form an $\mathcal{M}$-partition of $A$,

$$A=\bigcup_{n} B_{n}=\bigcup_{n, i} B_{n i}.$$

As $f$ is constant on each $B_{n i},$ it is elementary on $A .$

For measurable functions $f,$ slightly modify the method used in Corollary $2 . \square$

Corollary $\PageIndex{4}$

If $f : S \rightarrow\left(T, \rho^{\prime}\right)$ is measurable on $A$ in $(S, \mathcal{M}),$ so is the composite map $g \circ f,$ provided $g : T \rightarrow\left(U, \rho^{\prime \prime}\right)$ is relatively continuous on $f[A]$.

Proof

By assumption,

$$f=\lim _{m \rightarrow \infty} f_{m} \text { (pointwise) }$$

for some elementary maps $f_{m}$ on $A$.

Hence by the continuity of $g$,

$$g\left(f_{m}(x)\right) \rightarrow g(f(x)),$$

i.e., $g \circ f_{m} \rightarrow g \circ f$ (pointwise) on $A$.

Moreover, all $g \circ f_{m}$ are elementary on $A$ (for $g \circ f_{m}$ is constant on any partition set, if $f_{m}$ is).

Thus $g \circ f$ is measurable on $A,$ as claimed. $\square$

Theorem $\PageIndex{1}$

If the maps $f, g, h : S \rightarrow E^{1}(C)$ are simple, elementary, or measurable on $A$ in $(S, \mathcal{M}),$ so are $f \pm g, f h,|f|^{a}$ (for real $a \neq 0 )$ and $f / h$ (if $h \neq 0$ on $A ) .$

Similarly for vector-valued $f$ and $g$ and scalar-valued $h$.

Proof

First, let $f$ and $g$ be elementary on $A .$ Then there are two $\mathcal{M}$-partitions,

$$A=\bigcup A_{i}=\bigcup B_{k},$$

such that $f=a_{i}$ on $A_{i}$ and $g=b_{k}$ on $B_{k},$ say.

The sets $A_{i} \cap B_{k}$ (for all $i$ and $k )$ then form a new $\mathcal{M}$ -partition of $A(\text { why? })$, such that both $f$ and $g$ are constant on each $A_{i} \cap B_{k}(\text { why?); hence so is } f \pm g$.

Thus $f \pm g$ is elementary on $A .$ Similarly for simple functions.

Next, let $f$ and $g$ be measurable on $A ;$ so

$$f=\lim f_{m} \text { and } g=\lim g_{m} \text { (pointwise) on } A$$

for some elementary maps $f_{m}, g_{m}$.

By what was shown above, $f_{m} \pm g_{m}$ is elementary for each $m .$ Also,

$$f_{m} \pm g_{m} \rightarrow f \pm g (\text { pointwise }) \text { on } A,$$

Thus $f \pm g$ is measurable on $A$.

The rest of the theorem follows quite similarly. $\square$

If the range space is $E^{n}\left(\text { or } C^{n}\right),$ then $f$ has $n$ real (complex) components $f_{1}, \ldots, f_{n},$ as in Chapter 4,§3 (Part II). This yields the following theorem.

Theorem $\PageIndex{2}$

A function $f : S \rightarrow E^{n}\left(C^{n}\right)$ is simple, elementary, or measurable on a set $A$ in $(S, \mathcal{M})$ iff all its $n$ component functions $f_{1}, f_{2}, \ldots, f_{n}$ are.

Proof

For simplicity, consider $f : S \rightarrow E^{2}, f=\left(f_{1}, f_{2}\right)$.

If $f_{1}$ and $f_{2}$ are simple or elementary on $A$ then (exactly as in Theorem 1$)$, one can achieve that both are constant on sets $A_{i} \cap B_{k}$ of one and the same $\mathcal{M}$-partition of $A .$ Hence $f=\left(f_{1}, f_{2}\right),$ too, is constant on each $A_{i} \cap B_{k},$ as required.

Conversely, let

$$f=\overline{c}_{i}=\left(a_{i}, b_{i}\right) \text { on } C_{i}$$

for some $\mathcal{M}$-partition

$$A=\bigcup C_{i}.$$

Then by definition, $f_{1}=a_{i}$ and $f_{2}=b_{i}$ on $C_{i} ;$ so both are elementary (or simple) on $A .$

In the general case $\left(E^{n} \text { or } C^{n}\right),$ the proof is analogous.

For measurable functions, the proof reduces to limits of elementary maps (using Theorem 2 of Chapter 3, §15). The details are left to the reader. $\square$

Note 2. As $C=E^{2},$ a complex function $f : S \rightarrow C$ is simple, elementary, or measurable on $A$ iff its real and imaginary parts are.

By Definition $4,$ a measurable function is a pointwise limit of elementary maps. However, if $\mathcal{M}$ is a $\sigma$-ring, one can make the limit uniform. Indeed, we have the following theorem.

Theorem $\PageIndex{3}$

If $\mathcal{M}$ is a $\sigma$-ring, and $f : S \rightarrow\left(T, \rho^{\prime}\right)$ is $\mathcal{M}$-measurable on $A,$ then

$$f=\lim _{m \rightarrow \infty} g_{m} \text { (uniformly) on } A$$

for some finite elementary maps $g_{m}$.

Proof

Thus given $\varepsilon>0,$ there is a finite elementary map $g$ such that $\rho^{\prime}(f, g)<\varepsilon$
on $A$.

Theorem $\PageIndex{4}$

If $\mathcal{M}$ is a $\sigma$-ring in $S,$ if

$$f_{m} \rightarrow f(\text {pointwise}) \text { on } A$$

$\left(f_{m} : S \rightarrow\left(T, \rho^{\prime}\right)\right),$ and if all $f_{m}$ are $\mathcal{M}$ -measurable on $A,$ so also is $f$.

Briefly: $A$ pointwise limit of measurable maps is measurable (unlike continuous maps; cf. Chapter 4, §12).

Proof

By the second clause of Theorem $3,$ each $f_{m}$ is uniformly approximated by some elementary map $g_{m}$ on $A,$ so that, taking $\varepsilon=1 / m, m=1,2, \ldots$,

$$\rho^{\prime}\left(f_{m}(x), g_{m}(x)\right)<\frac{1}{m} \quad \text { for all } x \in A \text { and all } m.$$

Fixing such a $g_{m}$ for each $m,$ we show that $g_{m} \rightarrow f (\text { pointwise })$ on $A,$ as required in Definition $4 .$

Indeed, fix any $x \in A .$ By assumption, $f_{m}(x) \rightarrow f(x) .$ Hence, given $\delta>0$,

$$(\exists k)(\forall m>k) \quad \rho^{\prime}\left(f(x), f_{m}(x)\right)<\delta.$$

Take $k$ so large that, in addition,

$$(\forall m>k) \quad \frac{1}{m}<\delta.$$

Then by the triangle law and by $(1),$ we obtain for $m>k$ that

$$\begin{aligned} \rho^{\prime}\left(f(x), g_{m}(x)\right) & \leq \rho^{\prime}\left(f(x), f_{m}(x)\right)+\rho^{\prime}\left(f_{m}(x), g_{m}(x)\right) \\ &<\delta+\frac{1}{m}<2 \delta \end{aligned}.$$

As $\delta$ is arbitrary, this implies $\rho^{\prime}\left(f(x), g_{m}(x)\right) \rightarrow 0,$ i.e., $g_{m}(x) \rightarrow f(x)$ for any (fixed) $x \in A,$ thus proving the measurability of $f . \quad \square$

Note 3. If

$$\mathcal{M}=\mathcal{B} (=\text { Borel field in } S),$$

we often say "Borel measurable" for $\mathcal{M}$-measurable. If

$$\mathcal{M}=\left\{\text { Lebesgue measurable sets in } E^{n}\right\},$$

we say "Lebesgue (L) measurable" instead. Similarly for "Lebesgue-Stieltjes (LS) measurable."