
# 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."