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
\[
\begin{aligned} A(f \neq a) &=\{x \in A | f(x) \neq a\}, \\ A(f=g) &=\{x \in A | f(x)=g(x)\}, \\ A(f>g) &=\{x \in A | f(x)>g(x)\}, \text { etc. } \end{aligned}
\]
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.
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 \(\mathcal{P}^{\prime \prime}\).
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.
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 \(3 .(\mathrm{Why} \text { ? })\)
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 \(A\). \(square\)
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 \(B . \quad \square\)
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 \(2 . \square\)
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. \(\square\)
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.
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.
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 \(A\).
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."