8.2: Measurability of Extended-Real Functions
- Page ID
- 32365
( \newcommand{\kernel}{\mathrm{null}\,}\)
Henceforth we presuppose a measurable space (S,M), where M is a σ-ring in S. Our aim is to prove the following basic theorem, which is often used as a definition, for extended-real functions f:S→E∗.
Theorem 8.2.1
A function f:S→E∗ is measurable on a set A∈M iff it satisfies one of the following equivalent conditions (hence all of them):
(i∗)(∀a∈E∗)A(f>a)∈M;(ii∗)(∀a∈E∗)A(f≥a)∈M;(iii∗)(∀a∈E∗)A(f<a)∈M;(iv∗)(∀a∈E∗)A(f≤a)∈M.
We first prove the equivalence of these conditions by showing that (i∗)⇒ (ii∗)⇒(iii∗)⇒(iv∗)⇒(i∗), closing the "circle."
(i∗)⇒(ii∗). Assume (i∗). If a=−∞,
A(f≥a)=A∈M
by assumption. If a=+∞,
A(f≥a)=A(f=∞)=∞⋂n=1A(f>n)∈M
by (i∗). And if a∈E1,
A(f≥a)=∞⋂n=1A(f>a−1n).
(Verify!) By (i∗),
A(f>a−1n)∈M;
so A(f≥a)∈M( a σ-ring! ).
(ii∗)⇒(iii∗). For (i∗) and A∈M imply
A(f<a)=A−A(f≥a)∈M.
(iii∗)⇒(iv∗). If a∈E1,
A(f≤a)=∞⋂n=1A(f<a+1n)∈M.
What if a=±∞?
(iv∗)⇒(i∗). Indeed, (iv∗) and A∈M imply
A(f>a)=A−A(f≤a)∈M.
Thus, indeed, each of (i∗) to (iv∗) implies the others. To finish, we need two lemmas that are of interest in their own right.
Lemma 8.2.1
If the maps fm:S→E∗(m=1,2,…) satisfy conditions (i∗)−(iv∗), so also do the functions
supfm,inffm,¯limfm, and lim_fm,
defined pointwise, i.e.,
(supfm)(x)=supfm(x),
and similarly for the others.
- Proof
-
Let f=supfm. Then
A(f≤a)=∞⋂m=1A(fm≤a) (Why?)
But by assumption,
A(fm≤a)∈M
(fm satisfies (iv∗)). Hence A(f≤a)∈M( for M is a σ-ring ).
Thus sup fm satisfies (i∗)−(iv∗).
So does inf fm; for
A(inffm≥a)=∞⋂m=1A(fm≥a)∈M.
(Explain!)
So also do lim_fm and ¯limfm; for by definition,
lim_tfm=supkgk,
where
gk=infm≥kfm
satisfies (i∗)−(iv∗), as was shown above; hence so does sup gk=lim_fm.
Similarly for ¯limfm.◻
Lemma 8.2.2
If f satisfies (i∗)−(iv∗), then
f=limm→∞fm (uniformly) on A
for some sequence of finite functions fm, all M -elementary on A.
Moreover, if f≥0 on A, the fm can be made nonnegative, with {fm}↑ on A.
- Proof
-
Let H=A(f=+∞),K=A(f=−∞), and
Amk=A(k−12m≤f<k2m)
for m=1,2,… and k=0,±1,±2,…,±n,…
By(i∗)−(iv∗),
H=A(f=+∞)=A(f≥+∞)∈M,
K∈M, and
Amk=A(f≤k−12m)∩A(f<k2m)∈M.
Now define
(∀m)fm=k−12m on Amk,
fm=m on H, and fm=−m on K. Then each fm is finite and elementary on A since
(∀m)A=H∪K∪∞⋃k=−∞Amk(disjoint)
and fm is constant on H,K, and Amk.
We now show that fm→f (uniformly) on H,K, and
J=∞⋃k=−∞Amk,
hence on A.
Indeed, on H we have
limfm=limm=+∞=f,
and the limit is uniform since the fm are constant on H.
Similarly,
fm=−m→−∞=f on K.
Finally, on Amk we have
(k−1)2−m≤f<k2−m
and fm=(k−1)2−m; hence
|fm−f|<k2−m−(k−1)2−m=2−m.
Thus
|fm−f|<2−m→0
on each Amk, hence on
J=∞⋃k=−∞Amk.
By Theorem 1 of Chapter 4, §12, it follows that fm→f( uniformly ) on J. Thus, indeed, fm→f( uniformly ) on A.
If, further, f≥0 on A, then K=∅ and Amk=∅ for k≤0. Moreover, on passage from m to m+1, each Amk(k>0) splits into two sets. On one, fm+1=fm; on the other, fm+1>fm. (Why?)
Thus 0≤fm↗f( uniformly ) on A, and all is proved. ◻
Theorem 8.2.1 (Restated)
A function f:S→E∗ is measurable on a set A∈M iff it satisfies one of the following equivalent conditions (hence all of them):
(i∗)(∀a∈E∗)A(f>a)∈M;(ii∗)(∀a∈E∗)A(f≥a)∈M;(iii∗)(∀a∈E∗)A(f<a)∈M;(iv∗)(∀a∈E∗)A(f≤a)∈M.
- Proof
-
If f is measurable on A, then by definition, f=limfm (pointwise) for some elementary maps fm on A.
By Problem 4 (ii) in §1, all fm satisfy (i*)-(iv*). Thus so does fby Lemma 1, for here f=limfm=¯limfm.
The converse follows by Lemma 2. This completes the proof. ◻
Note 1. Lemmas 1 and 2 prove Theorems 3 and 4 of $1, for f:S→E∗. By using also Theorem 2 in §1, one easily extends this to f:S→En(Cn). Verify!
Corollary 8.2.1
If f:S→E∗ is measurable on A, then
(∀a∈E∗)A(f=a)∈M and A(f≠a)∈M.
Indeed,
A(f=a)=A(f≥a)∩A(f≤a)∈M
and
A(f≠a)=A−A(f=a)∈M.
Corollary 8.2.2
If f:S→(T,ρ′) is measurable on A in (S,M), then
A∩f−1[G]∈M
for every globe G=Gq(δ) in (T,ρ′).
- Proof
-
Define h:S→E1 by
h(x)=ρ′(f(x),q).
Then h is measurable on A by Problem 6 in §1. Thus by Theorem 1,
A(h<δ)∈M.But as is easily seen,
A(h<δ)={x∈A|ρ′(f(x),q)<δ}=A∩f−1[Gq(δ)].
Hence the result. ◻
Definition
Given f,g:S→E∗, we define the maps f∨g and f∧g on S by
(f∨g)(x)=max{f(x),g(x)}
and
(f∧g)(x)=min{f(x),g(x)};
similarly for f∨g∨h,f∧g∧h, etc.
We also set
f+=f∨0 and f−=−f∨0.
Clearly, f+≥0 and f−≥0 on S. Also, f=f+−f− and |f|=f++f−.
(Why?) We now obtain the following theorem.
Theorem 8.2.2
If the functions f,g:S→E∗ are simple, elementary, or measurable on A, so also are f±g,fg,f∨g,f∧g,f+,f−, and |f|a(a≠0).
- Proof
-
If f and g are finite, this follows by Theorem 1 of §1 on verifying that
f∨g=12(f+g+|f−g|)
and
f∧g=12(f+g−|f−g|)
on S. (Check it!)
Otherwise, consider
A(f=+∞),A(f=−∞),A(g=+∞), and A(g=−∞).
By Theorem 1, these are M-sets; hence so is their union U.
On each of them f∨g and f∧g equal f or g; so by Corollary 3 in §1, f∨g and f∧g have the desired properties on U. So also have f+=f∨0 and f−=−f∨0( take g=0).
We claim that the maps f±g and fg are simple (hence elementary and measurable) on each of the four sets mentioned above, hence on U.
For example, on A(f=+∞),
f±g=+∞( constant )
by our conventions (2∗) in Chapter 4, §4. For fg, split A(f=+∞) into three sets A1,A2,A3∈M, with g>0 on A1,g<0 on A2, and g=0 on A3; so fg=+∞ on A1,fg=−∞ on A2, and fg=0 on A3. Hence fg is simple on A(f=+∞).
For |f|a, use U=A(|f|=∞). Again, the theorem holds on U, and also on A−U, since f and g are finite on A−U∈M. Thus it holds on A=(A−U)∪U by Corollary 3 in §1. ◻
Note 2. Induction extends Theorem 2 to any finite number of functions.
Note 3. Combining Theorem 2 with f=f+−f−, we see that f:S→E∗ is simple (elementary, measurable) iff f+ and f− are. We also obtain the following result.
Theorem 8.2.3
If the functions f,g:S→E∗ are measurable on A∈M, then A(f≥g)∈M,A(f<g)∈M,A(f=g)∈M, and A(f≠g)∈M.