**I. **Henceforth we shall presuppose not just a measurable space (§1) but a measure space ((S, mathcal{M}, m),) where (m : mathcal{M}
ightarrow E^{*}) is a measure on a (sigma) -ring (mathcal{M} subseteq 2^{S}).

We saw in Chapter 7 that one could often neglect sets of Lebesgue measure zero on (E^{n}-) if a property held everywhere except on a set of Lebesgue measure zero, we said it held "almost everywhere." The following definition generalizes this usage.

Definition

We say that a property (P(x)) holds for almost all (x in A) (with respect to the measure (m )) or almost everywhere (a.e. ((m) )) on (A) iff it holds on (A-Q) for some (Q in mathcal{M}) with (m Q=0).

Thus we write

[

f_{n}
ightarrow f(a . e .) ext { or } f=lim f_{n}(a . e .(m)) ext { on } A

]

iff (f_{n} ightarrow f( ext { pointwise })) on (A-Q, m Q=0 .) Of course, "pointwise" implies (" a . e . "( ext { take } Q=emptyset),) but the converse fails.

Definition

We say that (f : S ightarrowleft(T, ho^{prime} ight)) is almost measurable on (A) iff (A in mathcal{M}) and (f) is (mathcal{M}) -measurable on (A-Q, m Q=0).

We then also say that (f) is (m) -measurable ((m) being the measure involved ) as opposed to (mathcal{M})-measurable.

Observe that we may assume (Q subseteq A) here (replace (Q) by (A cap Q )).

***Note 1. **If (m) is a generalized measure (Chapter 7, §11), replace (m Q=0) by (v_{m} Q=0left(v_{m}= ext { total variation of } m
ight)) in Definitions 1 and 2 and in the following proofs.

Corollary (PageIndex{1})

If the functions

[

f_{n} : S
ightarrowleft(T,
ho^{prime}
ight), quad n=1,2, ldots

]

are (m)-measurable on (A,) and if

[

f_{n}
ightarrow f(a . e .(m))

]

on (A,) then (f) is (m)-measurable on (A).

**Proof**By assumption, (f_{n} ightarrow f ( ext { pointwise })) on (A-Q_{0}, m Q_{0}=0 .) Also, (f_{n}) is (mathcal{M})-measurable on

[

A-Q_{n}, m Q_{n}=0, quad n=1,2, dots

](The (Q_{n}) need not be the same.)

Let

[

Q=igcup_{n=0}^{infty} Q_{n};

]so

[

m Q leq sum_{n=0}^{infty} m Q_{n}=0.

]By Corollary 2 in §1, all (f_{n}) are (mathcal{M})-measurable on (A-Q) (why?), and (f_{n} ightarrow f)

(pointwise) on (A-Q,) as (A-Q subseteq A-Q_{0} .)Thus by Theorem 4 in §1, (f) is (mathcal{M}) -measurable on (A-Q .) As (m Q=0) , this

is the desired result. (square)

Corollary (PageIndex{2})

If (f=g (a . e .(m))) on (A) and (f) is (m)-measurable on (A,) so is (g).

**Proof**By assumption, (f=g) on (A-Q_{1}) and (f) is (mathcal{M})-measurable on (A-Q_{2}), with (m Q_{1}=m Q_{2}=0).

Let (Q=Q_{1} cup Q_{2} .) Then (m Q=0) and (g=f) on (A-Q .) (Why? ())

By Corollary 2 of §1, (f) is (mathcal{M})-measurable on (A-Q). Hence so is (g), as claimed. (square)

Corollary (PageIndex{3})

If (f : S ightarrowleft(T, ho^{prime} ight)) is (m) -measurable on (A,) then

[

f=lim _{n
ightarrow infty} f_{n} ( ext {uniformly}) ext { on } A-Q(m Q=0),

]

for some maps (f_{n},) all elementary on (A-Q).

**Proof**Add proof here and it will automatically be hidden

(Compare Corollary 3 with Theorem 3 in §1).

Quite similarly all other propositions of §1 carry over to almost measurable (i.e., (m) -measurable) functions. Note, however, that the term "measurable" in §§1 and 2 always meant (" mathcal{M}) -measurable." This implies (m)-measurability (take (Q=emptyset ),) but the converse fails. (See Note (2,) however.)

We still obtain the following result.

Corollary (PageIndex{4})

If the functions

[

f_{n} : S
ightarrow E^{*} quad(n=1,2, ldots)

]

are (m)-measurable on a set (A,) so also are

[

sup f_{n}, inf f_{n}, overline{lim } f_{n}, ext { and } underline{lim} f_{n}

]

(Use Lemma 1 of §2).

Similarly, Theorem 2 in §2 carries over to (m)-measurable functions.

**Note 2.** If (m) is complete (such as Lebesgue measure and LS measures) then, for (f : S
ightarrow E^{*}left(E^{n}, C^{n}
ight), m-) and (mathcal{M})-measurability coincide (see Problem 3 below).

**II. Measurability and Continuity.** To study the connection between these notions, we first state two lemmas, often treated as definitions.

Lemma (PageIndex{1})

(A operatorname{map} f : S
ightarrow E^{n}left(C^{n}
ight)) is (mathcal{M})-measurable on (A) iff

[

A cap f^{-1}[B] in mathcal{M}

]

for each Borel set (equivalently, open set) (B) in (E^{n}left(C^{n}
ight)).

**Proof**See Problems (8-10) in §2 for a sketch of the proof.

Lemma (PageIndex{2})

(A operatorname{map} f :(S,
ho)
ightarrowleft(T,
ho^{prime}
ight)) is relatively continuous on (A subseteq S) iff for any open set (B subseteqleft(T,
ho^{prime}
ight),) the set (A cap f^{-1}[B]) is open in ((A,
ho)) as a subspace of ((S,
ho)).

(This holds also with "open" replaced by "closed.")

**Proof**By Chapter 4, §1, footnote (4, f) is relatively continuous on (A) iff its restriction to (A) (call it (g : A ightarrow T )) is continuous in the ordinary sense.

Now, by Problem 15((mathrm{iv})(mathrm{v})) in Chapter 4, §2, with (S) replaced by (A,) this means that (g^{-1}[B]) is open (closed) (i n(A, ho)) when (B) is so in (left(T, ho^{prime} ight) .) But

[

g^{-1}[B]={x in A | f(x) in B}=A cap f^{-1}[B].

]

(Why?) Hence the result follows. (square)

Theorem (PageIndex{1})

Let (m : mathcal{M} ightarrow E^{*}) be a topological measure in ((S, ho) .) If (f : S ightarrow) (E^{n}left(C^{n} ight)) is relatively continuous on a set (A in mathcal{M},) it is (mathcal{M}) -measurable on (A).

**Proof**Let (B) be open in (E^{n}left(C^{n} ight) .) By Lemma 2,

[

A cap f^{-1}[B]

]

is open (i n(A, ho) .) Hence by Theorem 4 of Chapter 3, §12,

[

A cap f^{-1}[B]=A cap U

]

for some open set (U) in ((S, ho)).

Now, by assumption, (A) is in (mathcal{M} .) So is (U,) as (mathcal{M}) is topological ((mathcal{M} supseteq mathcal{G})).

Hence

[

A cap f^{-1}[B]=A cap U in mathcal{M}

]

for any open (B subseteq E^{n}left(C^{n} ight) .) The result follows by Lemma 1. (square)

**Note 3.** The converse fails. For example, the Dirichlet function (Example ((mathrm{c})) in Chapter 4, §1) is L-measurable (even simple) but discontinuous everywhere.**Note 4.** Lemma 1 and Theorem 1 hold for a map (f : S
ightarrowleft(T,
ho^{prime}
ight),) too, provided (f[A]) is separable, i.e.,

[

f[A] subseteq overline{D}

]

for a countable set (D subseteq T) (cf. Problem 11 in §2).***III.** For strongly regular measures (Definition 5 in Chapter 7, §7), we obtain the following theorem.

*Theorem (PageIndex{2})

(Luzin). Let (m : mathcal{M}
ightarrow E^{*}) be a strongly regular measure in ((S,
ho)). Let (f : S
ightarrowleft(T,
ho^{prime}
ight)) be (m)-measurable on (A).

Then given (varepsilon>0,) there is a closed set (F subseteq A(F in mathcal{M})) such that

[

m(A-F)

and (f) is relatively continuous on (F).

(Note that if (T=E^{*},
ho^{prime}) is as in Problem 5 of Chapter 3, §11.)

**Proof**By assumption, (f) is (mathcal{M})-measurable on a set

[

H=A-Q, m Q=0;

]

so by Problem 7 in §1, (f[H]) is separable in (T). We may safely assume that (f) is (mathcal{M})-measurable on (S) and that all of (T) is separable. (If not, replace (S) and (T) by (H) and (f[H],) restricting (f) to (H,) and (m) to (mathcal{M})-sets inside (H ;) see also Problems 7 and 8 below.)

Then by Problem 12 of §2, we can fix globes (G_{1}, G_{2}, ldots) in (T) such that

[

ext { each open set } B eq emptyset ext { in } T ext { is the union of a subsequence of }left{G_{k} ight}.

]

Now let (varepsilon>0,) and set

[

S_{k}=S cap f^{-1}left[G_{k} ight]=f^{-1}left[G_{k} ight], quad k=1,2, ldots

]

By Corollary 2 in §2, (S_{k} in mathcal{M} .) As (m) is strongly regular, we find for each (S_{k}) an open set

[

U_{k} supseteq S_{k},

]

with (U_{k} in mathcal{M}) and

[

mleft(U_{k}-S_{k} ight)]

Let (B_{k}=U_{k}-S_{k}, D=igcup_{k} B_{k} ;) so (D in mathcal{M}) and

[

m D leq sum_{k} m B_{k} leq sum_{k} frac{varepsilon}{2^{k+1}} leq frac{1}{2} varepsilon

]

and

[

U_{k}-B_{k}=S_{k}=f^{-1}left[G_{k} ight].

]

As (D=igcup B_{k},) we have

[

(forall k) quad B_{k}-D=B_{k} cap(-D)=emptyset.

]

Hence by (left(2^{prime} ight)),

[

egin{aligned}(forall k) quad f^{-1}left[G_{k} ight] cap(-D) &=left(U_{k}-B_{k} ight) cap(-D) &=left(U_{k} cap(-D) ight)-left(B_{k} cap(-D) ight)=U_{k} cap(-D) end{aligned}.

]

Combining this with ((1),) we have, for each open set (B=igcup_{i} G_{k_{i}} operatorname{in} T),

[

f^{-1}[B] cap(-D)=igcup_{i} f^{-1}left[G_{k_{i}} ight] cap(-D)=igcup_{i} U_{k_{i}} cap(-D).

]

since the (U_{k_{i}}) are open in (S) (by construction), the set ((3)) is open in (S-D) as a subspace of (S .) By Lemma (2,) then, (f) is relatively continuous on (S-D,) or rather on

[

H-D=A-Q-D

]

(since we actually substituted (S) for (H) in the course of the proof). As (m Q=0) and (m D[

m(H-D)]

Finally, as (m) is strongly regular and (H-D in mathcal{M},) there is a closed (mathcal{M})-set

[

F subseteq H-D subseteq A

]

such that

[

m(H-D-F)]

since (f) is relatively continuous on (H-D,) it is surely so on (F .) Moreover,

[

A-F=(A-(H-D)) cup(H-D-F);

]

so

[

m(A-F) leq m(A-(H-D))+m(H-D-F)]

This completes the proof. (square)

*Lemma (PageIndex{3})

Given ([a, b] subset E^{1}) and disjoint closed sets (A, B subseteq(S, ho),) there always is a continuous map (g : S ightarrow[a, b]) such that (g=a) on (A) and (g=b).

**Proof**If (A=emptyset) or (B=emptyset,) set (g=b) or (g=a) on all of (S).

If, however, (A) and (B) are both nonempty, set

[

g(x)=a+frac{(b-a) ho(x, A)}{ ho(x, A)+ ho(x, B)}.

]

As (A) is closed, ( ho(x, A)=0) iff (x in A) (Problem 15 in Chapter 3, §14); similarly for (B .) Thus ( ho(x, A)+ ho(x, B) eq 0).

Also, (g=a) on (A, g=b) on (B,) and (a leq g leq b) on (S).

For continuity, see Chapter 4, §8, Example ((mathrm{e}) .) (square)

*Lemma (PageIndex{4})

(Tietze). If (f :(S,
ho)
ightarrow E^{*}) is relatively continuous on a closed set (F subseteq S,) there is a function (g : S
ightarrow E^{*}) such that (g=f) on (F),

[

inf g[S]=inf f[F], sup g[S]=sup f[F],

]

and (g) is continuous on all of (S).

(We assume (E^{*}) metrized as in Problem 5 of Chapter 3, §11. If (|f|

**Proof Outline**First, assume inf (f[F]=0) and (sup f[F]=1 .) Set

[

A=Fleft(f leq frac{1}{3} ight)=F cap f^{-1}left[left[0, frac{1}{3} ight] ight]

]

and

[

B=Fleft(f geq frac{2}{3} ight)=F cap f^{-1}left[left[frac{2}{3}, 1 ight] ight].

]

As (F) is closed (i n S,) so are (A) and (B) by Lemma (2 .) (Why? ())

As (B cap A=emptyset,) Lemma 3 yields a continuous map (g_{1} : S ightarrowleft[0, frac{1}{3} ight],) with (g_{1}=0) on (A,) and (g_{1}=frac{1}{3}) on (B .) Set (f_{1}=f-g_{1}) on (F ;) so (left|f_{1} ight| leq frac{2}{3},) and (f_{1}) is continuous on (F .)

Applying the same steps to (f_{1}) (with suitable sets (A_{1}, B_{1} subseteq F ),) find a continuous map (g_{2},) with (0 leq g_{2} leq frac{2}{3} cdot frac{1}{3}) on (S .) Then (f_{2}=f_{1}-g_{2}) is continuous, and (0 leq f_{2} leqleft(frac{2}{3} ight)^{2}) on (F).

Continuing, obtain two sequences (left{g_{n} ight}) and (left{f_{n} ight}) of real functions such that each (g_{n}) is continuous on (S),

[

0 leq g_{n} leq frac{1}{3}left(frac{2}{3} ight)^{n-1},

]

and (f_{n}=f_{n-1}-g_{n}) is defined and continuous on (F,) with

[

0 leq f_{n} leqleft(frac{2}{3} ight)^{n}

]

there (left(f_{0}=f ight)).

We claim that

[

g=sum_{n=1}^{infty} g_{n}

]

is the desired map.

Indeed, the series converges uniformly on (S) (Theorem 3 of Chapter 4, §12).

As all (g_{n}) are continuous, so is (g) (Theorem 2 in Chapter 4, §12). Also,

[

left|f-sum_{k=1}^{n} g_{k} ight| leqleft(frac{2}{3} ight)^{n} ightarrow 0

]

on (F( ext { why? }) ;) so (f=g) on (F .) Moreover,

[

0 leq g_{1} leq g leq sum_{n=1}^{infty} frac{1}{3}left(frac{2}{3} ight)^{n}=1 ext { on } S.

]

Hence inf (g[S]=0) and (sup g[S]=1,) as required.

Now assume

[

inf f[F]=a^{]Set[h(x)=frac{f(x)-a}{b-a}]so that inf (h[F]=0) and (sup h[F]=1 .) (Why?)As shown above, there is a continuous map (g_{0}) on (S,) with[g_{0}=h=frac{f-a}{b-a}]on (F,) inf (g_{0}[S]=0,) and (sup g_{0}[S]=1 .) Set[a+(b-a) g_{0}=g.]Then (g) is the required function. (Verify!)Finally, if (a, b in E^{*}(a}

*Theorem (PageIndex{3})

(Fréchet). Let (m : mathcal{M}
ightarrow E^{*}) be a strongly regular measure in ((S,
ho) .) If (f : S
ightarrow E^{*}left(E^{n}, C^{n}
ight)) is (m) -measurable on (A,) then

[

f=lim _{i
ightarrow infty} f_{i}(a cdot e .(m)) ext { on } A

]

for some sequence of maps (f_{i}) continuous on (S .) (We assume (E^{*}) to be metrized as in Lemma 4.)

**Proof**We consider (f : S ightarrow E^{*}) (the other cases reduce to (E^{1}) via components).

Taking (varepsilon=frac{1}{i}(i=1,2, ldots)) in Theorem (2,) we obtain for each (i) a closed (mathcal{M})-set (F_{i} subseteq A) such that

[

mleft(A-F_{i} ight)]

and (f) is relatively continuous on each (F_{i} .) We may assume that (F_{i} subseteq F_{i+1}) (if not, replace (F_{i}) by (igcup_{k=1}^{i} F_{k} )).

Now, Lemma 4 yields for each (i) a continuous map (f_{i} : S ightarrow E^{*}) such that (f_{i}=f) on (F_{i} .) We complete the proof by showing that (f_{i} ightarrow f) (pointwise) on the set

[

B=igcup_{i=1}^{infty} F_{i}

]

and that (m(A-B)=0).

Indeed, fix any (x in B .) Then (x in F_{i}) for some (i=i_{0},) hence also for (i>i_{0}) (since (left{F_{i} ight} uparrow ) .) As (f_{i}=f) on (F_{i},) we have

[

left(forall i>i_{0} ight) quad f_{i}(x)=f(x),

]

and so (f_{i}(x) ightarrow f(x)) for (x in B .) As (F_{i} subseteq B,) we get

[

m(A-B) leq mleft(A-F_{i} ight)]

for all (i .) Hence (m(A-B)=0,) and all is proved. (square)