If this condition is fulfilled, we may define recursively a sequence of pairwise disjoint compact sets $K_n \subset K$ such that $|\mu|(K - \bigcup_{i=1}^n K_i) \leq 1/n$ and such that the restriction of $f$ to each $K_n$ is continuous ($§ 4$, No. 6, Th. 4); the complement with respect to $K$ of the union of the $K_n$ is then negligible ($§ 4$, No. 5, Cor. of Prop. 7), thus $f$ is measurable. Conversely, suppose there exist a negligible set $N \subset K$ and a partition $(K_n)$ of $K - N$ formed of compact sets such that the restriction of $f$ to each $K_n$ is continuous; for every $\varepsilon > 0$ there exists an integer $n$ such that, if $H = \bigcup_{i=1}^n K_i$, then $|\mu|(K - H) \leq \varepsilon$ (§ 4, No. 5, Cor. of Prop. 7); the set $H$ is compact, the $K_i$ ($1 \leq i \leq n$) form a finite partition of $H$ into compact sets, and the restriction of $f$ to each $K_i$ is continuous; therefore the restriction of $f$ to $H$ is continuous.

For each integer $n \geq 1$ there exists a compact set $K_n \subset K$ such that $|\mu|(K - K_n) \leq \varepsilon / 2^n$ and the restriction of $f_n$ to $K_n$ is continuous. The set $K_0 = \bigcap_{n=1}^\infty K_n$ is compact, the restrictions to $K_0$ of all of the functions $f_n$ are continuous and, since $K - K_0$ is contained in the union of the $K - K_n$, $|\mu|(K - K_0) \leq \sum_{n=1}^\infty \varepsilon / 2^n = \varepsilon$.

The condition is necessary, because the union of a sequence of integrable sets $A_n$ is integrable when $\sum_n |\mu|(A_n)$ is finite (§ 4, No. 5, Cor. of Prop. 8). The condition is sufficient because, for every integrable set $B$, there exist a negligible set $N \subset B$ and a partition of $B - N$ into a sequence of compact sets (§ 4, No. 6, Cor. 2 of Th. 4).

In particular, the entire space $X$ is measurable.

By virtue of Prop. 3, it suffices to verify that every relatively compact Souslin set A is μ-integrable. Now, such a set A is capacitable for $|\mu|^*$ (GT, IX, §6, No. 9, Th. 5). Therefore, for every $\varepsilon > 0$ there exists a compact subset K of A such that $|\mu|^*(A) \leq |\mu|^*(K) + \varepsilon = |\mu|(K) + \varepsilon$. Let U be a relatively compact open set in X containing A such that

Then $|\mu|^*(U - K) = |\mu|(U) - |\mu|(K) \leq 2\varepsilon$, therefore $|\mu|^*(A - K) \leq 2\varepsilon$, which proves that A is μ-integrable (§4, No. 6, Cor. 1 of Th. 4).

Let K be a compact set; there exists a finite number of points $x_i \in K$ such that the $V_{x_i}$ form a covering of K. It follows at once (§4, No. 9, Lemma) that there exist a negligible set $N \subset K$ and a finite partition of $K - N$ formed of integrable sets $M_j$ such that each of the sets $K \cap V_{x_i}$ is the union of a subset of N and a certain number of the $M_j$, and such that on each of the $M_j$, $f$ is equal to one of the functions $g_{x_i}$. Now, for each $M_j$ there exist a negligible set $N_j \subset M_j$ and a partition of $M_j - N_j$ formed by a sequence of compact sets $K_{nj}\ (\mathbf{n} \in \mathbf{N})$; on the other hand, for each $K_{nj}$ there exist a negligible set $P_{nj} \subset K_{nj}$ and a partition of $K_{nj} - P_{nj}$ formed by a sequence of compact sets $K_{mnj}\ (\mathbf{m} \in \mathbf{N})$ such that the restriction of $f$ to each of the $K_{mnj}$ is continuous. Since the union of N, the $N_j$ and the $P_{nj}$ is negligible, $f$ is measurable.

The concept of measurable function is therefore a concept of local character.

The condition is obviously sufficient since every point of X has a compact neighborhood. It is necessary, because if, for every $x \in K$, there exists a neighborhood $V_x$ of $x$ such that $A \cap V_x$ is negligible, then there exists a finite number of points $x_i \in K$ such that the $V_{x_i}$ form a covering of K, and $A \cap K$ is contained in the union of the negligible sets $A \cap V_{x_i}$.

The condition is obviously necessary. Conversely, if it is satisfied then A is contained in an integrable open set G, which is the union of a negligible set N and a sequence $(K_n)$ of compact sets (§ 4, No. 6, Cor. 2 of Th. 4); since $A \cap N$ and the sets $A \cap K_n$ are negligible, the same is true of their union A.

For, the outer measure of an open set G is the supremum of the measures $|\mu|(K)$ of the compact sets $K \subset G$ (§ 4, No. 6, Cor. 4 of Th. 4); if G is locally negligible then $|\mu|(K) = 0$ for every compact set K contained in G, therefore $|\mu|^*(G) = 0$.

Since X is the union of a sequence $(K_n)$ of compact sets, every locally negligible set A is the union of the negligible sets $A \cap K_n$, hence is negligible.

One can give examples of locally compact spaces that are not countable at infinity, and of measures on such a space X such that there exist sets in X that are locally negligible but not negligible (§ 1, Exer. 5).

For every compact set $K \subset X$, $K \cap N$ is negligible (Prop. 5), therefore, for every $\varepsilon > 0$, there exists a compact set $K_1 \subset K - (K \cap N)$ such that $|\mu|(K - K_1) \leq \varepsilon$ (§ 4, No. 6, Th. 4), and by hypothesis the restriction of f to $K_1$ is continuous, whence the conclusion.

If $P\{x\}$ is a property, the property « P\{x\} locally almost everywhere (with respect to $\mu$)» is by definition equivalent to the property «the set of x such that* (x $\in$ X and not $P\{x\}$) is locally negligible (for the measure $\mu$)». If F is any set, the relation «$f(x) = g(x)$ locally almost everywhere» is an equivalence relation in the set of mappings of X into F. In particular, if F is a vector space, a mapping $f$ of X into F such that f(x) = 0\text{ locally almost everywhere is said to be } \textit{locally negligible}. We leave to the reader the task of establishing for these concepts most of the properties corresponding to those that have been enumerated in §2, Nos. 4, 5 and 6 for functions equal almost everywhere. We shall limit ourselves to observing that if two \emph{continuous} mappings $f, g$ of X into a Hausdorff topological space F are equal \emph{locally almost everywhere}, then they are equal \emph{almost everywhere} by virtue of Cor. 2 of Prop. 5 (hence are equal at every point of the support of $\mu$ (§ 2, No. 4, Prop. 9)); however, we state explicitly the following proposition, which is an immediate consequence of the principle of localization:

For every compact subset K of X and every number $\varepsilon > 0$, there exists a compact set $K_1 \subset K$ such that $|\mu|(K - K_1) \leq \varepsilon$ and such that the restrictions to $K_1$ of all the functions $f_n$ are continuous (No. 1, Prop. 2); it is clear that $u \circ f$ is continuous on $K_1$, whence the theorem.

The set of measurable mappings of X into a topological vector space F is thus a vector space.

In particular, if f is a measurable mapping of X into a real (resp. complex) normed space F, and g is a measurable mapping of X into R (resp. C), then gf is measurable. If F is a normed algebra and f, g are two measurable mappings of X into F, then fg is measurable.

The first assertion obviously follows from the second, which we are going to prove. There exists a compact set $K_0 \subset K$ such that $|\mu|(K - K_0) \leq \varepsilon/2$ and such that the restrictions to $K_0$ of all the functions $f_\alpha$ are continuous (No. 1, Prop. 2). Let $(A_n)$ be a decreasing countable base for the filter $\mathfrak{F}$; let d be a metric on F compatible with the topology. For every pair of integers $n > 0,\ r > 0$, let $B_{n,r}$ be the set of points $x \in K_0$ such that, for at least one pair of indices $\alpha, \beta$ belonging to $A_n$, $d(f_\alpha(x), f_\beta(x)) \geq 1/r$; for fixed $\alpha$ and $\beta$, the set of $x \in K_0$ such that $d(f_\alpha(x), f_\beta(x)) \geq 1/r$ is closed in $K_0$, hence is compact; consequently, $B_{n,r}$ is a countable union of compact sets contained in $K_0$, hence is integrable (§ 4, No. 5, Props. 6 and 8). If $r$ is fixed, the intersection of the decreasing sequence of sets $B_{n,r}$ ($n = 1, 2, \ldots$) has measure zero, since $f_\alpha(x)$ tends to $f(x)$ almost everywhere in $K_0$ with respect to the filter $\mathfrak{F}$; thus $\lim_{n \to \infty} |\mu|(B_{n,r}) = 0$ (§ 4, No. 5, Cor. of Prop. 7), consequently there exists an integer $n_r$ such that $|\mu|(B_{n_r,r}) \leq \varepsilon/2^{r+2}$. Let $B$ be the union (for $r = 1, 2, \ldots$) of the sets $B_{n_r,r}$; $B$ is integrable and

(§ 4, No. 5, Cor. of Prop. 8). Let $C$ be the complement of $B$ in $K_0$; by construction, $f_\alpha(x)$ converges uniformly to $f(x)$ in $C$ with respect to the filter $\mathfrak{F}$, and since the restrictions of the $f_\alpha$ to $C$ are continuous, so is the restriction of $f$ to $C$. It then suffices to take a compact set $K_1 \subset C$ such that $|\mu|(C - K_1) \leq \varepsilon/4$ in order to satisfy the conditions of the statement, since $|\mu|(K - K_1) = |\mu|(K - K_0) + |\mu|(B) + |\mu|(C - K_1) \leq \varepsilon$.

The conclusions of Th. 2 do not necessarily hold if $F$ is not metrizable (Exer. 1). If $F$ is metrizable, and the set $A$ is not countable but the filter $\mathfrak{F}$ has a countable base, then the first conclusion of Th. 2 is again valid; for, if $(A_n)$ is a countable base for $\mathfrak{F}$ and $\alpha_n$ is an element of $A_n$, then $f$ is the limit of the sequence $(f_{\alpha_n})$ locally almost everywhere, hence is measurable; however, the second conclusion of Th. 2 is no longer necessarily valid (cf. Exer. 4).

For, the extended real line $\overline{\mathbf{R}}$, being homeomorphic to a compact interval of $\mathbf{R}$, is metrizable. The function $\sup_n f_n$ is the pointwise limit of the increasing sequence of functions $g_n = \sup(f_1, f_2, \ldots, f_n)$, which are measurable (No. 3, Cor. 1 of Th. 1); similarly, $\limsup_{n \to \infty} f_n$ is the pointwise limit of the decreasing sequence of functions $h_n = \sup_{p \geq 0} f_{n+p}$, each of which is measurable by the foregoing. Finally, since $\inf_n f_n = -\sup(-f_n)$ and $\liminf_{n \to \infty} f_n = -\limsup_{n \to \infty} (-f_n)$, these functions are measurable.

For, if M and N are measurable then the functions

are measurable by No. 3, Cors. 3 and 5 of Th. 1. If $(M_n)$ is a sequence of measurable sets and M is their union, then $\varphi_M = \sup_n \varphi_{M_n}$ is measurable by Cor. 1 of Th. 2, whence the corollary.

In particular, since the open sets are measurable:

The condition is sufficient by Egoroff’s theorem and the principle of localization. Let us show that it is necessary: by hypothesis, there exist a negligible set $N \subset K$ and a partition $(K_m)$ of $K - N$ formed of compact sets such that the restriction of $f$ to each $K_m$ is continuous. To define the sequence $(g_n)$, it suffices to proceed in the following manner: let $d$ be a metric compatible with the topology of F; for each $K_i$ with index $i \leq n$, there exists a finite partition of $K_i$ into integrable sets $A_{ij}$ ($1 \leq j \leq q_i$) sufficiently small that the oscillation of $f$ on each of the $A_{ij}$ is $\leq 1/n$ (§ 4, No. 9, Lemma); take $g_n$ to be constant on each $A_{ij}$, equal to one of the values of $f$ in this set (for $1 \leq i \leq n,\ 1 \leq j \leq q_i$), and equal to a fixed element $a \in \mathbf{F}$ for every point of $X$ that does not belong to any of the $A_{ij}$. It is clear that the sequence $(g_n(x))$ converges to $f(x)$ at every point of $K$ not belonging to $N$.

With notations as in the proof of Th. 3, and denoting by $a_{ij}$ one of the values of $f$ in $A_{ij}$, it suffices to take, as the value of $g_n$ in $A_{ij}$, the point 0 if $|a_{ij}| \leq 1/n$ and the point $a_{ij}(1 - 1/(n|a_{ij}|))$ otherwise; finally, set $g_n(x) = 0$ on the complement of the union of the $A_{ij}$ ($1 \leq i \leq n,\ 1 \leq j \leq q_i$).

If $X$ is the union of an increasing sequence $(A_n)$ of compact sets, the sets $A_n - A_{n-1}$ that are nonempty form a partition of $X$ into integrable sets; consequently, there exist a negligible set $N \subset X$ and a partition of $\mathbf{C}N$ formed by a sequence $(K_n)$ of compact sets such that the restriction of $f$ to each $K_n$ is continuous; the proof can then be concluded as in Th. 3 without modification.

It suffices to carry out the proof for the inverse image $^{-1}f(A)$ of a closed set $A$ in $F$. Let $K$ be a compact subset of $X$; there exist a negligible set $N \subset K$ and a partition $(K_n)$ of $K - N$ formed of compact sets such that the restriction of $f$ to each $K_n$ is continuous. The intersection $K_n \cap ^{-1}f(A)$ is thus the inverse image of the closed set $A$ under the restriction of $f$ to $K_n$; it is therefore a closed set in $K_n$, hence is compact. The set $K \cap ^{-1}f(A)$ is therefore the union of the negligible set $N \cap ^{-1}f(A)$ and the compact sets $K_n \cap ^{-1}f(A)$, which proves that $^{-1}f(A)$ is measurable.

The condition a) is necessary by Prop. 7; on the other hand, in the notations of Th. 3, condition b) is satisfied by taking H to be the countable set formed by the values of all of the functions $g_n$.

Let us now show that the conditions a) and b) are sufficient. Let K be any compact subset of X; there exists a negligible subset N of K such that $f(K - N)$ is contained in the closure of a countable set of points of F, which we arrange in a sequence $(a_n)$. Let $A_{n,p}$ be the set of $x \in K - N$ such that $d(f(x), a_n) \leq 1/p$. It follows from condition a) that $A_{n,p}$ is measurable. For fixed $p$, define a sequence of sets $B_{n,p} \subset K - N$ recursively by setting

the sets $B_{n,p}$ are measurable, and those that are nonempty form a partition of $K - N$. Let $g_{m,p}$ be the function equal to $a_i$ on the set $B_{i,p}$ for $1 \leq i \leq m$ and equal to a constant $b \in F$ on the complement of the union of these sets; $g_{m,p}$ is a measurable step function; as $m$ tends to infinity, $g_{m,p}$ converges pointwise to the function $f_p$ equal to $a_n$ on $B_{n,p}$ ($n \geq 1$) and to $b$ on $N \cup C K$, therefore (Th. 2) $f_p$ is measurable. As $p$ tends to infinity, $f_p(x)$ tends to $f(x)$ for every $x \in K - N$, and to $b$ for $x \in N \cup C K$; the limit of the $f_p$ is therefore measurable, and the principle of localization proves that $f$ itself is measurable.

For, if this is so, for every $b \in \overline{\mathbf{R}}$ the set of $x$ such that $f(x) \geqslant b$ is measurable, being the intersection of the sets (which form a countable family) of the $x$ such that $f(x) \geqslant a$, as $a$ runs over the set of points of D that are $\leqslant b$. The set of $x$ such that $f(x) < b$ is measurable, being the complement of a measurable set. Next, if $b$ is finite then the set of $x$ such that $f(x) \leqslant b$ is measurable, being the intersection of the sets of the $x$ such that $f(x) < b + 1/n$; and $f(-\infty)^{-1}$ is measurable, being the intersection of the sets of the $x$ such that $f(x) < n$, as $n$ runs over $\mathbf{Z}$. Finally, the inverse image under $f$ of every closed interval of $\overline{\mathbf{R}}$ is measurable, being the intersection of two measurable sets, and Th. 4 may be applied.

One could show similarly that it suffices that for every $a \in \mathrm{D}$, the set of $x$ such that $f(x) > a$ be measurable.

For, if $f$ is lower semi-continuous, then the set of $x \in \mathrm{X}$ such that $f(x) \leqslant a$ is closed for every $a \in \overline{\mathbf{R}}$.

When F is metrizable and compact, Prop. 7 makes it possible to sharpen Th. 3 as follows:

Let $d$ be a metric compatible with the topology of F. For every positive integer $n$ there exists a finite number of points $a_k \in \mathrm{F}$ such that the closed balls $B_k$ with center $a_k$ and radius $1/n$ form a covering of F; the sets $A_k = f^{-1}(B_k)$ are therefore measurable (Prop. 7) and form a covering of X. Consequently (§ 4, No. 9, Lemma) there exists a partition $(C_i)$ of X into a finite number of measurable sets such that each $A_k$ is the union of certain of the $C_i$. Let $c_i$ be a point of $C_i$ and let $g_n$ be the measurable step function equal to $f(c_i)$ on $C_i$ (for each index $i$). It is clear that $d(f(x), g_n(x)) \leqslant 2/n$ for all $x \in \mathrm{X}$.

The condition being obviously necessary (No. 3, Th. 1), let us prove that it is sufficient; it suffices to verify condition a) of Th. 4 and, to this end,

¹ See the footnote at the end of § 2, No. 4.

it will suffice by the principle of localization to prove that, for every compact subset K of X and every closed ball $B \subset F$, with center $a$ and radius $r$, the set $A = K \cap f^{-1}(B)$ is measurable; now, for every $z \in F$,

$A$ is thus the intersection of $K$ and the sets defined by

as these sets are measurable by hypothesis, $A$ is measurable.

a) for every $a' \in F'$, the scalar function $x \mapsto \langle f(x), a' \rangle$ is measurable;

b) for every compact set $K \subset X$, there exists a countable subset H of F such that $f(x) \in \overline{H}$ for almost every $x \in K$.

The necessity of the conditions follows from No. 3, Th. 1, and Th. 4. To prove that the conditions are sufficient, it again suffices to verify condition a) of Th. 4. With notations as in the proof of Prop. 10, we may (on account of b)) suppose, after modifying $f$ if necessary on a negligible set, that $f(K) \subset \overline{H}$, where $H$ is a countable subset of $F$. If $V$ is the closed linear subspace of $F$ generated by the set $H \cup \{a\}$, then $V$ is a separable Banach space and every continuous linear form on $V$ is the restriction of a form $a' \in F'$; the same reasoning as in Prop. 10 then shows that $K \cap f^{-1}(B)$ is measurable.

We may regard F as a subspace of a countable product $\prod_n E_n$ of Banach spaces (TVS, II, §4, No. 3), and we can suppose that $\operatorname{pr}_n(F)$ is dense in $E_n$, which is therefore separable. For every $n$, the mapping $\operatorname{pr}_n \circ f$ is then measurable by Prop. 10, therefore $f$ is measurable by No. 3, Th. 1.

The condition is necessary by No. 3, Th. 1; let us prove that it is sufficient. Suppose first that F is metrizable and separable, and let D be a countable dense set in F. Let $(V_n)$ be a decreasing fundamental sequence of balanced convex open neighborhoods of 0 in F; the polar sets $V_n^\circ$ are equicontinuous and their union is all of $F'$. Let $X_n = f^{-1}(V_n^\circ)$; the sequence $(X_n)$ is increasing and $X = \bigcup_n X_n$; let us show that each $X_n$ is $\mu$-measurable. Indeed, $D \cap V_n$ is dense in $V_n$; for every $y \in D \cap V_n$ let $S_y$ be the set of $x \in X$ such that $|\langle y, f(x) \rangle| \leq 1$; the hypothesis implies that each of the $S_y$ is measurable, and $X_n$ is the intersection of the countable family of $S_y$ for $y \in D \cap V_n$. This being so, for every compact subset K of X and every $\varepsilon > 0$, there exists an integer n such that $|\mu|(K - (K \cap X_n)) \leq \varepsilon/4$, and then a compact subset $K_1$ of $K \cap X_n$ such that $|\mu|((K \cap X_n) - K_1) \leq \varepsilon/4$; finally, there exists a compact subset $K_2$ of $K_1$ such that $|\mu|(K_1 - K_2) \leq \varepsilon/2$ and such that the restrictions to $K_2$ of all of the functions $\langle y, f \rangle$, where $y \in D$, are continuous (No. 1, Prop. 2). Since the set $f(K_2) \subset f(X_n) \subset V_n^\circ$ is equicontinuous, the topology induced by $\sigma(F', F)$ on $f(K_2)$ is identical to the topology of pointwise convergence in D (GT, X, §2, No. 4, Th. 1); the restriction of f to $K_2$ is therefore continuous, whence our assertion in this first case.

Let us pass to the general case. If $z'$ is a continuous linear form on F, its restriction $z'_n$ to $F_n$ is continuous; since $F = \bigcup_n F_n$, the dual $F'$ of F may be identified (algebraically) with a linear subspace of the product $\prod_n F'_n$, and then $\operatorname{pr}_n(z') = z'_n$. Moreover, since each finite subset of F is contained in one of the $F_n$, the topology $\sigma(F', F)$ is none other than the topology induced by the product topology of the topologies $\sigma(F'_n, F_n)$. This being so, if $\langle a, f \rangle$ is measurable for every $a \in F$, then so is $\langle a_n, \operatorname{pr}_n \circ f \rangle$ for every n and every $a_n \in F_n$, since $\langle a_n, \operatorname{pr}_n \circ f \rangle = \langle a_n, f \rangle$; the first part of the proof then shows that $\operatorname{pr}_n \circ f$ is measurable for every n, therefore so is f (No. 3, Th. 1).

The condition is necessary: for, if $f \in \mathscr{L}_F^p$ then there exists a sequence $(g_n)$ of continuous functions with compact support that converges almost everywhere to f ($§ 3$, No. 4, Cor. 2 of Th. 3); by Th. 2 of No. 4, f is measurable.

To prove that the conditions are sufficient, we first establish a lemma:

Let $A_n$ be the set of points $x \in X$ such that $|g(x)| \geq 1/n$; A is the union of the $A_n$, and $\varphi_{A_n} \leq n|g|$, whence $|\mu|^*(A_n) \leq (n N_p(g))^p$; it follows that $A_n$ is contained in the union of a negligible set and a sequence of compact sets (§ 4, No. 6, Cor. 3 of Th. 4), therefore so is A.

The lemma proved, consider first the case that f has compact support K. By Cor. 1 of Th. 3 of No. 5, there exists a sequence $(g_n)$ of measurable step functions such that $|g_n(x)| \leq |f(x)|$ at every point $x \in X$ and such that $g_n(x)$ tends to $f(x)$ almost everywhere. Now, $g_n$ is a linear combination of characteristic functions of measurable sets contained in K; since these sets are integrable by Prop. 3 of No. 1, $g_n$ belongs to $\mathscr{L}_F^p$. Since $N_p(f) < +\infty$, Lebesgue’s theorem (§ 3, No. 7, Th. 6) shows that f belongs to $\mathscr{L}_F^p$.

In the general case, it follows from Lemma 1 that there exists an increasing sequence $(K_n)$ of compact sets such that $f(x)$ is zero almost everywhere in the complement of the union of the $K_n$. Let $f_n$ be the function equal to $f(x)$ on $K_n$ and to 0 elsewhere; $f_n$ is measurable by No. 3, Cor. 5 of Th. 1; since $|f_n| \leq |f|$, $f_n$ belongs to $\mathscr{L}_F^p$ by the first part of the argument. Since $f(x)$ is equal almost everywhere to the limit of the sequence $f_n(x)$, Lebesgue’s theorem again proves that $f \in \mathscr{L}_F^p$, which completes the proof.

One should take care to note that a function that is locally negligible but not negligible is not integrable; thus, a function equal locally almost everywhere to an integrable function is not necessarily integrable.

For, g is measurable (No. 3, Th. 1).

For, it follows from Th. 5 and No. 3, Cor. 5 of Th. 1 that $f \varphi_A$ is measurable, and $N_1(f \varphi_A) \leq N_1(f)$.

We write $\int_A f d\mu = \int f \varphi_A d\mu$ (or $\int_A f \mu$) for every integrable function $f$ and every measurable set $A$. We also write $\int^* f d|\mu|$ (or $\int^*_A f |\mu|$) in place of $\int^* f \varphi_A d|\mu|$ for every numerical function (finite or not) $f \geqslant 0$ (on setting $f(x)\varphi_A(x) = 0$ if $f(x) = +\infty$ and $\varphi_A(x) = 0$).