a) The mapping $p_i:f\mapsto f \varphi_i$ is a projector of $\mathscr{L}^p(X, \mu)$ and defines by passing to the quotients a projector $\widetilde{p}_i$ of $L^p(X, \mu)$. The latter is an orthoprojector of $L^2(X, \mu)$ if $p= 2$;

b) The mapping $f\mapsto f|X_i$ induces an isometric isomorphism of the image of $p_i$ onto the space $\mathscr{L}^p(X_i, \mu_i)$ and defines by passing to the quotients an isometric isomorphism of the image of $\widetilde{p}_i$ onto $L^p(X_i, \mu_i)$, through which these two spaces are identified;

c) The sum of the spaces $L^p(X_i, \mu_i)$ is total in the space $L^p(X, \mu)$. In the case $p= 2$, the space $L^2(X, \mu)$ is the Hilbert sum of the spaces $L^2(X_i, \mu_i)$.

Assertion a) is elementary. Assertion b) results from the scholium of INT, V, p. 84, § 7, n$^o1$.

Let $q\in ]1,+\infty ]$ be the conjugate exponent of $p$. We identify the dual of $L^p(X, \mu)$ with $L^q(X, \mu)$ (INT, V, p. 61, § 5, n$^o8$, th. 4). Let $f$ be a function in $\mathscr{L}^q(X, \mu)$ whose class $\widetilde{f}$ in $L^q(X, \mu)$ satisfies $\langle \widetilde{f} ,\widetilde{p}_i(\varphi )\rangle = 0$ for all $i\in I$ and all $\varphi \in L^p(X, \mu)$. Since the image of $p_i$ contains $\mathscr{K}(X_i)$, it follows that the measure $(f|X_i)\cdot \mu_i$ on $X_i$ is zero, which means that the restriction of $f$ to $X_i$ is locally $\mu_i$-negligible on $X_i$ (INT, V, p. 46, § 5, n$^o3$, cor. 2). Since the complement in X of the union of the $X_i$ is locally $\mu$-negligible, we conclude that the function $f$ is locally $\mu$-negligible on X (INT, IV, p. 190, § 5, n$^o9$ and p. 172, n$^o2$, prop. 5). The class of $f$ is then zero in $L^q(X, \mu)$ (using INT, V, p. 8, § 1, n$^o3$, lemma 1 and the corollary of proposition 9 when $p\not = 1$). By the Hahn–Banach theorem (EVT, II, p. 46, cor. 1), the first part of the assertion c) follows.

If $p= 2$, the image of $p_i$ is orthogonal to that of $p_j$ for all $i\not =j$, since $X_i$ and $X_j$ are then disjoint, whence the last assertion.

Let $(U_n)_{n\in\mathbf{N}}$ be a sequence of relatively compact open subsets of X whose union is equal to X and which satisfy $U_n\subset U_{n+1}$ for all $n\in \mathbf{N}$ (TG, I, p. 68, prop. 15). For all $n\in \mathbf{N}$, the space $\mathscr{K}(X,\overline{U}_n)$ identifies with a closed subspace of the Banach space $\mathscr{C}(U_n)$ (INT, III, p. 40, § 1, n$^o1$); since this latter space is of countable type (TG, X, p. 25, corollary), the same is true of $\mathscr{K}(X,\overline{U}_n)$ (TG, IX, p. 19, cor., (i)). Let $\mathscr{F}_n$ be a countable dense subset of $\mathscr{K}(X,\overline{U}_n)$.

Let $f\in \mathscr{L}^p(X, \mu)$ and let $\varepsilon >0$. There exists an integer $n\in \mathbf{N}$ such that $\int_{X-U_n}|f|^p< \varepsilon /2$, and there exists $g\in \mathscr{F}_n$ such that $\int_{\overline{U}_n}|f-g|^p< \varepsilon /2$. The union of the classes in $L^p(X, \mu)$ of the elements of the sets $\mathscr{F}_n$ is therefore dense in $L^p(X, \mu)$, which concludes the proof (TG, IX, p. 18, prop. 12).

Let $Z =\overset{-1}{g}(Y$ - S) be the set in question.

First suppose that X is compact and $g$ continuous. In this case, the image measure $g(\mu)$ is defined (INT, V, p. 69, § 6, n$^o1$, def. 1) since $\mu$ is a bounded measure. It follows from the definitions that the $\mu$-essential image of $g$ is the support of the measure $g(\mu) ($cf. INT, V, p. 70, § 6, n$^o2$, cor. 2), whence $\mu(Z) =g(\mu)(Y$ - S) = 0 (INT, IV, p. 118, § 2, n$^o2$, prop. 5).

Now consider the general case. Let C be a compact subset of X, and let $\varepsilon >0$ be a real number. Since $g$ is measurable, there exists a compact subset $C_1\subset C$ such that $\mu(C$- $C_1)\leqslant \varepsilon$ and such that $g|C_1$ is continuous (INT, IV, p. 169, § 5, n$^o1$, prop. 1). One then has

$\mu(Z\cap C)\leqslant \mu(C$ - $C_1) +\mu(Z\cap C_1)\leqslant \varepsilon +\mu(Z\cap C_1)$.

Let $\mu_1$ be the measure induced by $\mu$ on $C_1$ (INT, IV, p. 186, § 5, n$^o7$, def. 4). Let $S_1$ be the $\mu_1$-essential image of $g|C_1$. One has $Z\cap C_1\subset$ $(g|C_1)^{-1}(Y$ - $S_1)$. By the first case, the set $Z\cap C_1$ is therefore $\mu_1$-negligible, and consequently $\mu$-negligible (INT, IV, p. 187, § 5, n$^o7$, lemma 2, (i)). The above inequality becomes $\mu(Z\cap C)\leqslant \varepsilon$; since $\varepsilon$ and C are arbitrary, the set Z is locally $\mu$-negligible (INT, IV, p. 172, § 5, n$^o2$, prop. 5).

Let Y be the support of $\mu$. If $z\in \mathbf{C}$ is not adherent to $g$(Y), there exists an open neighbourhood U of $z$ such that the open set $\overset{-1}{g}(U)$ does not meet Y and is therefore locally $\mu$-negligible; this means that $z$ does not belong to the $\mu$-essential image of $g$.

Conversely, if $z\in \mathbf{C}$ is adherent to $g$(Y), then for every open neighbourhood U of $z$, the set $\overset{-1}{g}(U)$ is an open set in X which meets Y. It is therefore not $\mu$-negligible, and since it is open, it is not locally $\mu$-negligible (INT, IV, p. 172, § 5, n$^o2$, Cor. 2). Hence $z$ belongs to the $\mu$-essential image of $g$.

Put $h(x) = 1/g(x)$ if $g(x)\not = 0$ and $h(x) = 0$ otherwise. The function $h$ belongs to $\mathscr{L}^{\infty}(X, \mu)$. Let $\widetilde{h}$ be the class of $h$ in $L^{\infty}(X, \mu)$. Then we have the formula $\delta^{-1}=\|\widetilde{h}\|_{\infty}$.

Let U be an open neighbourhood of 0 such that the open set $Z =\overset{-1}{g}(U)$ is locally $\mu$-negligible, hence negligible (INT, IV, p. 172, § 5, n$^o2$, Cor. 2). Let Y be the support of $\mu$; we have $Y\subset X$ - Z. The restriction of the function $h$ to X - Z is continuous and bounded; therefore $h\in \mathscr{L}^{\infty}(X, \mu)$ and the norm of $\widetilde{h}$ in $\mathscr{L}^{\infty}(X, \mu)$ is equal to the norm of its restriction to X - Z. Moreover, for every $\alpha \in \mathbf{R}_+$, the set of $x\in X$ - Z such that $|h(x)|> \alpha$ is an open set in X - Z; therefore it is locally $\mu$-negligible if and only if it does not meet Y (INT, IV, loc. cit. and INT, III, p. 66, § 3, n$^o2$, Def. 1). Consequently, we have

$\|\widetilde{h}\|_{\infty}=$ sup$_{x\in Y}\frac{1}{|g(x)|}$,

whence

$\|\widetilde{h}1\|=$ inf$_{\infty x\in Y}|g(x)|=$ inf$_{\lambda\in g(Y)}|\lambda |=$ inf$_{\lambda\in\overline{g(Y)}}|\lambda |$

which is equal to $\delta$ by the preceding lemma.

Let $\mu$ be a positive measure with compact support on X and let C be its support. The restriction of $f$ to C is $(\mu|$C)-proper. The mapping $g$ is measurable with respect to the image measure $(f|C)(\mu|C)$, hence the mapping $(g\circ f)|C =g\circ (f|C)$ is $(\mu|$C)-measurable. Consequently, the mapping $g\circ f$ is $\mu$-measurable. The lemma follows.

We denote by $\mathscr{L}_u(X)$ the vector space of complex-valued functions which are universally measurable on X, and by $\mathscr{L}_u^{\infty}(X)$ the subspace of functions $f\in \mathscr{L}_u(X)$ which are bounded on X. For $f\in \mathscr{L}_u^{\infty}$(X), we put $\|f\|_{\infty}=$ sup$_{x\in X}|f(x)|$.

b) The set $\mathscr{L}_u^{\infty}(X)$ is a subalgebra of $\mathscr{L}_u(X)$ and the mapping defined by $f\mapsto \|f\|_{\infty}$ is a norm on $\mathscr{L}_u^{\infty}(X)$ for which $\mathscr{L}_u^{\infty}(X)$ is a unital stellar algebra;

c) The algebra $\mathscr{L}_u(X)$ contains $\mathscr{C}(X)$ and $\mathscr{L}_u^{\infty}(X)$ contains $\mathscr{C}_b(X)$;

d) Every Borel function from X into $\mathbf{C}$ (TG, IX, p. 61, def. 5) belongs to $\mathscr{L}_u(X)$;

e) For every sequence $(f_n)_{n\in\mathbf{N}}$ in $\mathscr{L}_u(X)$ which converges simply to a function $f$ from X into $\mathbf{C}$, one has $f\in \mathscr{L}_u(X)$.

Assertions a) and c) follow from the definitions (cf. INT, IV, p. 175, § 5, No$^o3$, cor. 3). Assertion d) is a consequence of INT, IV, p. 179, § 5, No$^o5$, th. 4, since the inverse image under a Borel function of every Borel subset of $\mathbf{C}$ is a Borel subset of X, which is universally measurable (INT, V, p. 28, § 3, No$^o4$). Finally, assertion e) follows from Egoroff's theorem (INT, IV, p. 175, § 5, No$^o4$, th. 2).

To prove assertion b), it is enough to remark that e) implies, a fortiori, that a uniform limit of universally measurable functions is universally measurable.

There may exist universally measurable functions on X which are not Borel. In fact, if X is metrisable, the characteristic function of a Souslin subset of X is universally measurable (cf. INT, IV, p. 171, § 5, No$^o1$, cor. 2); now, in every uncountable Polish space, there exists a Souslin subset which is not Borel ("Souslin's theorem"; this follows from TG, IX, p. 120, exercise 8 and from the fact that for every uncountable Polish space X, there exists a Borel bijection $X\rightarrow \mathbf{N}^{\mathbf{N}}$ whose inverse is Borel).

Let $x\in X$ and let U be a relatively compact open neighbourhood of $x$. The set N = U - $(U\cap \overset{-1}{g}(S))$ is $(\mu|$U)-negligible. For every integer $n\geqslant 1$, let $V_n$ be an open set such that $N\subset V_n\subset U$ and $\mu(V_n-N)<1/n$ (INT, IV, p. 116, § 1, n$^o4$, prop. 19).

The restriction $\widetilde{g}$ of $g$ to U- $V_n$ is $\mu|(U$- $V_n$)-proper, and its image is contained in S. Since $f$ is universally measurable, the mapping $x\mapsto f(g(x))$ from U - $V_n$ into $\mathbf{C}$ is measurable with respect to the measure $\mu|(U$ - $V_n)$ (INT, V, p. 71, § 6, n$^o2$, prop. 3). The mapping $h_n$ from X into $\mathbf{C}$ such that $h_n(x) = 0$ if $x \notin U$ - $V_n$ and $h_n(x) =f(g(x))$ if $x\in U$ - $V_n$ is therefore $\mu$-measurable (INT, IV, p. 193, § 5, n$^o10$, prop. 16).

Since $h_n(x)\rightarrow h(x)$ for $\mu$-almost all $x\in U$, the restriction of $h$ to U is $\mu$-measurable (INT, IV, p. 175, § 5, n$^o4$, th. 2). It follows that $h$ is $\mu$-measurable by the localisation principle (INT, IV, p. 171, § 4, n$^o2$, prop. 4).

Let us denote by $\widetilde{g}$ the class of $g$ in $L^{\infty}(X, \mu)$ and by S the $\mu$-essential image of $g$. Let $z\in \mathbf{C}-$ S. By definition, there exists an open neighbourhood U of $z$ such that $Y =\overset{-1}{g}(U)$ is locally $\mu$-negligible. The function $h$ defined by $h(x) = (g(x)-z)^{-1}$ if $x \notin Y$, and $h(x) = 0$ if $x\in Y$, then belong to $\mathscr{L}^{\infty}(X, \mu)$. Its class $\widetilde{h}$ in $L^{\infty}(X, \mu)$ satisfies $(\widetilde{g}-z)\widetilde{h}= 1$, since $(g(x)-z)h(x) = 1$ for every $x$ not belonging to the locally $\mu$-negligible set Y. Hence $z\in \mathbf{C}-$ Sp($\widetilde{g}$).

Conversely, let $z\in \mathbf{C}-$ Sp($\widetilde{g}$). Let $h\in \mathscr{L}^{\infty}(X, \mu)$ be a function whose class is the inverse of $\widetilde{g}-z$ in $L^{\infty}(X, \mu)$. There exists a real number $M>0$ such that $|h(x)|\leqslant M$ locally $\mu$-almost everywhere, and moreover one has $(g(x)-z)h(x) = 1$ locally $\mu$-almost everywhere. Let U be the open ball with center $z$ and radius $M^{-1}$ in $\mathbf{C}$; then $\overset{-1}{g}(U)$ is contained in the locally $\mu$-negligible set

therefore $z\in \mathbf{C}-$ S. The lemma is proved.

Let $f\in \mathscr{C}(S)$. The function $h=f\circ g$ is $\mu$-measurable (Lemma 5 of IV, p. 184) and bounded. Denote by $f\widetilde{\circ}g$ its class in $L^{\infty}(X, \mu)$. The mapping $f\mapsto f\widetilde{\circ}g$ is a continuous morphism of involutive algebras with unit of $\mathscr{C}(S)$ into $L^{\infty}(X, \mu)$. Since $g(x)$ belongs to S locally $\mu$-almost everywhere (Lemma 1 of IV, p. 181 and Lemma 6), this morphism associates with the identity mapping of S the class $\widetilde{g}$ of the function $g$. It therefore coincides with the functional calculus mapping of $\widetilde{g}$ (Prop. 7 of I, p. 111).

Let A denote the set of $x\in X$ such that $g(x) = 0$.

Suppose that A is locally $\mu$-negligible. Let $f\in \mathscr{L}^p(X, \mu)$ and $\widetilde{f}$ its class in $L^p(X, \mu)$. Suppose that $\widetilde{m}_g(\widetilde{f}) = 0$. By definition, this is the case if and only if the function $f g$ is $\mu$-negligible, so that the set B of $x\in X$ such that $f(x)g(x)\not = 0$ is $\mu$-negligible. The function $f$ is zero outside $A\cup B$, hence is $\mu$-negligible. This implies that the endomorphism $\widetilde{m}_g$ is injective.

Conversely, suppose that A is not locally $\mu$-negligible. Let C be a compact subset of X such that $A\cap C$ is not $\mu$-negligible, and let $\varphi$ be the characteristic function of $A\cap C$. The class of the function $\varphi$ in $L^p(X, \mu)$ is not zero, but that of $m_g(\varphi )$ is, hence $m_g$ is not injective.

Since the product of a locally $\mu$-negligible function and a $\mu$-negligible function is $\mu$-negligible, the endomorphism $\widetilde{m}_g$ depends only on the class of $g$ in $L^{\infty}(X, \mu)$. For $\widetilde{g}\in L^{\infty}(X, \mu)$, one also denotes by $\widetilde{m}_g$ the endomorphism $\widetilde{m}_g$ of $L^p(X, \mu)$ for every function $g\in \mathscr{L}^{\infty}(X, \mu\widetilde{)}$ whose class is $\widetilde{g}$. One says that $\widetilde{m}_g$ is the endomorphism of multiplication by $\widetilde{g}$ in $L^p(X, \mu)$.

For $a,b$ in $\mathbf{C}$ and $g_1,g_2$ in $L^{\infty}(X, \mu)$, one verifies at once that $\widetilde{m}_{ag_1+bg_2}=a\widetilde{m}g_1+b\widetilde{m}_{g_2}$, hence the mapping $\widetilde{m}$ is linear. Moreover, one has $\widetilde{m}_1= 1_{L^p(X,\mu)}$ and $\widetilde{m}_{g_1g_2}=\widetilde{m}_{g_1}\widetilde{m}_{g_2}$, hence the mapping $\widetilde{m}$ is a unital algebra morphism of $L^{\infty}(X, \mu)$ into $\mathscr{L}(L^p(X, \mu))$.

Formula (1) above proves that $\widetilde{m}$ has norm $\leqslant 1$.

Let $g\in \mathscr{L}^{\infty}(X, \mu)$ and $\widetilde{g}$ its class in $L^{\infty}(X, \mu)$. Let $\varepsilon >$ 0. The set Y of the $x\in X$ such that $|g(x)|>\|\widetilde{g}\|_{\infty}-\varepsilon$ is not locally $\mu$-negligible. There therefore exists a compact subset C of X such that $\mu(Y\cap C)>0$. Let $\varphi$ be the characteristic function of $Y\cap C$. Since

it follows that $\|\widetilde{m}_{\widetilde{g}}\|\geqslant \|\widetilde{g}\|_{\infty}-\varepsilon$. Since $\varepsilon$ is arbitrary, one deduces that $\|\widetilde{m}_{\widetilde{g}}\|\geqslant \|\widetilde{g}\|_{\infty}$, which proves that $\widetilde{m}$ is isometric.

Let $f\in \mathscr{L}^p(X, \mu)$. The sequence $(g_nf)$ is bounded in $\mathscr{L}^p(X, \mu)$ and converges pointwise $\mu$-almost everywhere to $gf$. By Lebesgue's theorem (INT, IV, p. 137, § 3, n$^o7$, th. 6), the sequence $(g_nf)$ converges to $gf$ in $\mathscr{L}^p(X, \mu)$, and the assertion follows.

We shall now consider the case where $p= 2$.

In particular, for every $g\in L^{\infty}(X, \mu)$, the multiplication endomorphism $\widetilde{m}_g$ is normal; it is hermitian if and only if $g$ is locally $\mu$-almost everywhere real-valued.

By Prop. 5, the mapping $\widetilde{m}$ is an injective isometric morphism of unital Banach algebras of $L^{\infty}(X, \mu)$ into $\mathscr{L}(L^2(X, \mu))$. Let $g\in \mathscr{L}^{\infty}(X, \mu)$. For $f_1$ and $f_2\in \mathscr{L}^2(X, \mu)$, one has

whence it follows that $\widetilde{m}^*_g=\widetilde{m}_{\overline{g}}$, which proves that $m$ is an involutive morphism. The final assertions follow from this (cf. I, p. 106, Prop. 5).

a) The spectrum of $\widetilde{m}_g$ in $\mathscr{L}(L^2(X, \mu))$ is the μ-essential image of $g$;

b) The endomorphism $\widetilde{m}_g$ is positive if and only if $g$ is locally μ-almost everywhere with positive values;

c) For every function $f\in \mathscr{C}$ (Sp($\widetilde{m}_g$)), one has $f(\widetilde{m}_g) =\widetilde{m}_{f\circ g}$.

Since $\widetilde{m}$ is injective, one has Sp($\widetilde{m}_g$) $=$ Sp($g$) by Prop. 5 of I, p. 106; the result then follows from Lemma 6 of IV, p. 185, from Proposition 4 of IV, p. 185 and from Definition 6 of I, p. 115.

It is enough to prove that if $u\in \mathscr{L}(L^2(X, \mu))$ is an endomorphism which commutes with $\widetilde{m}_g$ for every function $g\in \mathscr{L}^{\infty}(X, \mu)$, then $u$ belongs to the image of $\widetilde{m}$. Let $u$ be such an endomorphism.

We denote by $\widetilde{f}$ the class in $L^2(X, \mu)$ of a function $f\in \mathscr{L}^2(X, \mu)$. Let $v$ denote the linear mapping from $\mathscr{L}^2(X, \mu)$ into $L^2(X, \mu)$ defined by $f\mapsto u(\widetilde{f})$. One has $v\circ m_g=\widetilde{m}_g\circ v$ for every $g\in \mathscr{L}^{\infty}(X, \mu)$.

For every μ-measurable subset Y of X, let $\varphi_Y$ denote its characteristic function; the endomorphism $\widetilde{m}_{\varphi_Y}$ is an orthogonal projector of $L^2(X, \mu)$.

Suppose first that X is compact, so that the class of the constant function 1 belongs to $L^2(X, \mu)$. Let $g$ be a function in $\mathscr{L}^2(X, \mu)$ whose class in $L^2(X, \mu)$ is $v(1)$.

Let $c$ be a positive real number and Y the set of $x\in X$ such that $|g(x)|\geqslant c$; this is a μ-measurable set. One has in $L^2(X, \mu)$ the equalities $\varphi_{\widetilde{Y}}g=\widetilde{m}_{\varphi_Y}(v(1)) =v(m_{\varphi_Y}(1)) =v(\varphi_Y)$.

Since moreover $|c\varphi_Y|\leqslant |g\varphi_Y|$, one obtains

This inequality implies that $\mu(Y) = 0$ if $c >\|u\|$, so that the function $g$ is bounded μ-almost everywhere by $\|u\|$. Finally, for every function $f\in \mathscr{C}$(X), one has

whence the equality $u=\widetilde{m}_g$, and in particular $N_{\infty}(g) =\|u\|$.

Let us now consider the general case. There exists a locally countable family $(X_i)_{i\in I}$ of pairwise disjoint compact subsets of X such that Z = X $-\bigcup_{i\in I}X_i$ is locally $\mu$-negligible (INT, IV, p .190, § 5, n$^o9$, prop. 14). Let $\mu_i$ denote the measure induced by $\mu$ on $X_i$ (INT, IV, p. 186, § 5, n$^o7$, def. 4).

By prop. 1 of IV, p. 179, for every $i\in I$, the image $E_i$ of $\widetilde{m}_{\varphi_{Xi}}$ is identified with $L^2(X_i, \mu_i)$ by $f\mapsto f|X_i$. For every function $g\in \mathscr{L}^{\infty}(X, \mu)$, the multiplication endomorphism $\widetilde{m}_g$ commutes with $\widetilde{m}_{\varphi_{Xi}}$, hence leaves $E_i$ stable, and the endomorphism of $E_i$ deduced from $\widetilde{m}_g$ by passing to subspaces coincides with the multiplication endomorphism by $g|X_i$ on $L^2(X_i, \mu_i)$.

Since $u$ commutes with $m_{\varphi_{Xi}}$, it likewise leaves the subspace $E_i$ stable. Let $u_i$ denote the endomorphism of $L^2(X_i, \mu_i)$ deduced from $u$ by passing to subspaces.

The map from $\mathscr{L}^{\infty}(X, \mu)$ into $\mathscr{L}^{\infty}(X_i, \mu_i)$ defined by $g\mapsto g|X_i$ being surjective, the hypothesis implies that $u_i$ commutes with $\widetilde{m}_g$ for every function $g\in \mathscr{L}^{\infty}(X_i, \mu_i)$. By the first part of the proof, there exists a function $g_i\in \mathscr{L}^{\infty}(X_i, \mu_i)$ such that $u_i=\widetilde{m}_{g_i}$ and $N_{\infty}(g_i)\leqslant \|u_i\|\leqslant \|u\|$.

The function $g$ on X which coincides with $g_i$ on $X_i$ and which is zero on Z is bounded and $\mu$-measurable (INT, IV, p. 193, § 5, n$^o10$, prop. 16), hence $g\in \mathscr{L}^{\infty}(X, \mu)$. For every $i\in I$, the restriction of $u$ to $E_i$ coincides with that of $\widetilde{m}_g$; therefore one has $u=m_g$ by prop. 1, c) of IV, p. 179.