How does professor Amann conclude that "Corollary 1.5 remains true for incomplete measure spaces"?

Question

I'm reading Remark 1.6 at page 64 Amann's Analysis III. You can click on the link to see this page from Google book. The remark states that "Corollary 1.5 remains true for incomplete measure spaces", i.e.,

1.3 Remarks

• (a) Exercise IX.1.6 shows that the set of $\mathcal{A}$-measurable functions coincides with the set of $\mathcal{A}$-$\mathcal{B}(E)$-measurable functions.
• (b) Every subspace of a separable metric space is separable. Proof By Proposition IX.1.8, separability amounts to having a countable basis. But by restriction, a basis of a topological space yields a basis (of no greater cardinality) for any given subspace; see Proposition III.2.26.
• (c) Suppose $E$ is separable and $f \in E^{X}$. Then $f$ is $\mu$-almost separable valued. Proof This follows from (b).
• (d) Every finite-dimensional normed vector space is separable. ${ }^{5}$

1.4 Theorem A function in $E^{X}$ is $\mu$-measurable if and only if it is $\mathcal{A}$-measurable and $\mu$-almost separable valued.

1.5 Corollary Suppose $E$ is separable and $f \in E^{X}$. The following statements are equivalent:

• (i) $f$ is $\mu$-measurable.
• (ii) $f$ is $\mathcal{A}$-measurable.
• (iii) $f^{-1}(\mathcal{S}) \subset \mathcal{A}$ for some $\mathcal{S} \subset \mathfrak{P}(E)$ such that $\mathcal{A}_{\sigma}(\mathcal{S})=\mathcal{B}(E)$.
• (iv) $f^{-1}(\mathcal{S}) \subset \mathcal{A}$ for any $\mathcal{S} \subset \mathfrak{P}(E)$ such that $\mathcal{A}_{\sigma}(\mathcal{S})=\mathcal{B}(E)$.

Proof This follows from Theorem 1.4, Remark 1.3(c), and Exercise IX.1.6.

1.6 Remark The proof of Theorem 1.4 and Remark 1.3(c) show that Corollary 1.5 remains true for incomplete measure spaces.

My question: In the proof of direction $\implies$ in Theorem 1.4, i.e., $f \in E^{X}$ is $\mu$-measurable implies $f$ is $\mathcal{A}$-measurable, the author uses the completeness of the measure space as follows

"...Furthermore, the completeness of $\mu$ shows that $f^{-1}(O) \cap N$ is a $\mu$-null set..."

Could you explain how 1.5(i) implies 1.5(ii) in the case of non-complete measure space?

I attach all related definitions and the proof of Theorem 1.4 below for reference and clarity.

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Let $(X, \mathcal{A}, \mu)$ be a complete, $\sigma$-finite measure space; $(E,|\cdot|)$ a Banach space.

• Suppose $\mathrm{E}$ is a property that is either true or false of each point in $X$. We say that $\mathrm{E}$ holds ${\mu}$-almost everywhere, or for ${\mu}$-almost every $x \in X$, if there exists a $\mu$-null set $N \in \mathcal A$ such that $\mathrm{E}(x)$ is true for every $x \in N^{c}$. "Almost every" and "almost everywhere" are both abbreviated "a.e.".
• We say $f \in E^{X}$ is ${\mu}$-simple if $f(X)$ is finite, $f^{-1}(e) \in \mathcal{A}$ for every $e \in E$, and $\mu\left(f^{-1}(E \backslash\{0\})\right)<\infty$. We denote by $\mathcal{S}(X, \mu, E)$ the set of all $\mu$-simple functions from $X$ to $E$.
• A function $f \in E^{X}$ is said to be ${\mu}$-measurable if there is a sequence $\left(f_{j}\right)$ in $\mathcal{S}(X, \mu, E)$ such that $f_{j} \rightarrow f \mu$-almost everywhere as $j \rightarrow \infty$. We set $$ \mathcal{L}_{0}(X, \mu, E):=\left\{f \in E^{X} ; f \text { is } \mu \text {-measurable }\right\} . $$
• A function $f \in E^{X}$ is said to be $\mathcal{A}$-measurable if the inverse images of open sets of $E$ under $f$ are measurable, that is, if $f^{-1}\left(\mathcal{T}_{E}\right) \subset \mathcal{A}$, where $\mathcal{T}_{E}$ is the norm topology on $E$.
• If there is a $\mu$-null set $N \in \mathcal A$ such that $f\left(N^{c}\right)$ is separable, we say $f$ is ${\mu}$-almost separable valued.

Proof of Theorem 1.4.

"$\implies$" Suppose $f \in \mathcal{L}_{0}(X, \mu, E)$.

• (i) There exist a $\mu$-null set $N$ and a sequence $\left(\varphi_{j}\right)$ in $\mathcal{S}(X, \mu, E)$ such that $$ \varphi_{j}(x) \rightarrow f(x) \quad(j \rightarrow \infty) \quad \text { for } x \in N^{c} . \quad \quad (1.1) $$

By Proposition I.6.8, $F:=\bigcup_{j=0}^{\infty} \varphi_{j}(X)$ is countable and therefore the closure $\bar{F}$ is separable. Because of $(1.1)$ we have $f\left(N^{c}\right) \subset \bar{F}$. Remark 1.3(b) now shows that $f$ is $\mu$-almost separable valued.

• (ii) Let $O$ be open in $E$ and define $O_{n}:=\left\{y \in O\right.$; $\left.\operatorname{dist}\left(y, O^{c}\right)>1 / n\right\}$ for $n \in \mathbb{N}^{\times}$.

Then $O_{n}$ is open and $\bar{O}_{n} \subset O$. Also let $x \in N^{c}$. By (1.1), $f(x)$ belongs to $O$ if and only if there exist $n \in \mathbb{N}^{\times}$and $m=m(n) \in \mathbb{N}^{\times}$such that $\varphi_{j}(x) \in O_{n}$ for $j \geq m$. Therefore $$ f^{-1}(O) \cap N^{c}=\bigcup_{m, n \in \mathbb{N}^{\times}} \bigcap_{j \geq m} \varphi_{j}^{-1}\left(O_{n}\right) \cap N^{c} . \quad \quad (1.2) $$ But $\varphi_{j}^{-1}\left(O_{n}\right) \in \mathcal{A}$ for $n \in \mathbb{N}^{\times}$and $j \in \mathbb{N}$, because $\varphi_{j}$ is $\mu$-simple. Hence (1.2) says that $f^{-1}(O) \cap N^{c} \in \mathcal{A}$.

Furthermore, the completeness of $\mu$ shows that $f^{-1}(O) \cap N$ is a $\mu$-null set, and altogether we obtain $$ f^{-1}(O)=\left(f^{-1}(O) \cap N\right) \cup\left(f^{-1}(O) \cap N^{c}\right) \in \mathcal{A} . $$

"$\impliedby$" Suppose $f$ is $\mu$-almost separable valued and $\mathcal{A}$-measurable.

• (iii) We consider first the case $\mu(X)<\infty$.

Take $n \in \mathbb{N}$. By assumption, there is a $\mu$-null set $N$ such that $f\left(N^{c}\right)$ is separable. If $\left\{e_{j} ; j \in \mathbb{N}\right\}$ is a countable dense subset of $f\left(N^{c}\right)$, the collection $\left\{\mathbb{B}\left(e_{j}, 1 /(n+1)\right) ; j \in \mathbb{N}\right\}$ covers the set $f\left(N^{c}\right)$, and thus $$ X=N \cup \bigcup_{j \in \mathbb{N}} f^{-1}\left(\mathbb{B}\left(e_{j}, 1 /(n+1)\right)\right) . $$ Since $f$ is $\mathcal{A}$-measurable, $X_{j, n}:=f^{-1}\left(\mathbb{B}\left(e_{j}, 1 /(n+1)\right)\right)$ belongs to $\mathcal{A}$ for every $(j, n) \in \mathbb{N}^{2}$. The continuity of $\mu$ from below and the assumption $\mu(X)<\infty$ then imply that there are $m_{n} \in \mathbb{N}^{\times}$and $Y_{n} \in \mathcal{A}$ such that $$ \bigcup_{j=0}^{m_{n}} X_{j, n}=Y_{n}^{c} \quad \text { and } \quad \mu\left(Y_{n}\right)<\frac{1}{2^{n+1}} . $$ Now define $\varphi_{n} \in E^{X}$ through $$ \varphi_{n}(x):=\left\{\begin{array}{cl} e_{0} & \text { if } x \in X_{0, n}, \\ e_{j} & \text { if } x \in X_{j, n} \backslash \bigcup_{k=0}^{j-1} X_{k, n} \text { for } 1 \leq j \leq m_{n}, \\ 0 & \text { otherwise }. \end{array}\right. $$

Obviously $\varphi_{n} \in \mathcal{S}(X, \mu, E)$ for $n \in \mathbb{N}$, and $$ \left|\varphi_{n}(x)-f(x)\right|<1 /(n+1) \text { for } x \in Y_{n}^{c} . $$ The decreasing sequence $Z_{n}:=\bigcup_{k=0}^{\infty} Y_{n+k}$ satisfies $$ \mu\left(Z_{n}\right) \leq \sum_{k=0}^{\infty} \mu\left(Y_{n+k}\right) \leq \frac{1}{2^{n}} \quad \text { for } n \in \mathbb{N}. $$ It therefore follows from the continuity of $\mu$ from above that $Z:=\bigcap_{n \in \mathbb{N}} Z_{n}$ is $\mu$-null. We now set $$ \psi_{n}(x):=\left\{\begin{array}{cl} \varphi_{n}(x) & \text { if } x \in Z_{n}^{c}, \\ 0 & \text { if } x \in Z_{n}. \end{array}\right. $$ Then $\left(\psi_{n}\right)$ is a sequence in $\mathcal{S}(X, \mu, E)$. Also there is for every $x \in Z^{c}=\bigcup_{n} Z_{n}^{c}$ an $m \in \mathbb{N}$ such that $x \in Z_{m}^{c}$. Since $Z_{m}^{c} \subset Z_{n}^{c}$ for $n \geq m$, it follows that $$ \left|\psi_{n}(x)-f(x)\right|=\left|\varphi_{n}(x)-f(x)\right|<1 /(n+1) . $$ Altogether, $\lim \psi_{n}(x)=f(x)$ for every $x \in Z^{c}$. Therefore $f$ is $\mu$-measurable.

• (iv) Finally, we consider the case $\mu(X)=\infty$.

Remark IX.2.4(c) shows there is a disjoint sequence $\left(X_{j}\right)$ in $\mathcal{A}$ such that $\bigcup_{j} X_{j}=X$ and $\mu\left(X_{j}\right)<\infty$. By part (iii), there exist for each $j \in \mathbb{N}$ a sequence $\left(\varphi_{j, k}\right)_{k \in \mathbb{N}}$ in $\mathcal{S}(X, \mu, E)$ and a $\mu$-null set $N_{j}$ such that $\lim _{k} \varphi_{j, k}(x)=f(x)$ for every $x \in X_{j} \cap N_{j}^{c}$. With $N:=\bigcup_{j} N_{j}$ and $$ \varphi_{k}(x):=\left\{\begin{array}{cl} \varphi_{j, k}(x) & \text { if } x \in X_{j}, \quad j \in\{0, \ldots, k\}, \\ 0 & \text { if } x \notin \bigcup_{j=0}^{k} X_{j} \end{array}\right. $$ for $k \in \mathbb{N}$, we have $\varphi_{k} \in \mathcal{S}(X, \mu, E)$ and $\lim _{k} \varphi_{k}(x)=f(x)$ for $x \in N^{c}$. The result follows because $N$ is $\mu$-null.

Answer 1

Items (i) and (ii) are not equivalent.

Take $N \subset X$ non $\mathcal{A}$-measurable such that $\mu^*(N) = 0$. That is, $N$ is contained in some $M \in \mathcal{A}$ with $\mu(M) = 0$.

Then, in $M^c$, \begin{equation*} 0 \rightarrow I_N. \end{equation*} That is, $I_N$ is $\mu$-measurable, but not $\mathcal{A}$-measurable.

I suggest you try to understand the difference between $\mu$-measurable and measurable on the $\mu$-completion of $\mathcal{A}$. When are those concepts the same and when they are not? Which one is easier to handle and in what circumstances?