Let $(X, \mathcal A, \mu)$ be a measure space and $(E, | \cdot |)$ a Banach space. Let $f \in E^X$.
$f$ is called $\mu$-simple if $f = \sum_{k=1}^n e_k 1_{A_k}$ where $e_k \in E$ and $(A_k)_{k=1}^n$ is a finite sequence of pairwise disjoint sets with finite measure in $\mathcal A$. Let $\mathcal S (X, \mu, E)$ be the space of such $\mu$-simple functions.
$f$ is called $\mu$-measurable if $f$ is a $\mu$-a.e. limit of a sequence $(f_n)$ in $\mathcal S (X, \mu, E)$.
$f$ is called $\mathcal A$-measurable if $f^{-1}(O) \in \mathcal A$ for all open sets $O \subseteq E$.
$f$ is called $\mu$-almost separable valued if there is a $\mu$-null set $A$ such that $f(A^c)$ is separable.
There is a theorem relating different kinds of measurability notions, i.e.,
Theorem 1: Let $(X, \mathcal A, \mu)$ be complete and $\sigma$-finite. Then a function $f \in E^X$ is $\mu$-measurable if and only if $f$ is $\mathcal A$-measurable and $\mu$-almost separable valued.
The assumptions completeness and $\sigma$-finiteness are both essential in Theorem 1. At page 47 of Folland's Real Analysis: Modern Techniques and Their Applications, we encounter the following paragraph.
If $\mu$ is a measure on $(X, \mathcal{M})$, we may wish to except $\mu$-null sets from consideration in studying measurable functions. In this respect, life is a bit simpler if $\mu$ is complete.
Proposition 2.11 The following implications are valid iff the measure $\mu$ is complete:
- a. If $f$ is measurable and $f=g$ $\mu$-a.e., then $g$ is measurable.
- b. If $f_{n}$ is measurable for $n \in \mathbb{N}$ and $f_{n} \rightarrow f \mu$-a.e., then $f$ is measurable.
Previously, Folland defines measurability of a function as follows.
If $(X, \mathcal{M})$ and $(Y, \mathcal{N})$ are measurable spaces, a mapping $f: X \rightarrow Y$ is called $(\mathcal{M}, \mathcal{N})$-measurable, or just measurable when $\mathcal{M}$ and $\mathcal{N}$ are understood, if $f^{-1}(E) \in \mathcal{M}$ for all $E \in \mathcal{N}$.
Folland's Proposition 2.11 can be expressed in my notations of measurability as follows.
Proposition 2.11 The following implications are valid iff the measure $\mu$ is complete:
- a. If $f$ is $\mathcal{M}$-measurable and $f=g$ $\mu$-a.e., then $g$ is $\mathcal{M}$-measurable.
- b. If $f_{n}$ is $\mathcal{M}$-measurable for $n \in \mathbb{N}$ and $f_{n} \rightarrow f \mu$-a.e., then $f$ is $\mathcal{M}$-measurable.
A detailed proof of this proposition can be found here. On the other hand, I proved below related theorem.
Theorem 2: Let $(X, \mathcal A, \mu)$ be $\sigma$-finite, $(f_n)$ a sequence of $\mu$-measurable functions, and $f \in E^X$ the $\mu$-a.e. limit of $(f_n)$. Then $f$ is also $\mu$-measurable.
I'm surprised that different notions of measurability lead to strikingly different (if not opposite) results. In particular,
- Proposition 2.11 uses the measurability criterion that depends only on the $\sigma$-algebra. However, its conclusion depends completely on the completeness of $\mu$.
- Theorem 2 uses the measurability criterion that depends on both the $\sigma$-algebra and $\mu$. However, its conclusion is independent of the completeness of $\mu$.
Could you confirm if my understanding is correct?
Update: In this thread, with $\mu$-measurability notion I proved the Proposition 2.11.a without completeness assumption.
