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I'm trying to prove that working with a completion of a measure space causes no harm to the strong measurability of a function.

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 $\mathcal S (X, \mu, E)$.

Theorem: Let $(X, \mathcal A', \mu')$ be the completion of $(X, \mathcal A, \mu)$. A function $f \in E^X$ is $\mu$-measurable if and only if $f$ is $\mu'$-measurable.

Could you have a check on my attempt?

My attempt: Let $\mathcal Z$ be the collection of all subsets of the $\mu$-null subsets of $X$.

Because $\mathcal A \subset \mathcal A'$ and $\mu' \restriction \mathcal A = \mu$, the direction $\implies$ then follows. Let's prove the reverse direction. Fix a $\mu'$-measurable function $f$. Then there is a sequence $(f_n)$ of $\mu'$-measurable functions such that $f_n \to f$ $\mu'$-a.e. Assume $f_n = \sum_{i=1}^{\varphi (n)} e_{ni} 1_{A_{ni}}$ with $(A_{ni})_{i=1}^{\varphi (n)}$ is a finite sequence of pairwise disjoint sets with finite measure in $\mathcal A'$. By construction of $\mathcal A'$, there exist $B_{ni} \in \mathcal A$ and $C_{ni} \in \mathcal Z$ such that $A_{ni} = B_{ni} \cup C_{ni}$ and $\mu(B_{ni}) = \mu'(A_{ni}) < \infty$.

Let $g_n := \sum_{i=1}^{\varphi (n)} e_{ni} 1_{B_{ni}}$. Then $(g_n)$ is a sequence of $\mu$-measurable functions. It remains to prove that $g_n \to f$ $\mu$-a.e. Let $N'$ be a $\mu'$-null set such that $f_n \to f$ on $(N')^c$. Notice that $\{x \in X \mid f_n (x) \neq g_n (x)\} \subset C_n :=\bigcup_{i=1}^{\varphi (n)} C_{ni}$. Let $N'' := N' \cup \bigcup_n C_n$. Then $N''$ is a $\mu'$-null set. It follows that $g_n \to f$ on $A :=(N'')^c$. There exist $B \in \mathcal A$ and $C \in \mathcal Z$ such that $A = B \cup C$ and $\mu(B) = \mu'(A) = 1$. It follows that $g_n \to f$ on $B$. This completes the proof.

Akira
  • 18,439
  • In a rigorous way, the concept of strong measurability doesn't apply to individual functions, but to equivalence classes of functions (defined by $=$a.e.). Strong measurability is different from measurability. If $(X, \mathcal A, \mu)$ is not complete and $(X, \mathcal A', \mu')$ is its completion, then there is $C \in \mathcal A' \setminus \mathcal A$. So the function $1_C$ is a $\mu'$-measurable, but it is not $\mu$-measurable. However, $1_C$ is $\mu'$-strong measurable and it is also $\mu$-strong measurable. In fact, there is $B \in \mathcal A$ such that $1_C=1_B$ a.e. – Ramiro Oct 07 '22 at 03:13

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