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

I'm trying to prove below result which does not assume the completeness of $\mu$. This is opposite to this exercise from Folland's Real Analysis: Modern Techniques and Their Application. The difference is due to that of measurability notion.

Theorem: Let $f,g \in E^X$ such that $f$ is $\mu$-measurable and $f=g$ $\mu$-a.e. Then $g$ is $\mu$-measurable.

Could you have a check on my attempt?

My attempt: Because $f$ is $\mu$-measurable, there is a sequence $(f_n)$ of $\mu$-simple functions such that $f_n \to f$ $\mu$-a.e. This means there is a $\mu$-null set $N$ such that $f_n \to f$ on $N^c$. Because $f=g$ $\mu$-a.e., there is a $\mu$-null set $N'$ such that $f=g$ on $(N')^c$. Let $M = N \cup N'$. Then $M$ is a $\mu$-null set. Clearly, $f_n \to g$ on $M^c$. This completes the proof.

Akira
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    @OP you have been asking questions about integration of $E$-valued functions. What is the reference you are using? – Snoop Aug 08 '22 at 05:39
  • it depends on the notion you are using here of "$\mu$-a.e.", as this generally means that there is a difference of a null set, where null here is respect to the completion of $\mathcal{A}$ – Masacroso Aug 08 '22 at 06:02
  • @Snoop My reference is Amann's Analysis III. – Akira Aug 08 '22 at 07:28
  • @Masacroso I use the same definition of "$\mu$-a.e." as in above textbook. Let $(X, \mathcal A, \mu)$ be a measure 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 $N$ such that $\mathrm{E}(x)$ is true for every $x \in N^{c}$. "Almost every" and "almost everywhere" are both abbreviated "a.e." – Akira Aug 08 '22 at 07:37
  • @Akira in this section of the book the measure space is assumed as complete, see page 62 – Masacroso Aug 08 '22 at 08:47
  • @Masacroso But it has nothing to do with my theorem. You can see that the authors do not appeal to completeness in his definitions of being $\mu$-simple, $\mu$-measurable, and $\mu$-integrable. They only use the completeness in this result. – Akira Aug 08 '22 at 08:52
  • @Masacroso My definition of $\mu$-measurability is exactly the same as in the standard reference Vector Measures by Diestel/Uhl. Please see here for a screenshot of the definition I take from this textbook. As you can see they define $\mu$-measurability for arbitrary finite measure space. – Akira Aug 08 '22 at 08:59

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