1

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 $0 \neq e_k \in E$ 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 $\mathcal S (X, \mu, E)$.

Several months ago, I posted a proof of below theorem

Theorem 1: Let $(X, \mathcal A, \mu)$ be complete and $\sigma$-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.

The proof of Theorem 1 relies on both assumptions completeness and $\sigma$-finiteness. Now I want to prove Theorem 2 which generalizes Theorem 1 by removing the requirements completeness.

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.

Could you have a check on my attempt?

My attempt: The proof relies on Lemma 2.

Lemma 2: 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.

Let $(f_n)$ be a sequence of $\mu$-measurable functions and $f \in E^X$ the $\mu$-a.e. limit of $(f_n)$. Let $(X, \mathcal A', \mu')$ be the completion of $(X, \mathcal A, \mu)$. Then $(X, \mathcal A', \mu')$ is complete and $\sigma$-finite. Because $\mathcal A \subset \mathcal A'$ and $\mu' \restriction \mathcal A = \mu$, we get

  • $(f_n)$ is a sequence of $\mu'$-measurable functions,
  • and $f \in E^X$ the $\mu'$-a.e. limit of $(f_n)$.

By Theorem 1, $f$ is $\mu'$-measurable. The claim then follows from Lemma 2. This completes the proof.

Akira
  • 18,439
  • 2
    Theorem 2 is not true. Take for example $(\mathbb R, \mathcal B(\mathbb R),m)$ where $m$ is the Lebesgue measure. Let $\mathcal N$ be a non Borel nul set. Then you have that $m-$a.e $f_n(x):=\frac{1}{n}\to \boldsymbol 1_{\mathcal N}(x)$, but $\boldsymbol 1_{\mathcal N}$ is not measurable. – Surb Aug 07 '22 at 14:00
  • @Surb I could not see how $f_n \to 1_\mathcal N$ $\mu$-a.e. Could you provide a Borel set $A \in \mathcal B (\mathbb R)$ such that $m(A) = 0$ and $f_n \to 1_\mathcal N$ on $A^c$. I also could not see how $1_\mathcal N$ is not $m$-measurable. Please notice that my definition of $\mu$-measurability is not the same as $(\mathcal A, \mathcal B (\mathbb R))$-measurability. – Akira Aug 07 '22 at 14:13
  • because $\boldsymbol 1_{\mathcal N}=0$ a.e. Therefore $\frac{1}{n}\to 0=\boldsymbol 1_{\mathcal N}$ a.e. Your definition of $\mu-$measurability doesn't work if the space is not complete. The example I provide previously is a good counter-example (at least you can easily adapt it to find a counter example w.r.t. your definition). – Surb Aug 07 '22 at 14:53
  • 1
    @Surb It seems to me that $1_{\mathcal N}$ is $m$-measurable in my definition. It follows from this thread that there is a Borel $m$-null set $N$ such that $\mathcal N \subset N$. Let $g_n := 1_N$ for all $n \in \mathbb N$. Then $g_n$ is $m$-simple and that $g_n \to 1_{\mathcal N}$ on $N^c$. Hence $g_n \to 1_{\mathcal N}$ $m$-a.e. Hence $1_{\mathcal N}$ is $m$-measurable. – Akira Aug 07 '22 at 15:18
  • 1
    Seems all correct, although I have to say that I didn't check your proof of the lemma. Which book (if any) are you using, BTW? – PhoemueX Aug 07 '22 at 15:52
  • Thank you so much @PhoemueX. The proof of Lemma 2 is actually very short and straightforward. I'm using Analysis III by Amann. Theorem 1 actually comes from the same book. – Akira Aug 07 '22 at 15:56
  • $\frac{1}{n}\to\boldsymbol 1_{\mathcal N}$ when $n\to \infty $$m-$a.e. (where $\mathcal N$ is a non borel nul set) but $\boldsymbol 1_{\mathcal N}$ is not Borel measurable... so, I don't see how your theorem 2 can be true without completness of the $\sigma -$algebra... – Surb Aug 07 '22 at 20:08
  • @Surb My definition of $\mu$-measurability does not require a function to be Borel measurable... On the other hand, being $\mu$-measurable and being Borel measurable are equivalent under some conditions. Please see this theorem. I'm pretty sure that you know it :v – Akira Aug 08 '22 at 00:01
  • You take at the very beginning an unspecified space $(X,\mathcal A,\mu)$. The example I took in my previous comment is $(X,\mathcal A,\mu)=(\mathbb R,\mathcal B(\mathbb R),m)$ for $m$ being the Lebesgue measure. Therefore, in this space, being measurable means being Borel measurable. So, no, your theorem 2 is wrong in general, at least if you don't have completeness of the $\sigma -$algebra. – Surb Aug 08 '22 at 04:08
  • @Surb In my definition of $\mu$-measurability, $f$ is called $\mu$-measurable if $f$ is a $\mu$-a.e. limit of a sequence $(f_n)$ of $\mu$-simple functions. I already defined what being a $\mu$-simple function means. As you can see, I do not require $f^{-1} (O) \in \mathcal A$ for $O \in \mathcal B(E)$. As such, I don't think your sentence "...being measurable means being Borel measurable..." makes sense. – Akira Aug 08 '22 at 04:14
  • That's exactly why your definition of being measurable is wrong without completeness. Your definition of being measurable only makes sense if the $\sigma -$algebra is complete. – Surb Aug 08 '22 at 04:42
  • @Surb 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 05:17
  • 1
    As written on the first link of your document : $(\Omega ,\Sigma,\mu)$ is a finite measure space, and thus, obviously $\sigma -$finite. Therefore, the definition of $f$ being measurable if it's the $\mu-$a.e. limit of a simple sequence $(f_n)$ works. Moreover, seeing again your theorem 2, you add the hypothesis of $\sigma -$finitude, and thus, as I said from the very beginning, the result holds. However, the $\sigma -$finitude was not mention at the moment where I saw your post, and this hypothesis was crucial... but now, indeed, things are completely correct. – Surb Aug 08 '22 at 12:21
  • @Surb Thank you so much for your elaboration. It seems to me that the $\sigma$-finiteness is already there from the beginning. You can check this by looking at the edit history. – Akira Aug 08 '22 at 15:39
  • @Surb I have just posted this question about how the completeness of a measure space affects the Borel measurability of a function. I hope that you can help me answer it. Thank you so much for your help! – Akira Aug 09 '22 at 09:27
  • 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 03:11
  • @Ramiro In Diestel-Uhl's Vector Measures, the definition of $\mu$-measurability applies to all functions, not just to their equivalence classes. Please see their definition here. – Akira Oct 07 '22 at 03:16
  • @Akira The issue here is just a confusion of terminology. The definition of $\mu$-measurability in Diestel-Uhl's Vector Measures, does not distinguish between functions that are a.e. equal. Also that definition is not equivalent to the standard definition of measurability used for real or complex valued measures. – Ramiro Oct 07 '22 at 03:26
  • @Ramiro Yess, I agree. There is a theorem connecting them... – Akira Oct 07 '22 at 03:31

1 Answers1

1

I will use a social argument to disprove your Lemma 2.

First, notice that being $\mu'$-measurable is just the same as being measurable in the completed $\sigma$=algebra, as you have stated in another post.

If Lemma 2 was true, that would kind of make the whole $\mu$-measurable concept a little useless as it would simply be the same as measurable over the completed $\sigma$-algebra. If Lemma 2 is true, I would reinforce my comment that this is a very horrible definition of $\mu$-measurable. :-)

André Caldas
  • 6,283
  • In contrast, I feel that Lemma 2 makes the definition of $\mu$-measurability more appealing. First, adding more null sets to your $\sigma$-algebra is meaningless, which is consistent with the fact that values of a function on a null set does not affect its integral. Second, the space of "measurable" functions is enlarged and is strictly bigger than that of Borel measurable functions. Finally, it gives us some kind of invariance property :) – Akira Aug 09 '22 at 16:34
  • I would like to mention a previous comment by @Surb about my proof, i.e., "...but now, indeed, things are completely correct", to convince you that Lemma 2 is indeed correct. – Akira Aug 09 '22 at 16:37
  • @Akira , 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 03:12