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$( \Omega,F, P )$: a measurable space equipped with a finite measure

$(B , \Vert \cdot \Vert) $ : a Banach space with $\mathcal{B}$ as its borelian $\sigma$-algebra

$p$ : a constant bigger than $1$

Define $L^p(\Omega, B )$ the vector space that contain all $( F, \mathcal{B})$-measurable function $f$ such that :

$ \vert \Vert f \Vert \vert = \sqrt[p]{ \int_{\Omega} \Vert f \Vert ^p \cdot dP } < \infty$

I'm looking for a version of Riesz-Fischer theorem which affirms that:

Proposition: $\left( L^p(\Omega, B ) , \vert \Vert \cdot \Vert \vert \right)$ is a Banach space

With some quick calculations, I have the feeling that this proposition is quite easy to be proved. But as we all know, it's always better to have a reliable reference.

So my question is: "Is the above proposition true? And does anyone have references to this matter?"

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    $\begingroup$ You may be interested also in this book springer.com/gp/book/9783540637455 $\endgroup$ – Tomek Kania Oct 25 '18 at 17:02
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    $\begingroup$ NO! Unless $B$ is separable, or $(\Omega, F, P)$ is a special sort of measurable space, this can fail. In general, if you restrict to functions with almost all values in a separable subspace of $B$, then you get the Bochner spaces, which are, indeed, complete. $\endgroup$ – Gerald Edgar Oct 25 '18 at 18:49
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    $\begingroup$ As the issue pointed out by @GeraldEdgar shows, the "right" definition of measurable vector-valued functions is not via the Boral $\sigma$-algebra on $B$ but via approximation by simple functions. See for instance Section 1.1 of the book by Hytönen et. al. quoted in my comment below. $\endgroup$ – Jochen Glueck Oct 25 '18 at 21:48
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With the definitions in the OP, this is false. It is OK if the Banach space $B$ is separable and $(\Omega,\mathcal F, P)$ is an arbitrary probability space. It is OK if the Banach space $B$ is arbitrary and $(\Omega,\mathcal F,P)$ is a perfect measure space. But for arbitrary $B$ and $(\Omega,\mathcal F, P)$, it can fail. It can fail in many different ways.

(A theorem of Charles Stegall: if $(\Omega,\mathcal F,P)$ is a perfect probability space, $B$ is a metric space, and $f : \Omega \to B$ is $(\mathcal F, \mathcal B)$-measurable, then there is a set $\Omega_1 \subseteq \Omega$ of measure $1$ such that $f(\Omega_1)$ is separable.)

Here is the simplest way in which it may fail. Write $\mathcal B = \mathrm{Borel}(B)$. Let $L^p(\Omega,B)$ be the set of all functions $f : \Omega \to B$ such that $f$ is $(\mathcal F, \mathcal B)$-measurable, and $$ \int_\Omega \|f(\omega)\|^p\;dP(\omega) < \infty . $$

It is possible that there are $f,g \in L^p(\Omega,B)$ such that $f+g \notin L^p(\Omega,B)$ because $f+g$ is not even $(\mathcal F , \mathcal B)$-measurable.

Example I
Let $T$ be a discrete space with cardinal $\frak{a} > 2^{\aleph_0}$. Let $B = l^2(T)$, that is, a Hilbert space with orthonormal basis of cardinal $\frak{a}$. For each $t \in T$ let $e_t \in l^2(T)$ be defined by: $e_t(t) = 1$ and $e_t(s) = 0$ if $t\ne s$. This system of "unit vectors" is an orthonormal basis of the space $B$.

Let $\Omega = T \times T$ be the Cartesian square. Let $\mathrm{Borel}(T)$ be the Borel sigma-algebra on $T$, which is of course the power set of $T$. Let the sigma-algebra $\mathcal{F} = \mathrm{Borel}(T) \otimes \mathrm{Borel}(T)$, the product sigma-algebra. The reason for requiring that $\mathrm{card}(T) > 2^{\aleph_0}$ is so that the diagonal $$ \Delta := \{(t,t) \in \Omega : t \in T\}, $$ although closed, is not in $\mathcal F$. See HERE.

We do not care what the probability measure $P$ is. (In an extreme case it could even be the point mass at a single point.)

Finally we are ready. Define $f : \Omega \to B$ by $$ f\big((u,v)\big) = e_u, $$ That is: Given $\omega = (u,v)$ in $\Omega$, we take its first component, and use the corresponding unit vector. Similarly, define $g : \Omega \to B$ by $$ g\big((u,v)\big) = -e_v, $$ using the second component and a minus sign.

I claim that $f, g \in L^p(\Omega,B)$ but $f+g$ is not.

First: $f$ is $(\mathcal F, \mathcal B )$-measurable. Indeed, if $Q \in B$ is Borel, then $f^{-1}(Q) \in \mathcal F$ because $f^{-1}(Q) = \widetilde{Q} \times T \in \mathcal F$ where $\widetilde{Q} = \{t \in T : e_t \in Q\}$. So $f$ is $(\mathcal F, \mathcal B )$-measurable. Similarly $g$ is $(\mathcal F, \mathcal B )$-measurable.

Next, $$ \int_\Omega \|f(\omega)\|^p\,dP(\omega) = 1 < \infty. $$ (Regardless of what the probability measure $P$ is, the integral of the constant $1$ is $1$.) So $f \in L_p(\Omega,B)$. Similarly, $g \in L_p(\Omega,B)$.

Now we claim the sum $f+g$ is not measurable. Indeed, even more, we claim that $\{\omega\in \Omega : f(\omega)+g(\omega) = 0\} \notin\mathcal F$. (Since $\{0\}$ is closed, this shows $f+g$ is not measurable.) Indeed, $$ \{\omega : f(\omega) + g(\omega) = 0\} = \{(u,v) : e_u-e_v = 0\} = \{(u,v) : u=v\} = \Delta. $$ As noted above, $\Delta \notin \mathcal F$

End of Example I

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  • $\begingroup$ This looks right. $\endgroup$ – Nik Weaver Oct 25 '18 at 23:20
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These are called Bochner spaces. Under mild assumptions (see Gerald's post), they are Banach spaces.

It is sufficient to assume that $B$ is separable, or that $L^p(\Omega, B)$ is defined to include only functions with almost every value in a separable subspace. Without some assumptions, it is possible that your $L^p(\Omega, B)$ is not even a vector space.

Given such assumptions, then at least one of the standard proofs that $L^p$ is complete goes through basically without change:

Let $f_n$ be Cauchy in this norm. Pass to a subsequence so that $|\|f_n - f_{n+1}\|| \le 4^{-n}$. By Chebyshev's inequality, we then have $P(\|f_n - f_{n+1}\| \ge 2^{-n}) \le 2^{-pn}$. Then the Borel-Cantelli lemma implies that for almost every $\omega \in \Omega$, we have $\|f_n(\omega) - f_{n+1}(\omega)\| \le 2^{-n}$ for all but finitely many $n$. In particular, for such $\omega$, the sequence $\{f_n(\omega)\}$ is Cauchy in $B$, so it converges to some $f(\omega) \in B$.

Now you have that $f$ is the a.e. limit of the $f_n$. Let $\epsilon > 0$. Since $f_n$ is Cauchy in $|\|\cdot\||$-norm, choose $N$ so large that $|\|f_n - f_m\|| < \epsilon$ for all $n,m > N$. Letting $m \to \infty$ and using Fatou's lemma on the integrals $\int_\Omega \|f_n - f_m\|\,dP$, conclude that $|\|f_n - f\|| < \epsilon$ as well. Thus the subsequence $f_n$ converges to $f$ in norm. Now use the Cauchy assumption one more time to see that the original sequence converges to $f$ as well.

I think that Evans's PDE book has some basic results about these spaces. There should be lots of other functional analysis texts that discuss them in more detail.

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    $\begingroup$ Just to add a concrete example of quite a recent reference: see for instance "T. Hytönen, J. van Neerven, M. Veraar, L. Weis: Analysis in Banach Spaces, Volume I (2016)". Bochner spaces are treated in Chapter I in a rather general setting (for instance, without assuming $\sigma$-finiteness in general). $\endgroup$ – Jochen Glueck Oct 25 '18 at 16:59
  • $\begingroup$ I think the first sentence still has the potential to mislead someone who doesn't already know Bochner spaces. Specifically, Bochner spaces can still be defined where $B$ is inseparable, but then part of the definition is that the range of each function lies in a separable subspace (but that subspace varies from function to function). The problem, as pointed out by Gerald Edgar, is that the OP does not add this condition. $\endgroup$ – Robert Furber Oct 26 '18 at 7:04
  • $\begingroup$ @RobertFurber: How about now? $\endgroup$ – Nate Eldredge Oct 26 '18 at 13:44
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A beautiful treatment of vector valued $L^p$ spaces is in the book:

J. Diestel, J. J. Uhl, Vector measures. Mathematical Surveys, No. 15. American Mathematical Society, Providence, R.I., 1977.

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