Is there a version of Fischer-Riesz theorem for Banach space?

$$( \Omega,F, P )$$: 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?"

• You may be interested also in this book springer.com/gp/book/9783540637455 – Tomek Kania Oct 25 '18 at 17:02
• 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. – Gerald Edgar Oct 25 '18 at 18:49
• 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. – Jochen Glueck Oct 25 '18 at 21:48

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

• This looks right. – Nik Weaver Oct 25 '18 at 23:20

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.

• 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). – Jochen Glueck Oct 25 '18 at 16:59
• 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. – Robert Furber Oct 26 '18 at 7:04
• @RobertFurber: How about now? – Nate Eldredge Oct 26 '18 at 13:44

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.