# 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 Oct 25, 2018 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. Oct 25, 2018 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. Oct 25, 2018 at 21:48
• One usually distinguishes between $\mathcal{L}^p(\Omega,B)$ (a space of measurable functions, what you denote $L^p(\Omega,B)$, and $L^p(\Omega,B)$, its quotient modulo a.e. equality of functions, thus its element are classes of functions, not functions $\Omega\to B$. Strictly speaking $\mathcal{L}^p(\Omega,B)$ fails to be a Banach space just because $\||\cdot\||$ is not a norm (just a seminorm), whenever $(\Omega, \mathcal{F}, P)$ has non-empty set of null measure. Jan 5, 2021 at 14:36

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. Oct 25, 2018 at 23:20
• A very nice example. Yet I don't see clearly what happens to the quotient of the space (which I would rather denote $\mathcal{L}_p$) of functions, modulo equivalence a.e. (i.e. the space of the classes of functions, which I would denote $L_p:=\mathcal{L}_p/\sim$ ). In the space $(\Omega,\mathcal{F},P)$ here, whatever is the probability $P$, the sigma-algebra $\mathcal{F}$ has uncountably many singletons, so some of them are null sets, and $L_p$ is not $\mathcal{L}_p$. Can we adapt the example to show $L^p(\Omega)$ (quotient) is not a vector space? Jan 5, 2021 at 14:41
• (Or is it just the fact f+g is not equal a.e. to any measurable function? ) Jan 5, 2021 at 15:02

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.

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). Oct 25, 2018 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. Oct 26, 2018 at 7:04
• @RobertFurber: How about now? Oct 26, 2018 at 13:44