TL;DR: Is there a version of the Bochner integral which allows for the integration of isometric embeddings $\phi:X\to H$ from a metric space to a Hilbert space, satisfying $\int_X \|\phi\| d\mu < \infty$ for finite Borel measures $\mu$?

I'm reading the article Distance covariance in metric spaces. The author considers (p. 9-11) an isometric embedding $\phi:(X,d^{1/2})\to H$, where $(X,d)$ is a metric space and $H$ is a Hilbert space. That is $\phi$ is an injective and continuous mapping satisfying $$ d(x,y) = \| \phi(x)-\phi(y)\|^2 \quad \quad \forall x,y\in X. $$ Then the author uses the following integral without further explanation $$ \int_X \phi(x) \,d\mu (x), $$ where $\mu$ is a finite signed Borel measure on $(X,\mathcal{B}(X))$ satisfying $$ \int_X d(x,o) \,d|\mu|(x)< \infty. $$ for some $o\in X$.

I have not encountered the integration of Hilbert space valued mappings before, but after further invesigation I stumbled upon the theory of integration of Banach space valued mappings by the Bochner integral. Hence we want to show that both $$ \int_X \phi(x) \,d\mu^- (x) \quad \text{ and } \quad \int_X \phi(x) \,d\mu^+ (x), $$ exists. In order for these integrals to exist the mapping $\phi$ needs to be $\mu^{\pm}$-Bochner integrable and satisfy $$ \int_X \|\phi(x)\| \, d \mu^{\pm}(x) < \infty. $$ Now it is easily verified by the triangle inequality that $$ \int_X \|\phi(x)\| \,d |\mu|(x) \leq \int d(x,o) \,d|\mu|(x) + \|\phi(o)\|\, |\mu|(X) < \infty, $$ so it suffices to show that $\phi$ is $\mu^{\pm}$-Bochner integrable which by the Pettis measurability theorem holds if $\phi$ is $\mu^{\pm}$-almost surely separably valued and $\mathcal{B}(X)/\mathcal{B}(H)$-measurable. Continuity of $\phi$ yields the Borel-measurability but as far as i can see there is no way to guarantee that an isometric embedding $\phi$ into a Hilbert space is $\mu^{\pm}$-almost surely separable valued for all finite signed Borel measures satisfying the above condition.

Hence the question: Are my findings correct, in that there is no way to guarantee $\mu^{\pm}$-Bochner measurability of $\phi$, and in that case is there an extension/version of the Bochner integral for Hilbert space valued mappings which does not require such strict measurability conditions?

The only requirement for the integral is that it should satisfy the condition $$ \int f\circ \phi \, d\mu = f \left( \int \phi \,d\mu \right), $$ for $f\in H^*$, that is $f:H\to\mathbb{R}$ is a continuous linear functional. Or at the very least $$ \left\langle \int \phi \, d\mu ,h \right\rangle = \int \langle \phi(x) ,h \rangle \,d\mu(x), $$ for any $h\in H$, since this is a property needed in the article.

  • $\begingroup$ Are you sure that your Hilbert space is not tacitly assumed to be separable? That's often the case $\endgroup$ Jul 20, 2016 at 15:05
  • $\begingroup$ Yes I'm pretty sure that the Hilbert space is not assumed to be separable. In fact the above $\phi$ is the mapping from the theorem "A metric space $(X,d)$ is a space of negative type if and only if there exists an isometric embedding $\phi:(X,d^{1/2})\to H$ where $H$ is a Hilbert space. This is a result is also found in Embeddings and Extensions in Analysis by Wells & Williams (§1-2), and here it is not assumed that either $X$ or $H$ is separable. $\endgroup$
    – John
    Jul 22, 2016 at 15:51
  • $\begingroup$ I also found this answer math.stackexchange.com/questions/941415/… . He almost states what i want, except he restricts $X=[0,1]$ and $\mu=\lambda$, and he does not provide any proof or reference to how he constructs the integral. $\endgroup$
    – John
    Jul 22, 2016 at 15:55

1 Answer 1


I found that the Pettis integral (see "Topics in Banach space integration" by Stefan Schwabik) was exactly what i neeeded. The Pettis integral does not need such strong measurability conditions of the integrand. In fact whenever the value-space is Hilbert one only needs that $\phi$ is scalarly $\mu^\pm$-integrable.


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