# Existence of the Pettis integral

This is related to a question of MO: A question on the integral of Hilbert valued functions. I'm sure it's easy, but I cannot think right now, so I thought I'd ask.

Let $f:[0,1]\rightarrow H$ be a map to a Hilbert space, such that for each $x\in H$, the map $t\mapsto (f(t)|x)$ is Lebesgue measurable, and in $L^1([0,1])$. Thus we can define a map $$F:H\rightarrow\mathbb C; \quad F(x) = \int_0^1 (x|f(t)) \ dt$$ This is linear, but it's not obviously bounded. The MO question seems to suggest that it should be bounded; if so, then there is $y\in H$ with $\int_0^1 (x|f(t)) \ dt = (x|y)$ for all $x$, and we denote $y = \int_0^1 f(t) \ dt$ the Pettis integral.

What bothers me is that all the sources I can Google (for "Pettis integral") seem to assume that $y$ exists (or put strong conditions, e.g. continuity, on $f$). My guess is that a Hilbert space is sufficiently nice that I can somehow prove that $y$ exists (equivalently, that $F$ is continuous). This is obviously true if additionally $t\mapsto \|f(t)\|$ is measurable and in $L^1$, but does it hold more generally?

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"All the sources you can Google" probably are defining the Pettis integral for values in a Banach space, but (as we see in Theo's answer) unless the space is reflexive scalar integrability is not enough. –  GEdgar May 26 '11 at 12:45
@GEdgar: A question on terminology: Is "scalar integrability" the accepted term nowadays? I actually prefer this, as weakly integrable can be interpreted quite ambiguously. –  t.b. May 26 '11 at 14:16
I think the term "scalarly integrable" is well-known nowadays. As you note, "weakly integrable" has many possible meanings. Related term: scalarly measurable. –  GEdgar May 26 '11 at 21:39

This holds for an arbitrary reflexive Banach space $X$ (essentially by definition of the Pettis integral). This follows from the closed graph theorem (Baire to the rescue!). Let me first treat the Dunford integral on an arbitrary Banach space (it takes values in the bidual of $X$) and then specialize to the Pettis integral later on. Reflexivity is then only used to ensure that a weakly measurable function is actually Pettis integrable.

Added: The argument I'm presenting here is quite standard, but I should point out that I learned about it in R.A. Ryan's book Introduction to tensor products on Banach spaces.

So assume that $(\Omega,\mu)$ is a ($\sigma$-finite) measure space, that $f: \Omega \to X$ is weakly $\mu$-integrable, that is: for all $\phi \in X^{\ast}$ the function $\phi \circ f$ is ($\mu$-measurable and) integrable. We then have a linear map $T_{f}: X^{\ast} \to L^{1}(\Omega,\mu)$, which I claim to be bounded. So suppose that $\phi_{n} \to \phi$ in $X^{\ast}$ and $T_f(\phi_{n}) \to g$ in $L^{1}$. Passing to a subsequence, we may assume that $T_f(\phi_{n}) \to g$ a.e. On the other hand, $\phi_{n} \circ f \to \phi \circ f$ pointwise everywhere, so $\phi \circ f = g$ a.e., in other words $T_f \phi = g$ in $L^1$, so $T_{f}$ is indeed bounded by the closed graph theorem.

Now the adjoint operator $T^{\ast}: L^{\infty} \to X^{\ast\ast}$ satisfies for $h \in L^{\infty}$ $$\langle \phi, T_{f}^{\ast}h \rangle = \int_{\Omega} h \cdot T_f\phi = \int_{\Omega} h \cdot \phi \circ f,$$ where I used $\sigma$-finiteness in order to ensure that the duality $(L^{1})^{\ast} = L^{\infty}$ works nicely.

If $E \subset \Omega$ is measurable, then $T_{f}^{\ast} \chi_{E} =: \int_{E} f\,d\mu \in X^{\ast\ast}$ is called the Dunford integral of $f$ over $E$ and it is a tautology that every weakly $\mu$-integrable function $f: \Omega \to X$ has a Dunford integral.

In general, the function $f$ is called Pettis-integrable if $T_{f}^{\ast}\chi_{E} \in X \subset X^{\ast\ast}$ for every measurable $E \subset \Omega$. If $f$ is Pettis-integrable then $T_{f}^{\ast}$ actually defines an operator $S_{f}: L^{\infty} \to X$, as the characteristic functions span a norm-dense subspace of $L^{\infty}$ and $X$ is closed in $X^{\ast\ast}$. Now what you call the Pettis integral of $f$, is simply $S_{f} \chi_{\Omega} = T_{f}^{\ast} \chi_{\Omega} = \int_{\Omega} f\,d\mu \in X$.

Finally, if $X$ is reflexive, then every weakly $\mu$-integrable $f: \Omega \to X$ is Pettis integrable (by definition of reflexivity), and of course, putting $y = \int_\Omega f\,d\mu = T_{f}^{\ast}\chi_{\Omega} \in X$ we get the identity and existence you asked about for free: for $\phi \in X^{\ast}$ we have $$\langle \phi, y \rangle_{X^\ast, X} = \langle \phi, T_{f}^{\ast}\chi_{\Omega}\rangle_{X,X^{\ast\ast}} = \int_{\Omega} \chi_{\Omega} \cdot \phi \circ f\,d\mu = \int_{\Omega} \langle \phi, f(x)\rangle_{X^{\ast},X}\,d\mu$$ as we wanted.

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Yes, very nice. The weird thing is that somehow considering the map $E^*\rightarrow L^1$ (and not composing with $1\in L^\infty$ to give a map $E^*\rightarrow\mathbb C$) seems to make it easier, as then we can use that convergence in norm $L^1$ allows us to move to a.e. convergence. –  Matthew Daws May 26 '11 at 9:21
@Matt: Yes, it's a bit of magic, but you just wouldn't dare to do such a thing when speaking of scalar functions. If I see it correctly, you could appeal to weak$^{\ast}$-measurability to conclude norm continuity (at least in case $X$ is separable), but it seems more of a stretch and less natural (at least less familiar). –  t.b. May 26 '11 at 9:28