# Pettis integrability condition

Let $$B$$ be a Banach Space, $$(X,\mu)$$ a measure space and $$f:X\longrightarrow B$$ a weak measurable function, that is a function such that: $$\phi\circ f:X \longrightarrow \mathbb C$$ is measurable forall $$\phi\in B^*$$ (the topological dual).

If $$||f(x)||\leq M(x)$$ with $$M\in L^1$$, can we deduce that $$f$$ is Pettis integrable?

• If $B$ is separable and $f$ is strongly measurable $\big(f^{-1}(U)$ is measurable for every onen subset $U\subseteq B\big)$ then $f$ is actually Bochner integrable and hence also Pettis integrable.
– Ruy
Jan 8, 2021 at 14:44
• Thanks. I know that but I want to allow $B$ or $f(X)$ be non-separable. My motivation are the unitary group representations $p:G\longrightarrow U(H)$, (and I don't want to suppose $H$ to be separable). Jan 9, 2021 at 22:51
• I guess only representations of compact groups would satisfy your last condition, in which case they will be Bochner integrable.
– Ruy
Jan 10, 2021 at 0:13
• Do you mean that only the compact group representation are strongly measurable (Borel measurable and $p(G)$ dense)? Sorry, I didn't understand you Jan 10, 2021 at 0:30
• I meant the last condition, namely $\|f(x)\|\leq M(x)$, with $M$ in $L^1$. The point is that the total Haar measure of non-compact groups is infinite.
– Ruy
Jan 10, 2021 at 1:09

$$\newcommand{\bf}[1]{\mathbb #1}\newcommand{\sc}[1]{\mathscr #1}$$The answer is affirmative for separable Banach spaces.
In order to prove it, consider the linear functional $$\Lambda$$ defined on the (topological) dual space $$B'$$ by $$\Lambda (\varphi )=\int_X \varphi \big (f(x)\big )\, d\mu (x), \quad\forall \varphi \in B'.$$ Notice that the integrand satisfies $$|\varphi \big (f(x)\big )| \leq \|\varphi \| \|f(x)\| \leq \|\varphi \| M(x),$$ so $$|\Lambda (\varphi )|\leq \|\varphi \|\|M\|_1,\quad\forall \varphi \in B',$$ and hence $$\Lambda$$ is seen to be continuous. We next claim that $$\Lambda$$ is also continuous relative to the weak-star topology on $$B'$$.
Using V.12.8 in Conway's "A Course in Functional Analysis", a Corollary of Krein-Smulian's Theorem, it is enough to show that $$\Lambda$$ is weak-star sequentially continuous. So let us suppose that $$\{\varphi _n\}_n$$ is a sequence in $$B'$$ converging to some $$\varphi \in B'$$ relatively to weak-star topology.
If tollows that $$\varphi _n\circ f \to \varphi \circ f$$ pointwise everywhere on $$X$$. Moreover, by the Banach-Steinhauss Theorem, we have that the $$\varphi _n$$ are uniformly bounded, whence $$K := \sup_{n\in {\bf N}}\|\varphi _n\| <\infty ,$$ and hence $$|\varphi _n(f(x))| \leq \|\varphi _n\| \|f(x)\| \leq K M(x),\quad \forall x \in X.$$ We may then use Lebesgue's Dominated Convergence Theorem to conclude that $$\lim_{n\to \infty } \Lambda (\varphi _n) = \lim_{n\to \infty } \int_X \varphi _n(f(x))\, d\mu (x) =$$$$= \int_X \varphi (f(x))\, d\mu (x) = \Lambda (\varphi ),$$ hence concluding the proof of the weak-star continuity of $$\Lambda$$.
As it is well known, every weak-star continuous linear functional on $$B'$$ must be a point-evaluation, so there exists $$b\in B$$ such that $$\Lambda (\varphi )= \varphi (b),\quad\forall \varphi \in B'.$$ Consequently $$\int_X \varphi (f(x))\, d\mu (x) = \varphi (b),$$ so $$f$$ is Pettis integrable!