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?

  • 1
    $\begingroup$ 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. $\endgroup$
    – Ruy
    Jan 8, 2021 at 14:44
  • $\begingroup$ 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). $\endgroup$ Jan 9, 2021 at 22:51
  • $\begingroup$ I guess only representations of compact groups would satisfy your last condition, in which case they will be Bochner integrable. $\endgroup$
    – Ruy
    Jan 10, 2021 at 0:13
  • $\begingroup$ Do you mean that only the compact group representation are strongly measurable (Borel measurable and $p(G)$ dense)? Sorry, I didn't understand you $\endgroup$ Jan 10, 2021 at 0:30
  • $\begingroup$ 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. $\endgroup$
    – Ruy
    Jan 10, 2021 at 1:09

1 Answer 1


$\newcommand{\bf}[1]{\mathbb #1}\newcommand{\sc}[1]{\mathscr #1}$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!

Thanks to MathOverflow user @Mikael de la Sale for pointing me towards Krein-Smulian's Theorem!


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .