Pettis measurability theorem

Remarks

The direct implication is easy: $\varphi \circ s_n \to \varphi \circ f$ pointwise, and $f(X)$ lies in the closure of the countably many values of the approximating simple functions $s_n$. For the converse, one may replace $E$ by the closed linear span of $f(X)$, which is separable. In a separable Banach space there is a sequence $(\varphi_k)$ in $E^*$ with $\norm{e} = \sup_k \abs{\varphi_k(e)}$ for all $e$. So $x \mapsto \norm{f(x) - e}$ is measurable for each $e$, and one builds simple approximations by sending $x$ to the nearest point of a countable dense set among its first $n$ points.

When $E$ is separable, condition 2 is automatic, so weak and strong measurability coincide. Without condition 2 the equivalence fails. Take $X = [0, 1]$ with the Borel σ-algebra and the map $f(t) = \mathbf{1}_{\set{t}}$ into the non-separable Hilbert space $\ell^2([0, 1])$. It is weakly measurable, since $\varphi \circ f$ vanishes outside a countable set for each $\varphi$. But $f(X)$ is not separable, so $f$ is not strongly measurable.

There is also a version relative to a measure $\mu$. Then $f$ is strongly $\mu$-measurable if and only if it is weakly $\mu$-measurable and takes values in a separable subset of $E$ outside a $\mu$-null set.

Going deeper

For a student

The theorem is the main tool to check the hypotheses of the Bochner integral. With Bochner's criterion, it shows when $f$ is Bochner integrable. This holds as soon as $f$ is weakly measurable, essentially separably valued and $\int \norm{f} \, d\mu < \infty$. For a separable space $E$, weak measurability alone suffices. The theorem also explains why the Pettis integral, which only asks for weak measurability, covers more functions than the Bochner integral in non-separable spaces.