---
tag: 1P9ZR606JN
kind: statement
role: theorem
name: Pettis measurability theorem
msc: [28B05, 46G10]
uses: [JDM2M1FRY6, 0M856Q1AMZ, 0GKJ78948V, YYNCJSR0PJ]
about: [JDM2M1FRY6]
sources:
  - url: https://en.wikipedia.org/wiki/Pettis_integral
---

## Statement

Let $(X, \Sigma)$ be a measurable space and let $E$ be a Banach space over $\R$ or $\C$ with [[dual space|0M856Q1AMZ]] $E^*$.
Let $f \colon X \to E$ be a function.
Then $f$ is [[strongly measurable|JDM2M1FRY6]] if and only if both of the following hold.

1. The function $\varphi \circ f$ is [[measurable|YYNCJSR0PJ]] for every continuous linear functional $\varphi \in E^*$, that is, $f$ is *weakly measurable*.
2. The image $f(X)$ is a [[separable|0GKJ78948V]] subset of $E$.

## 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.

## Explanations

### L1 human

Testing measurability of a vector-valued function directly is hard.
The theorem says that it is enough to test each linear functional, one at a time.
This works as long as the values of the function stay inside a separable part of the space.
So questions about a function in a large space reduce to questions about ordinary real or complex functions.

### L2 human

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.