Let $(\Omega, \Sigma)$ be a measurable space and $X$ a Banach space. Let $f: \Omega \rightarrow X$.

  • $f$ is called measurable if every the preimage of every Borel set in $X$ is an element of $\Sigma$.
  • $f$ is called strongly measurable if $f$ is the pointwise limit of a sequence of simple functions.

It is known that strongly measurable and measurable are equivalent when $X$ is separable. For this reason, the notion of strong measurability is only relevant when dealing with Bochner integration in full generality. What is an example of a function $f$ taking values in a non-separable Banach space $X$ such that $f$ is measurable but not strongly measurable?

  • 1
    $\begingroup$ Probably something like $F:[0,1]\to\ell^2([0,1]):x\mapsto\chi_{\{x\}}$. That one has isolated points only so should be measurable but same time countably many so should be nonseparable... By the way this is also the standard example for not Bochner integrable but sort of Riemann integrable. $\endgroup$ Aug 30, 2014 at 10:53
  • $\begingroup$ @Freeze_S: You can make this even more obvious (but somewhat artificial), by considering the counting measure (with the whole power set as sigma algebra) on the unit interval. $\endgroup$
    – PhoemueX
    Aug 30, 2014 at 11:36
  • $\begingroup$ @Freeze_S, by $\ell^2([0,1])$, what exactly do you mean? I'm assuming this is not the same as $L^2[0,1]$, since that would be separable. $\endgroup$ Aug 30, 2014 at 20:09
  • $\begingroup$ @ChristopherA.Wong: My favorite definition of $\ell^2(S)$ is as the space of square integrable families indexed by $S$. This definition highlights the fact that every Hilbert space $\mathcal{H}$ is actually isomorphic to these with $\dim\mathcal{H}=\# S$. In this manner they become the canonical Hilbert spaces and especially every separable Hilbert space is precisely isomorphic to those with countable index set. But now $[0,1]$ is uncountable hence $\ell^2([0,1])$ is nonseparable. $\endgroup$ Aug 31, 2014 at 1:14
  • $\begingroup$ @Freeze: your example is not measurable. If $Q\subset [0,1]$ is non-measurable, then $\{\chi_{\{x\}} : x \in Q\}$ is a closed set. $\endgroup$
    – GEdgar
    Aug 31, 2014 at 13:30

2 Answers 2


If you take $(\Omega,\Sigma)$ as $(X,\mathcal{B}(X))$, where $X$ is a non-separable Banach space, then the identity function $I:X\longmapsto X$ is not strongly measurable (but is continuous so is Borel measurable).


Let me try to show that there is no counterexample when $(\Omega, \Sigma) = ([0, 1], \Sigma)$ where $\Sigma$ is the class of Lebesgue measurable subsets of $[0, 1]$.

Claim: Suppose $X$ is a metric space and $f:[0, 1] \rightarrow X$ is measurable - This means preimages of Borel sets are Lebsgue measurable. Then, there is Lebesgue null set $N \subset [0, 1]$ such that the image of $f \upharpoonright [0, 1] \backslash N$ is separable.

Proof: Suppose not. Let $\{A_n : n \geq 1\}$ be a maximal family of pairwise disjoint positive measure subsets of $[0, 1]$ such that the image of $f \upharpoonright A_n$ is separable. Let $B = [0, 1] \backslash \bigcup_{n \geq 1} A_n$. If $B$ is null, we are done. So assume otherwise. WLOG, assume $B = [0, 1]$. Let $m$ be a Borel measure on $X$ defined by $m(E) = \mu(f^{-1}(E))$. Note that whenever $E$ is separable, $m(E) = 0$. Let $\mathcal{F}$ be the family of all open balls in $X$ which are $m$-null. Then, we claim that $U = \bigcup \mathcal{F}$ is $m$-null. Let us assume this and finish the proof. Let $Y = X \backslash U$. Then every open ball in $Y$ has positive $m$-measure. If $Y$ is not separable, then we can find uncountably many pairwise disjoint open balls in $Y$ each with positive $m$-measure which is impossible. Now, towards a contradiction, suppose $m(U) > 0$. Then the family $\mathcal{N} = \{f^{-1}[V]: V \in \mathcal{F}\}$, consists of pairwise disjoint Lebesgue null subsets of $[0, 1]$ whose union $W$ is not null and the union of every subfamily of $\mathcal{N}$ is Lebesgue measurable. Let $T \subset [0, 1]$ be a set of reals of size continuum which does not contain any perfect set (so you can take $T$ to be a Bernstein set). Let $h: W \rightarrow T$ be a function such that the set of preimages of points in $T$ is the family $\mathcal{N}$. Then $h$ is Lebesgue measurable. By Lusin's theorem, the restriction of $h$ to a positive measure compact subset $K \subseteq W$ is continuous. But $h[K]$ is an uncountable compact subset of $T$ which is impossible.


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