I wonder if we can characterize weak measurability of a function taking values in a Banach space using sequence of step functions (functions that have finite range) just like how we define strong measurability?

More specifically, a function $f:\Omega\mapsto X$ defined on a measure space $(\Omega,\Sigma,\mu)$ and taking values in a Banach space $X$ is strongly measurable if there exists a sequence of step functions $\{ \phi_n \}$ such that $\phi_n\rightarrow f$ in norm a.e.. Could we analogously say that $f$ is weakly measurable iff there exists a sequence of step functions $\{ \phi_n \}$ such that $\phi_n\rightarrow f$ weakly a.e.? One direction is obviously true, but I can't figure out the other direction.

For reference, here is the definition of weak measurability: A function $f:\Omega\mapsto X$ is weakly measurable if $\langle f(\omega), x \rangle$ is measurable for each $x\in X'$, the norm dual of $X$.


2 Answers 2


If there is a sequence of step functions such that $\phi_n\to f$ weakly a.e., then $f$ is almost separably valued. But if it is weakly measurable and almost separably valued, it is strongly measurable.

  • $\begingroup$ @ Michael Renardy: Yes, then $f$ is almost separably valued in the space $X$ equipped with the weak topology, but not necessarily with the norm topology which is needed to apply Pettis' theorem. While I was spending time to recover my account, Gerald Edgar posted his example which I too had in my mind. $\endgroup$
    – TaQ
    Jun 11, 2011 at 16:39
  • $\begingroup$ Weak and strong closure of a subspace are the same. This follows from the Hahn-Banach theorem $\endgroup$ Jun 11, 2011 at 17:42
  • $\begingroup$ @ Michael Renardy. Indeed, taking the vector subspace spanned by the countable weakly dense set $S$, then the set of rational linear combinations of points in $S$ is a countable dense set in the norm topology. Thanks for pointing out my error. $\endgroup$
    – TaQ
    Jun 11, 2011 at 21:24
  • $\begingroup$ @ Michael Renardy: Thank you for the great explanation! @ TaQ: Thank you for your comments, I was about to ask the same question! $\endgroup$
    – Rhymer
    Jun 13, 2011 at 6:45

Example 1: $f : [0,1] \to l^2[0,1]$ that is not almost separably valued: $f(x) = \delta_x$, the function equal to $1$ at $x$ and zero elsewhere. At least this one is scalarly equivalent to the constant zero.

Example 2: (page 672 of [1] where details are found) $f : [0,1] \to L^\infty[0,1]$ with $f(x) = 1_{[0,x]}$, the characteristic function of $[0,x]$. Then $f$ is scalarly measurable but not scalarly equivalent to a Bochner measurable function.

my references on measurability in Banach space:

[1] Indiana Univ Math J. 26 (1977) 663--667

[2] Indiana Univ Math J. 28 (1979) 559--579

Here I have used the terms "scalarly measurable" and "Bochner measurable" in place of weakly and strongly measurable.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.