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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 $X$.

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up vote 4 down vote accepted

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.

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@ 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. –  TaQ Jun 11 '11 at 16:39
    
Weak and strong closure of a subspace are the same. This follows from the Hahn-Banach theorem –  Michael Renardy Jun 11 '11 at 17:42
    
@ 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. –  TaQ Jun 11 '11 at 21:24
    
@ Michael Renardy: Thank you for the great explanation! @ TaQ: Thank you for your comments, I was about to ask the same question! –  Rhymer Jun 13 '11 at 6:45
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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.

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