3
$\begingroup$

Given a map from $\big([0,1], \mathcal{B}[0,1], m\big)$ to a Banach space $(X, \|\cdot \|)$. There are strong measurable functions (they are the point wise a.e. limit of simple functions) and weak measurable functions (for each $u^* \in X^*$, we have $t\mapsto \langle u^*, f(t)\rangle$ is a measurable as a function from $[0,1] \rightarrow \mathbb{R}$).

I have two questions:

  1. Why don't we use the normal measurable definition here? That is a function is measurable if the pre-image of all Borel sets $U\in \mathcal{B}(X)$ are in $\mathcal{B}[0,1]$.
  2. Now suppose $X= L^1(\Omega, \mathcal{F}, \mu)$, given a measurable $f:[0,1]\rightarrow L^1$ (either strong, weak or the normal definition), can we say that the function $f(t,\omega)$ is measurable as a function from $[0,1] \times \Omega \rightarrow \mathbb{R}$ with respect to the product sigma algebra on the domain?

Here, assume elements of $L^1$ are just measurable functions, not equivalent classes, so that $f(t,x)$ is well defined.

Thank you for your time.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
4
$\begingroup$

(1) If the Banach space $X$ is separable; and if you use the Lebesgue-measurable sets on $[0,1]$ not the Borel sets; then all three definitions are equivalent.

But of course the main thing of interest is not "measurable function" but "integrable function". When $X$ is not separable, you probably want the Bochner integral, using the definition pointwise a.e. limit of simple functions. There is also the Pettis integral, using weak measurability, but its properties are much worse than the Bochner integral.

For (2), it is clear with the Bochner definition of measurable: pointwise a.e. limit of simple functions. Again you want a complete measure (like Lebesgue) so that your sequnce can do weird things on a set of measure zero.

Plug:

Improve this answer
$\endgroup$
2
  • $\begingroup$ I think the claim that (2) is clear can't be quite right. Let $f : [0,1] \to L^1(\Omega)$ and call $r : [0,1] \times \Omega \to \mathbb{R}$ a representation of $f$ if for (almost) every $t \in [0,1]$, $f(t) = r(t,\cdot)$ a.e. (this is as uniquely as we can define things as $L^1(\Omega)$ is only defined up to a.e. equivalence). Then one can show there are non-measurable representations of $f$ even if $f(t) =0$ (say with $\Omega = (0,1)$). Your comment suggests to me that this shouldn't be possible, but I may be misinterpreting. $\endgroup$
    – Keefer Rowan
    Commented Jun 21, 2021 at 19:37
  • $\begingroup$ @KeeferRowan I think the intended interpretation is that there is at least one such measurable representation. Dear professor Edgar, could you confirm if my understanding is correct? $\endgroup$
    – Akira
    Commented Aug 9, 2023 at 21:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .