1
$\begingroup$

I am now studying Bayesian inverse problems. In the note of Dashti and Stuart https://arxiv.org/abs/1302.6989, they mentioned that "... when considering a non-separable Banach space $B$, it is not clear what the ''natural" $\sigma$-algebra on $B$ is ...", where the usual Borel $\sigma$-algebra and the cylindrical $\sigma$-algebra are considered.

My questions are:

  1. What is the intuition behind the cylindrical $\sigma$-algebra? (More precisely, why do we need this $\sigma$-algebra?)
  2. Why do we need them to coincide (as in the case $B$ is a separable Banach space)? Is it a problem if we just consider the usual $\sigma$-algebra on a Banach space $B$ to study Bayesian inverse problems?

Thank you very much.

$\endgroup$
1
  • $\begingroup$ Do you have a response to the answer below? $\endgroup$ Commented Jan 31, 2022 at 19:13

1 Answer 1

2
$\begingroup$

$\newcommand\B{\mathscr B}\newcommand\C{\mathscr C}$There is hardly any particular intuition behind the concept of the cylindrical $\sigma$-algebra. This is just the smallest $\sigma$-algebra with respect to which all continuous linear functionals (which are arguably the simplest functionals) are measurable.

So, the cylindrical $\sigma$-algebra (say $\C$) appears to be the smallest $\sigma$-algebra of natural interest. It is therefore great to have a situation when this smallest $\sigma$-algebra is already big enough to coincide with the bigger Borel $\sigma$-algebra (say $\B$) generated by the strong topology; in particular, this happens when the Banach space $B$ is separable.

However, $\C$ may be strictly smaller than $\B$; see e.g. this answer.

It may be easier to first define a probability measure on the smallest $\sigma$-algebra of interest, $\C$ -- this can be done just via (the characteristic functional and) the Kolmogorov extension theorem. (In the usual setting of the Kolmogorov extension theorem, the cylindrical $\sigma$-algebra is naturally generated by the so-called "finite-dimensional" measurable sets.)

More work will then be needed to extend the probability measure from $\C$ to $\B$ -- so that one be able to define, in particular, the probability measure of balls.

$\endgroup$

You must log in to answer this question.

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