mathoverflow.net/questions/67434/…

Yes, the measurability of $f$ implies the existence of a product measurable $g$ whenever $X$ is an ideal space ($X=L_\infty(\Bbb R^d)$ is such a space). The converse may fail for non-regular ideal spaces (like $X=L_\infty(\Bbb R^d$): For product-measurable $g$ the function $f$ is not (strongly Bochner) measurable, in general.

I realized that my example is not correct without additional work: The boundary is normally not $C^1$ at $x=0$. It might be possible to save the example by rounding the corners in a very special way, but it needs more consideration.

Take a $C^1$ function on $(0,1]$ which has a continuous but not differentiable extension to $[0,1]$. Now let the manifold be the open “rounded” square (the corners rounded smoothly), and move its $z$ coordinate by that function w.r.t. the $x$-coordinate: The normal has a continuous extension to the boundary, the boundary is a $C^1$-curve, but the manifold is not smoother than continuous at the boundary

Thanks: So far, I was only aware that integration by parts and substitution rule hold for absolutely continuous (substitution) functions. Thanks for the link to the general case.

Nice argument, but the problem does not seem to be solved completely: For absolutely continuous $f$ with $f(0)=0$ and $f(1)=1$, it seems to be rather straightforward to see that the latter integral is $1/2$. Can this be shown for the general case as well? (In fact, it seems plausible for the devli's staircae, but I do not see a rigorous proof.)

Your main estimate is not uniform in $t$ (as you claim). In fact, the smaller $t>0$ is, the larger becomes the right-hand side (going to $\infty$ as $t\to0$). For this reason, it is not possible to conclude from this estimate that you can exchange the limits. Roughly speaking,your function $a$ must depend on $t$ to get a good estimate for $\lvert g(t)-T(a)(t)\rvert$ (because you have no uniform estimate), but if $a$ depends on $t$, you cannot estimate $\lvert T(a)(t)-h(a)\rvert$. In fact, my intuition is that the whole approach with Weierstraß does not work.

Another often-used sufficient condition for unbounded $\Omega$ is that $\lVert K(x)\rVert/\lVert x\rVert\to\infty$ as $\lVert x\rVert\to\infty$ (which implies that $(I-K)^{-1}(B)$ is bounded for bounded $B$).

A sufficient condition is of course that $K(\Omega)$ is relatively compact. If you use the usual definition of compact map (maps bounded sets into relatively compact sets), it is thus sufficient that $\Omega$ is bounded,

I do not really understand forcing, so I also do not understand whether "Martin's Maximum" has anything to do with what I called "maximal comprehension".

"large cardinals ... is not compatible with “minimal” ordinals.". This is not what I mean by minimality principle for ordinals. Instead I mean by it: In the same way as there are no “nonstandard” natural numbers in the "standard" $\mathbb N\cong\omega_0$, there should also be no “nonstandard” ordinals in any other limit ordinal in a "standard" model, so that all must be “standard” in a sense. I have no strong intuition whether or not this allows large cardinals, but I guess that this is independent.

"Saying that axiomatic NSA somehow "misses" $\omega_0$ is missing the point of axiomatic NSA.". That's a misunderstanding: I claimed that NSA "misses" the "standard copy" of $\omega_0$ ("standard copy" what it would be in Robinson's approach). Of course, there is some nonstandard entity $\omega_0$ in IST, but this is much larger - actually must be much larger, because it must contain nonstandard elements.

"There is a consensus among a majority of set theorists against assuming V=L". I did not mean this part of Elliott's reply (V=L clearly contradicts "maximal comprehension"): I meant the first part of the explanation, how non-standard natural numbers are ruled out by higher order logic. I assume that the same idea can be used for sets to make sure that $\omega_0$ is standard.

Concerning the collection of metalanguage integers and maximal comprehension: If I understand Jessie Elliott's comment correctly, it might be possible to formalize at least this special case in higher order logic. But even if it is not possible to formalize this special case or the general case, at least it can help to exclude some models (like IST) from having this property. Whether there actually are models with this property is probably hard to prove, though intuitively plausible (as I believe that this is what many mathematicians visualize when they think about sets).

I stand to the (admissibly sloppy) claim that any model of IST "misses" the "standard copy" $\omega_0$, that is, the collection of those elements of $\omega_0$ which are standard. Yes, this is not a set in the IST language, but this is exactly what I mean by the sloppy formulation that IST "misses" it. And I do not agree that $\omega_0$ has the same meaning as in ZFC, since most models of ZFC (in which $\omega_0$ is minimal) cannot be extended to a model of IST: You just cannot see the difference from inside the model which is something else than being the "same".