Mathematics - 2024-09-15 (page 1 of 2)

Joe

12:13 AM

@BenSteffan I think it's possible to turn this is into a rigorous proof, as follows. (It's quite late where I am right now, so no guarantees this is correct...) Let $\mathfrak p$ be a prime ideal of $A$ which does not meet $S$. Then, the extension of $\mathfrak p$ consists of the quotients $x/y$, where $x\in\mathfrak p$ and $y\in S$. If $\frac{a}{s}\frac{b}{t}\in\mathfrak p^e$, then $ab/1\in\mathfrak p^e$, so $ab/1=c/t$ for some $c\in\mathfrak p$ and $t\in S$.

Then, $tab-c$ is annihilated by an element of $S$, say $s$, so that $stab=sc$. Hence, $stab\in\mathfrak p$. Since $\mathfrak p$ does not meet $S$, it follows that $ab\in\mathfrak p$, so $a\in\mathfrak p$ or $b\in\mathfrak p$. If for example $a\in\mathfrak p$, then $a/1\in\mathfrak p$, hence $a/s\in\mathfrak p^e$.

Ben Steffan

looks good, but no guarantees from me either at this hour :)

Bml

@BenSteffan (and also @SineoftheTime) I posted a question on the main site with all the information. I hope my effort this time is sufficient so that, if you want, you can offer me help.

0

Q: How to solve (or verify) the following inequality?

The following question is directly related to this one and involves the resolution (or verification) of an inequality. Again, the first part of my question is dedicated to the creation of the (necessarily physical) context, the second part to the transition to mathematical language, since my goal...

Soham Saha

4:15 AM

Does anyone know what this shape is called? Kinda like two large intersecting tetrahedrons, or equivalently, 8 small tetrahedrons attached to the faces of an octahedron…

Jakobian

Stellated octahedron

The stellated octahedron is the only stellation of the octahedron. It is also called the stella octangula (Latin for "eight-pointed star"), a name given to it by Johannes Kepler in 1609, though it was known to earlier geometers. It was depicted in Pacioli's De Divina Proportione, 1509. It is the simplest of five regular polyhedral compounds, and the only regular compound of two tetrahedra. It is also the least dense of the regular polyhedral compounds, having a density of 2. It can be seen as a 3D extension of the hexagram: the hexagram is a two-dimensional shape formed from two overlappi...

Soham Saha

4:41 AM

@Jakobian Thanks a lot!

Sine of the Time

7:43 AM

@Bml +1 for the effort

Ryder is not Rude.

Can ChatGPT o1-preview Solve PhD-level Physics Textbook Problems? (Part 2)

►

Ryder is not Rude.

7:56 AM

PSQs preferred.

Soumik Mukherjee

8:50 AM

tl;dw no it can't

Hi Sine

Sine of the Time

Hi

Mats Granvik

Can someone remind me. If one Riemann zeta zero is off the line, then automatically its conjugate is also off the line, but in what rule/pattern for the real part?

Like if one zeta zero is 1/3+I*t then its conjugate is 2/3-I*t?
Or is it worse than that, like we get also 2/3+I*t and 1/3-I*t as well?

I saw this years ago in a youtube video by Giuseppe Mussardo but I could not find it now on youtube.

Sine of the Time

is $I$ the imaginary unit?

Mats Granvik

yes

Sine of the Time

the conjugate of $a+ib$ is $a-ib$

Ryder is not Rude.

9:02 AM

@SoumikMukherjee fwd to the last minute.

Mats Granvik

yes. But do we also get $1-a+ib$ and $1-a-ib$?

I mean if one zeta zero is off the line then we suddenly have 4 zeros in total off the line?

Sine of the Time

If I recall correclty, $\zeta(s)=0\iff \zeta(1-s)=0$ in the strip $0<\Re s < 1$

Mats Granvik

9:16 AM

You are right. I also found the youtube video now.

Palestra Especial: Generalised Riemann Hypothesis - Giuseppe Mussardo (2019)

►

Soumik Mukherjee

10:03 AM

@RyderisnotRude. watched, but that doesn't change the fact that it can't solve

Ryder is not Rude.

90% is a pretty good grade.

And a lot faster.

Soumik Mukherjee

Finding a solution on the internet is not same as solving

Ryder is not Rude.

Yes.

Mats Granvik

11:54 AM

Is it correct to say that repeated integration by parts is Abelian?

Ben Steffan

12:09 PM

probably not

the word you want is "commutative," I think

but even then it needs context

Mats Granvik

12:24 PM

Ok.

Mad

1:13 PM

I understand intutively why the middle inequality is true, however rigorosly, i can not seem to construct it. The Function $f^+ $ has a finite integral. And i know that the Integral is linear. So i thought i would write:
$ \int_X f^+ \chi_A d\mu \leq \int_X \infty \chi_A $ and then pull out infty , but in my notes, it is only possible to do this for finite scalars, not infty

so i am not sure how to replicate this inequality

Ben Steffan

pull out the other scalar, i.e. 0 :)

this inequality strikes me as... not a great way to write it

Mad

i do not understand what you mean? my question regards the middle inequality.

Ben Steffan

$\int_X f^+ \chi_A d\mu \leq \int_X f^+ \cdot 0 d\mu = 0 \int_X f^+ d \mu$

something like that

Mad

Das stimmt aber nicht, weil die charakteristische Funktion größer als 0 sein kann.

Schöne Grüße an Bonn.

Ben Steffan

Thank you, but let's stick to english.

but the assumption here is $\mu(A) = 0$

otherwise the last inequality is false

Mad

1:25 PM

Well yes, but i do not see how you can write the first inequality from this fact.

psie

I have a fundamental doubt. If $\lim_{x\to a}f(x)$ exists, i.e. the limit of a function at a point $a$ where $a$ is an accumulation/cluster/limit point, then $f$ is defined in a neighborhood of $a$, right?

Jakobian

No

psie

@Jakobian Hmm. Rudin states in his PMA that for $f^{(n)}$ to exist at $x$, $f^{(n-1)}$ has to exist in a neighborhood around $x$. The reason I was asking was because the derivative is a limit, and so I thought Rudin's claim stems from what I was claiming above.

Derso

@psie I guess it depend on your definition. I think it would make sense to ask $f$ to be defined around $a$ in advance in order for the limit to make sense at the first place

Jakobian

@Derso it would, but this is not necessary

Derso

1:36 PM

Alternatively, if you talk about induced topology on $f$'s domain... Well, then $f$ is defined on a neighborhood of that type almost by definition I guess

Jakobian

@psie well this isn't true

We can ask for existence of $f'$ regardless. The issue is what it would mean for $f$ to be continuously differentiable at $x$

I suppose I understand Rudin. How can you claim $f"$ exists at $x$ if $f'$ is undefined at some point in all neighbourhoods of $x$? You might think of $f'$ as a function which is $f'(x)$ when $x$ exists and, say, $\Omega$ when it doesn't, a symbol not representing any real number

Derivative is supposed to be taken of a function in this interpretation

@psie Still, this is just false

psie

ok, Rudin has a point then at least

Jakobian

Not if we think about one-sided derivatives

According to Rudin's definition, if $f:[0, 1]\to \mathbb{R}$ is a function then it can't be differentiable at $0, 1$

psie

1:53 PM

one-sided neighborhoods?

Jakobian

Rudin didn't need to assume that $f^{(n-1)}$ exists in a neighbourhood of $x$ in order to define $f^{(n)}$

Just that its a limit point

psie

ah ok

Jakobian

But there might be some intuitive reasons for why you want neighbourhood. Or perhaps just to make sense of a theorem

Its not useful to always be the most general

psie

true

Vladimir Lysikov

I'm looking at my copy of Rudin and technically he defines derivatives only for functions defined on a segment or an interval

And he also says this:

Derso

2:03 PM

Again, for me it's all on what "$\lim_{x\to a}$" means. This will say if $f$ must, by definition, be defined on a neighborhood or not.

Vladimir Lysikov

Limits he defines at a limit point of a subset of a metric space

Derso

In order for $\lim_{x\to a}$ even make sense at the first place (then the limit may exist or not)

Jakobian

@Derso you do want such limits for $\mathbb{Q}\subseteq\mathbb{R}$ I'd say

Derso

@VladimirLysikov Exactly, that's what I was trying to say. Because otherwise, you simply can't evaluate $f(x)$ if it isn't defined on $x$, right?

Vladimir Lysikov

Right

Derso

2:08 PM

So you use induced topology on $f$ domain, asking also $x$ to be an accumulation point

Otherwise if it is isolated then things get trivial (I guess)

Jakobian

We want deleted neighbourhood if anything

Vladimir Lysikov

yes, it makes sense for the limit to be defined in this generality

Derso

Sorry, also maybe I'm using the same letter $x$ for the limit point and points around it lol

$f$ surely doesn't need to be defined on the target limit point, but on a neighborhood around it yes (at least on a induced topology neighborhood on the domain)

Jakobian

Yeah. So it makes sense to speak of such limits for $f:A\to Y$ relative to that $A\subseteq X$ for some larger $X$

Vladimir Lysikov

For limits in general not, but for derivatives we require $f$ to be defined at the point where we take derivative

I don't know enough analysis to say what happens if we define derivatives as $\lim_{x \to a, y\to a} \frac{f(x) - f(y)}{x - y}$
Probably nothing goes wrong too much

one potato two potato

2:13 PM

at a certain point, mathematics becomes super hard.

psie

@Derso the definition of a limit point $p$ of a set $E$ is that for every neighborhood of $p$ it contains a point $q\neq p$ such that $q\in E$. I still don't quite understand why it follows from the definition of $\lim_{x\to a}$ that $f$ needs to be defined around the limit point.

Vladimir Lysikov

@psie Are we talking in general or about Rudin?

psie

I'd say Rudin only.

Jakobian

@psie because you want something to hold for $0<|x-a|<\delta$ and if $a$ is isolated this becomes empty so its like everything should be a limit

You want uniqueness of this types of limits though

psie

ah ok, makes sense

Vladimir Lysikov

2:18 PM

Rudin only defines derivatives for functions defined on a segment $[a,b]$ or on an interval $(a,b)$
So when he defines higher derivatives, he wants the previous derivative to be defined at an interval around the point, or at a segment starting from this point

so a neighborhood or a one-sided neoghborhood

psie

ah ok, makes sense too. He needs to obey his definitions of course :)

Derso

@psie I guess this limit point really doesn't imply $f$ is defined on every point around the limit point (but it's reasonable to ask $f$ to be defined on some points around).

Then if we speak of the induced topology on the domain, the confusion disappears because, of course $f$ is defined on the points of its domain, right? (tautology)

psie

@Derso hmm, how do you reason your way to "$f$ defined on some points around $p$ ($p$ being a limit point)" to "there is a neighborhood around $p$ where $f$ is defined at every point"? You mention induced topology; what is its significance here?

nickbros123

The definition of a "connected set in a metric space X" that I know of is that if is not the union of two separated sets in X, i.e, it is not $A \cup B$ where $\overline{A} \cap B= \overline{B} \cap A= \emptyset$. There was another one I saw where it says the only clopen sets are $\emptyset$ and set $S$ itself. I am now thinking clopen with respect to ambient space or the relative space S?

Vladimir Lysikov

Relative

So for example $[0;1]\cup [2;3]$ is not connected because $[0;1]$ and $[2;3]$ are clopen in the induced topology

nickbros123

2:28 PM

Ok, now in the definition using separated sets, the "closure" is done in the ambient space (right?), so I am struggling to prove the following: if A and B are separated, and S=A $\cup$ B, then there exists a non trivial clopen set in $S$ relative to $S$

Thorgott

is this Rudin's doing

nickbros123

@Thorgott the departed set thing is Rudin's definition. The clopen thing is something I came across

Separated*

Derso

@psie That's the point: there isn't (a "full" neighborhood around $p$ on which $f$ is defined).

Ben Steffan

"departed set" lol :D

nickbros123

@BenSteffan pitfalls of typing on phone

Thorgott

2:32 PM

ok, my recommendation is not trying to learn any topology from Rudin, it's a terrible source for that

now, all this worrying about relative vs ambient is completely obfuscatory

define what it means for a metric space $X$ to be connected and stick with that

Vladimir Lysikov

The idea is that if $\bar{A} \cap B = \varnothing$, then for every point of $B$ there exists a neighborhood disjoint from $A$, and by taking the union of these neighborhoods there exists an open $U \supseteq B$ such that $A \cap U = \varnothing$

Thorgott

whenever you have a subset $S$ of $X$, you consider it as a metric space in its own right with the induced metric and apply the definition to that

Sine of the Time

what's the best book of general topology in your opinion?

Ben Steffan

you might want to think about closures relative to something in relation to closures in an ambient space

and convince yourself that what Thorgott is suggesting is actually allowed

Vladimir Lysikov

And because of this set $U$ in the induced topology $B = X \cap U$ is open, and the same for $A$

Ben Steffan

2:35 PM

I don't think you need to do that

Derso

If topology is making things confusing, I would simply suggest the equivalent statement for this context:
Let $f:X\subset \mathbb{R}\to \mathbb{R}$ be a function and $a\in X$ be a limit point of $X$ (namely, for every $\epsilon>0$ there is some $x\in X$ such that $0<|x-a|<\epsilon$. Notice this implies $x\neq a$)

Then we say $\lim_{x\to a} f(x) = L$ if, and only if, for every $\epsilon>0$ there is some $\delta>0$ such that $x\in X$ and $0<|x-a|<\delta$ implies $|f(x)-L|<\epsilon$.

So we evaluate $f(x)$ only on the points that make sense (this is expressed by asking $x\in X$).

Ben Steffan

Just note $\bar{A}$ is closed and disjoint from $B$, so $S = A \cup B$ already implies that $B = S \setminus \bar{A}$ is open, and vice versa

Vladimir Lysikov

Oh right

didn't do any topology for a long time

Ben Steffan

This also implies $A = \bar{A}$, so $A$ is clopen, q.e.d. :)

nickbros123

@BenSteffan oh damn nice

Derso

2:37 PM

Notice in my definition that I'm not asking $a\in X$.

Oh, I am by the way lol hahahaha

My mistake

It seems I can't edit it anymore, but I meant to ask $a$ only a limit point of $X$

@psie Anyways, do you see now? We can't ask such a thing as $|f(x)-L|<\epsilon$ if $x\notin X$. This evaluation only makes sense on the domain of $f$.

So in the end we are looking at the induced topology on $X\subset \mathbb{R}$.

psie

@Derso thanks! it makes sense and the rewording helped. If I understood you correctly, you did not mean for $a\in X$, right? A limit point of $X$ does not have to belong to $X$ if I'm not mistaken.

Derso

Exactly, yes

psie

👍

Jakobian

2:56 PM

@SineoftheTime Rings of continuous functions by Gillman and Jerison

I just love that book

Sine of the Time

noted

Jakobian

@nickbros123 no this is standard definition

Ben Steffan

the standard GT text people use nowadays is Munkres', I believe

Jakobian

The definition of connected is less so but it exists among authors

@BenSteffan I understood the question as the book you like the most. Is Munkres your favourite?

Ben Steffan

It might as well be

I haven't spend much time with GT books in my lifetime

Jakobian

3:00 PM

I see. Munkres is good for introduction but I didn't read it. And it gives me bad vibes from ppl who learn from it

Ben Steffan

I'm not a huge fan of the non-GT things I've seen from Munkres' books

but yeah, I can't say that it's the best, or even particularly good; I just wanted to drop the name for reference since it's so popular :)

Jakobian

In particular they think closed sets are those which contain their limit points... but this just conflicts with the mentality I think ppl have in general topology

It's true of course but not something I would be thinking of first and foremost

Ben Steffan

sequences and limits don't play as big a role in GT, for GT reasons

and nets haven't really caught on that much

Jakobian

Nets are better for topological vector spaces, not topological spaces in general

Thorgott

GT? my favorite geometric topology textbook is... yeah no, I don't like any

Ben Steffan

3:05 PM

@Jakobian yeah, the only real use in general topology I've seen is proving Tychonov's theorem

and even that people do with ultrafilters instead nowadays

Thorgott

I've appreciated Munkres as a reference in the past, it has a decent treatment of paracompactness (at least for the standard applications) and even covers some of the dimension theory that's necessary if you're a weird person dealing with non-compact, non-smooth manifolds

Ben Steffan

oof

Jakobian

Actually... I think Cauchy filters for top. vector spaces might be viable too

Yeah idk, preference

I was thinking of completeness first and foremost

nickbros123

@BenSteffan our place apparently uses GF Simmons

Ben Steffan

never heard of it

Jakobian

3:11 PM

On the other hand the construction of completion of a top. vector space probably uses nets... so it's better... yeah. I think nets are better for this.

Although I never needed that - knowledge that weak and weak* topology on a Banach space is complete is enough

nickbros123

@BenSteffan archive.org/details/…

Ben Steffan

that looks heavily analysis flavored

Jakobian

I bet Bourbaki constructs completion of a topological vector space using filters

Ben Steffan

probably

@nickbros123 I flipped through it and the book doesn't seem to even so much as mention things like path-connectedness at all lol

that's bizarre

Jakobian

@SineoftheTime if you meant something best to learn, or best reference, then standards are Munkres for learning and Engelking for reference

There's almost everything in Engelking

Thorgott

3:23 PM

I own a set of Bourbaki, General Topology cause our library threw it out

don't really have a use for it, though

Ben Steffan

at least it's not the Algebraic Topology text

Jakobian

The French have a very abstract way of writing from what I recall

Joe

I always got the impression that Bourbaki influenced the Definition-Theorem-Proof style of mathematics the most, but perhaps I am mistaken

nickbros123

3:42 PM

@BenSteffan thats weird, ive seen metric space books talk about that

@Thorgott for what its worth ime done with the topology section of that book, so theres actually not more topology to learn from that book :)

nice problems though, in the 2nd chapter

a lot of talk about "perfect" sets though in the 2nd chapter which I havent seen other books talk about a lot

Jakobian

@nickbros123 I checked just to make sure and yeah, no path-connectedness it seems, lol. I guess it's just not that important for them

nickbros123

well that is worrysome

Jakobian

why worrysome though?

nickbros123

isnt that something one must know as an undergrad?

Jakobian

The book seems to be pretty good

Ben Steffan

3:50 PM

must know? no

Jakobian

@nickbros123 I wouldn't say must know, but you might now that, it's pretty basic

Ben Steffan

but if you wanted to learn e.g. algebraic topology then this would not be a great base to start off of

Jakobian

Yeah. Homotopies are based on the ideas of paths between maps... it's pretty important for algebraic topology

Ben Steffan

but I find it a little strange because it's not like analysis doesn't care about paths/curves

Jakobian

I guess just not this type of analysis

Ben Steffan

3:51 PM

yeah

Jakobian

They seem to be focusing on operators and not on something like say, integration

nickbros123

@Jakobian yeah, ive seen it for R^n in the book hubbard and hubbard.

Thorgott

@nickbros123 it's not a particularly relevant concept

I'll refrain from going into another tirade on how much I dislike Rudin Ch. 2

Jakobian

@nickbros123 as in sets for which $A' = A$?

Cantor-Bendixon theorem is a pretty cool topic

nickbros123

@Jakobian yeah

@Thorgott haha, I would like to hear that, cuz I personally liked 2nd chapter (but Jakobian earlier alluded to the possibility of it being some sort of stockholm syndrome xD )

Xander Henderson

3:58 PM

@nickbros123 "Must"? Meh...

Mad

Can someone help me understand the Lemma of Fatou:
In it, the talk is about Limit inferior and superior of $f_n $ function sequence.

Hwoever, as far as i understand, this limit it self is just a number, not a function.

but it seems like the lemma is treating as a function that delivers different values for different x's. i am confused

Sine of the Time

you're considering the limit of a sequence of functions

Xander Henderson

But I think that anyone completing an undergraduate degree in mathematics should know the basic vocabulary of analysis, topology, and algebra. You should know what groups and rings are, you should have a basic understanding of limits and continuity (via epsilons and deltas), and you should know what an open set is.

Ben Steffan

@XanderHenderson yeah, but what this "basic vocabulary" comprises exactly is up for debate

Xander Henderson

@BenSteffan Sure.

But I think that any introductory class in real analysis is going to cover about the same ground.

nickbros123

4:01 PM

@Mad the limit is a function though, or rather, u evaluate $f_n(x)$ at each point (on the set in which it converges) and call it $f(x)$

u can also call it :)(x)

or xD(x)

Xander Henderson

@nickbros123 Be careful about that---$f_n(x)$ may not converge, even if $f_n$ converges (depending on the mode of convergence being described).

Jakobian

path-connectedness isn't that necessary because if you have connectedness you will deduce anything you need about path-connectedness from that

nickbros123

@XanderHenderson yh, thats why I restrict to the set in which they do converge

Jakobian

especially from an exposition from that book

Xander Henderson

@nickbros123 That doesn't work, really, if you are considering, for example, convergence in $L^p$.

Mad

4:04 PM

@nickbros123 so for each different x i have a different lim superior because f_n(x) takes different values ahhh i see!

got it.

Xander Henderson

@Mad Yes, that is the basic idea.

Sine of the Time

if you're studying Fatou's lemma, you should be familiar with pointwise convergence (?)

Jakobian

@XanderHenderson yeah. For that you need to take a subsequence

Mad

4:33 PM

Wierdly, in the requirements, these functions can have integrals equal to infinity, right?

Xander Henderson

@Mad For Fatou? Yes.

Mad

@XanderHenderson okay

Also, it is very interesting, but apparently with the lesbegue definitions, you can integrate over a single point.

Xander Henderson

@Mad Sure, but it will be zero.

(Assuming the usual Lebesgue measure on the real line.)

Mad

Xander Henderson

(or any non-atomic measure)

Mad

4:42 PM

maybe i wrote this wrong?

Xander Henderson

Your measure is not Lebesgue measure...

Mad

@XanderHenderson Oh.

Well neverthelss, it is interesting for me, because i still have that riemanian image in my head.

In that example, how would the rational be, why Lim inf f_n = 0 ?

that n will go to infinity and thus you can never plug in the value infinity?

i mean, infinity is part of N

Oh nvm this is just N not N unity infinity

Jakobian

5:54 PM

@BenSteffan @Joe Not sure if you're interested, but the appropriate generalization to the "polynomial" question (it was about algebraic curves, really) is this

Ben Steffan

I see

thanks for sharing!

Jakobian

So it's just coincidence that for curves of degree $3$, the one that passes through all but one is supposed to be of degree $3+3-3 = 3$

Soumik Mukherjee

6:32 PM

Bezout, Pappus, Pascal

Related to Caylay-Bacharach theorem

Tapi

8:00 PM

can someone tell what exactly i need to prove to show whats asked?

Ben Steffan

there's an equation with a left hand side and a right hand side. You're supposed to show these are equal :)

more seriously, what do you mean?

what's unclear about the problem statement?

Jakobian

8:15 PM

@Tapi I don't understand the notation. What's $Q^X$?

the question seems to be about conditional probability?

Tapi

@Jakobian It's the regular conditional probability

Jakobian

@Tapi with respect to what

Tapi

@BenSteffan I mean, what would be the first line you write, or, what is an equivalent statement

Jakobian

the random variable $\omega\mapsto \omega$?

Tapi

@Jakobian yes

Jakobian

8:27 PM

Okay so let $Y(\omega) = \omega$. Then $Q^X(M, \omega) = P(X\in M | Y = \omega)$

And you want to show that 1) For each fixed $M$, $\omega\mapsto Q^X(M, \omega)$ is a candidate for $E[1_{X\in M}|Y]$, 2) for almost all $\omega$, $M\mapsto Q^X(M, \omega)$ is a probability measure, 3) for all Borel $M$, $\omega\mapsto Q^X(M, \omega)$ is measurable

If we want to be formal

Tapi

okay so all the conditions mentioned in the definition of rcp right?

Jakobian

Yes, this is what it means to be a regular conditional probability

Ben Steffan

that is very hard to read

Tapi

@Jakobian so whatever has been done in the above image is incorrect right?

okay I'll write it

Jakobian

@Tapi I can't zoom it in, who knows

Ben Steffan

8:34 PM

I can, but it's too blurry

Jakobian

oh I guess if I use ctrl and +

$B\in \sigma(X)$ is, I don't know. Its just Borel sets on $\mathbb{R}$ here

Ben Steffan

ctrl-+ oder ctrl & scroll

Tapi

better pictures

Jakobian

It'd be much better if you weren't doing pictures at all and just wrote it in LaTeX.

tinyurl.com/cfqcvpc

^ for LaTeX in chat

its a bookmark

Tapi

okay, I didn't write this, but will keep in mind the next time

Jakobian

8:40 PM

I don't know what $B$ is supposed to be, like what is its purpose, what are you trying to show

are you trying to show that fixing $M$, $\omega\mapsto Q^X(M, \omega)$ is measurable?

oh. Maybe you are trying to show that its a candidate for $E[1_{X\in M} | Y]$

Tapi

@Jakobian yes i think so

Jakobian

Yeah. I think you are trying to show that $E[1_B E[1_M\circ X|Y]] = E[1_B\cdot 1_M\circ X]$, where $B\in \sigma(Y)$ condition here

Tapi

I didn't do this.. I just wanted to make sure that doing this is not enough, and I don't get why if $\omega$ is in some $\mathcal{B}$ why should $-\omega$ be in $\mathcal{B}$

Jakobian

so first issue is that you want to take $B\in \sigma(Y)$ and not $B\in \sigma(X)$

so $B$ is just a random Borel set, since $\sigma(Y) = \mathcal{B}_\mathbb{R}$

Tapi

yes it's a borel set

what's $\sigma(Y)$?

Jakobian

8:45 PM

@Tapi you mean $B$?

@Tapi the sigma-algebra generated by $Y$

you are taking conditional expectation with respect to $Y$, so you want $\sigma(Y)$ and not $\sigma(X)$

@Tapi by the way in case if you meant $B$, this is just false is all

Tapi

@Jakobian yeah I meant that B is a borel set

Jakobian

I mean sure maybe its true if you take $B\in\sigma(X)$.

but here $B$ is supposed to come from $\sigma(Y)$ which is all Borel sets

Tapi

okay understood

Jakobian

if you were to take $\sigma(X)$, then those would be preimages of $\omega\mapsto \omega^2$ of Borel sets, which are symmetric

but thats not what you should be doing from what you told me

@Tapi here and in the comment above it is my question

Tapi

sorry but im really confused

Jakobian

8:51 PM

this is complicated so you better alleviate that confusion, or at least try to tell me what you are confused about

Tapi

Let $(Ω, \mathcal{F}, P)$ be a probability space, $\mathcal{B} \subset \mathcal{F}, Q^\mathcal{B}(A, \omega) : \mathcal{F} \times \Omega \to [0,1]$ is a regualar conditional probability on $\mathcal{F}$ given $\mathcal{B}$ under P if
$(i) \forall A \in \mathcal{F}$, it is a version of $E(\mathbb{1_A})$
$(ii) \forall \omega \in \Omega$,as a function of A it is a probability on $(\Omega, \mathcal{F})$.

this is the definition I was given

Jakobian

probability theory is one of the most complex subjects among the undergrad courses

Tapi

how does $X(\omega)=\omega^2$ (with reference to the question) come in the picture here

Derso

@Jakobian Why do you think so?

Jakobian

@Tapi this definition feels incomplete

@Derso why would I use words. See for yourself why

@Tapi this definition is wrong I'm pretty sure

26 mins ago, by Jakobian

And you want to show that 1) For each fixed $M$, $\omega\mapsto Q^X(M, \omega)$ is a candidate for $E[1_{X\in M}|Y]$, 2) for almost all $\omega$, $M\mapsto Q^X(M, \omega)$ is a probability measure, 3) for all Borel $M$, $\omega\mapsto Q^X(M, \omega)$ is measurable

I've described correct definition here

I use "candidate" instead of "version"

Tapi

8:58 PM

that's alright

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