why is there nobody at 18pm

18pm?

1800 ~ 6:00pm

@infinitesimal isn't it 18pm in the (east) us ?

"military time"

yes sorry, I meant 6pm of course...

:)

often times in america, we use mod12 instead of mod24., but both should be understood

@JMoravitz not by eveyone apparently -__-

@JMoravitz Mind if I bug you again?

Hello!! Could you give me informations about functional analysis?? What is it about??

I'm trying to type something up, but if I can easily answer it without too much distraction, I'll try to help

can "information" be written with an "s" ?

@MaryStar en.wikipedia.org/wiki/Functional_analysis

@JMoravitz Nevermind -- I'll let you be. I have received three completely disparate "hints" for this problem and understand none of them.

I need to churn a bit more on it all.

@JMoravitz What knowledge is required?? Is it difficult??

@Ramanewbie written with an "s" in place of the t? or at the end., like how there is "Math" and "Maths"?

Informasion?

@JMoravitz at the end... I thought it was incorrect, but maybe I'm wrong

@infinitesimal no, at the end

@MaryStar I recommend at least a course in real analysis to get some solid proof writing, and to see at least some notions of continuity and basic topology before hand. Functional analysis could probably be taken alongside real analysis.

@MaryStar ok then... I think British don't write the "s"

It does sound "wierd" to my ear :-)

I've never seen it as "I have many informations to share" or "I have much informations to share". The term information already has with it a sense of plurality to begin with unless specified "I have a single piece of information"

@JMoravitz What about the subject: An Introduction to Manifolds ??

As for the difference between "Math" in the US and "Maths" in the UK, I believe their argument is that "Math(s)" is short for "Mathematics", which is itself a plural noun.

Informations is like saying datas

I think

@infinitesimal Not quite, because "data" is already a plural, while "information" is an uncountable noun.

I could be wrong :-/

@DanielFischer Any information about the subject An Introduction to Manifolds ??

@MaryStar Seems to be the title of a book. By whom?

@DanielFischer What are manifolds about??

@MaryStar Generalisations of curves and surfaces.

@Daniel @Mary I believe Mary is looking to get suggested course prerequisites for various subjects and broad overviews. Most overviews can be found easily on wikipedia with much greater detail and clarity than can be expressed in a chatroom however.

@DanielFischer And about partial differential equations- theory of weak solutions?

>_< 10 figures to prove this result... I wonder if there was an easier way to go about the proof...

Not my cup of tea at all.

@DanielFischer I'm sorry to be dense, but I don't really understand your comments on my question at all.

I'm trying to reconcile what you've said with some comments my professor made, and I just can't make sense of anything.

@Arkamis You want to show that $\mu^\ast$ is the completion of $\mu$. That means you want to show that a set is $\mu^\ast$-measurable if and only if it is sandwiched between two $\mu$-measurable sets of the same measure, or, put differently, if and only if it is the union of a $\mu$-measurable set and a subset of a $\mu$-null set, or, third way, if and only if it is the set difference between a $\mu$-measurable set and a subset of a $\mu$-null set.

Is someone familiar with measure theory??

...

@Arkamis And the lemma gives you half of that, it tells you $E$ is $\mu^\ast$-measurable if and only if there is a $\mu$-measurable $B$ with $B\supset E$ and $\mu^\ast(B\setminus E) = 0$.

@MaryStar I don't mean to be rude, but I highly recommend going to wikipedia or an academic advisor with those sorts of questions.

@JMoravitz I have a specific question about an exercise.

@JMoravitz Find a $f \in L^1(\mathbb{R})$ such that $Mf \notin L^1(\mathbb{R})$. $Mf$ is the maximal function.

@DanielFischer Well, I'll try to make sense of it.

Now what you need is that there is a $\mu$-null set $D$ containing $B\setminus E$ - which follows easily from the fact that $\mathcal{M}$ is a $\sigma$-algebra.

I mean, I feel like intuitively I get it, but mechanically I can't make the connection.

Like, I don't see how the lemma gets me half of what you're saying.

@Arkamis ... then $E$ is $\mu^\ast$-measurable iff there exists ...

So how is that only halfway? It seems to me that the lemma gives me just about everything.

Since $\mathcal{M}$ is a $\sigma$-algebra here, your $\mathcal{A}_{\sigma\delta}$ is just $\mathcal{M}$.

Right.

I'm sorry, I just have no sense for this material. These all just seem like disconnected facts in my head.

I can't bring them together.

@Arkamis Well, I should probably have said it gives you half directly.

Then you can use it again to get the other half.

I don't even know what the halves are at this point. I'm sorry.

I will be seeing my professor tomorrow. I need a step by step on this. I can't extrapolate the connections right now

I should probably just withdraw from this class.

i'm trying to remember what the usual toolbox is for establishing upper bounds for (the modulus of) complex contour integrals. the estimation lemma is obvious, but i figure there's results more specialized than that

Ask your prof @Arkamis

@Arkamis You want an approximation from the outside - that's given to you directly by the lemma - and one from the inside. For the latter, you look at $B\setminus E$ instead of at $E$ directly.

Ok, so I can show that $B\setminus E$ is $\mu^*$-measurable

Right?

@Arkamis Right. Every $\mu^\ast$-null set is $\mu^\ast$-measurable.

(Proof: ...)

Easy proof, I got that earlier.

Now, given that $B\setminus E$ is $\mu^*$-measurable, then... I have no idea.

(i can give the specific kind of example i'm interested in, but in that case i'd rather do it as a proper MSE question rather than a hey-does-anyone-remember chat)

Then some $D \supset B\setminus E$?

Yep. So we have a $D \supset (B\setminus E)$ with $D\in \mathcal{M}$ such that $$\mu^\ast\bigl(D\setminus(B\setminus E)\bigr) = 0.$$

Which is a re-application of the lemma?

Yes.

Ok

So $\mu^*(D\setminus E) = 0$

wait no

Then $D\setminus (B\setminus E) = D\cap (B\cap E^c)^c = D\cap (B^c \cup E) = D\cap B^c \cup D\cap E$.

Right

So by monotonicity, $\mu^\ast(D\setminus B) = \mu(D\setminus B) = 0$.

Wait wait hang on

Well, that wasn't so helpful yet.

Ok got it

$D\setminus B \subset D\setminus (B\setminus E)$. The latter has measure 0, so the former must as well

Then likewise $\mu^*(D\setminus E) = 0$?

That argument seems wrong.

We have $\mu^\ast(D\cap E) = 0$.

er yes

that's what i meant to type

But $E\subset D,$ so $D\cap E = E$.

Therefore, $\mu^*(E) = 0$?

So we have $\mu(D) = \mu^\ast(D) = \mu(D\setminus E)$.

@Arkamis No, we (generally) don't have $E\subset D$.

ok

it is at least in the guidelines --->, but sure

@DanielFischer But $B \setminus E \subset D$, and $E \subset B$. Ohh... I see

@anon Is it? I must have missed it.

$D$ could be some kind of annulus.

Alright, so $\mu(D) = \mu^*(D) = \mu(D\setminus E)$.

Now, @Arkamis, $D = (D\setminus B) \cup (D\cap B)$.

@anon I don't see it in these guidelines. Is there one I am missing?

We saw above that $\mu(D\setminus B) = 0$

Therefore, $\mu(D\setminus E) = \mu(D) = \mu(D\cap B)$.

@robjohn sorry, not guidelines, I mean room description in the upper right

@anon Ah, right... forgot about that.

Hello everyone!

Alright, @DanielFischer How is this going to get me a set in $E$? that's what I can't figure out

@Juan Just imagine the cacophony if everyone replied ;-)

I don't know how to approximate from below.

@Arkamis It must be $\mu^\ast(D\setminus E)$, since $E$ is generally not $\mu$-measurable. But thus we saw that $\mu(D) = 0$.

@robjohn !! How are you?

@Arkamis Now look at $A = B\setminus D$.

you're the best. Are you Euler's son? haha

@Juan doing fine

so it was a while ago

Can anyone help me with this (Fourier Cosine Transform): math.stackexchange.com/questions/1138711/…

@anon Yeah, I just forgot that that was there... I am getting old.

I don't know what can I do

@Juan people will look...

I cannot envision such a set being a subset of $E$.

Let me re-walk the chain of events here

@robjohn OK! I read some books but I couldn't solve it

Sorry if my English is not perfect, in my country we speak in Spanish

Ok, i can envision such a set.

Why again do we have $\mu(D) = 0$?

@Arkamis What is $B\setminus (B\setminus E)$? Now, since $D\supset B\setminus E$, it follows that $B\setminus D \subset \dotsc$

@Arkamis $$D = (D\setminus B) \cup (D\cap B) = (D\setminus B) \cup (D\cap (B\setminus E)) \cup (D\cap E)$$ and each of the three is a $\mu^\ast$-null set.

@robjohn I think you/we should limit the guidelines to seven :-)

So $\mu^\ast(D) = 0$. But that means $\mu(D) = 0$ since $D\in \mathcal{M}$.

Alright

So now we take $A = B \setminus D$.

@Arkamis No.

$A = B\setminus D$

Right

What i meant.

Then $A \subset E$.

@infinitesimal Perhaps we should have a ninth: Don't ask to think, just think.

Then $A\subset E\subset B$, and $\mu(A) \leqslant \mu(B) \leqslant \mu(A) + \mu(D)$.

But $\mu(D) = 0$

So $\mu(A) = \mu(B)$.