00:00 - 21:0021:00 - 00:00

12:00 AM
It might seem like I'm jumping onto GMT now, but its actually just that Bartle ordered me to see proof of Banach-Zarecki theorem and I figured out this would be the best method to prove that if $f'$ exists on some measurable set $E$ then $f'$ is measurable (this is needed for the proof).

12:11 AM
@oscarmetalbreak Um, NO.

Plausible.
Maybe not in torus. How about circle?

@oscarmetalbreak If by "circle" you mean "disk", sure, why not?

yes a disk

(Assuming you mean "A rectangle is homeomorphic to a disk". Maybe you mean some other form of isomorphism?---I just assumed topology....)

yes a homeomorphism from rectangle to a disk
Is it because you can find a continuous function from one point to another so it is true?

12:24 AM
Two spaces are homeomorphic if there is a homeomorphism between them. What is the definition of a homeomorphism?

I guess I will encounter the same problem in reading dynamical systems latter...Thanks

4 hours later…
5:00 AM
0

As shown in the answer you linked, the Lindemann–Weierstrass theorem implies that $\cos 1$ is transcendental. It turns out that $\dfrac{\sin n}{\sin 1}$ is a polynomial in $\cos 1$; in particular, $$\frac{\sin n}{\sin 1}=U_{n-1}(\cos 1)$$ where $U$ is the Chebyshev polynomial of the second kind. ...

@TedShifrin

Why you pinging me, DogAteMy?

Because you interacted with this question earlier

Not I.

5:52 AM
DogAteMy???

Where have you been? That goes back 6 or 7 years.
Remember the old “the dog at my homework” excuse?

Is that a label for someone in chat or a generic term you use?
"left my copy book at home, Sir"
"3 slaps for you, hold your hand out"

6:12 AM
Akiva had a dog-something name years ago.

if you have a former nickname on this chat, ted knows it and still uses it.

thankfully i don't have a nickname

you are legally copper.hat. it's on your passport.

card carrying member of the electronic frontier foundation
clear & tsa don't even know i'm there

@AkivaWeinberger this is cool.

6:55 AM
😎

7:12 AM
Suppose that $f:X\to \bar R$ is $\scr F$ measurable, $\scr F$ is some sigma field on X. Let $\mu$ be a finite measure on $(X,\scr F)$. Given that $\int_A f d\mu=0$ for all $E\in \scr F$, can it be said that f=0 a.e.?
if f were non negative, then sure.
I think yes one can say that.
I take $E^+=\{x\in X: f(x)\ge 0\}$. Since, $\int_{E^+} f =0,$ it follows that f=0 for a.e. x in E^+.
Similarly, f = 0 on a.e. x in E^-.
$X= E^+\cup E^-$ so f is zero for a.e. x in X.

8:03 AM
anyone can help me with this ?
https://chat.stackexchange.com/transcript/message/64807719#64807719

2 hours later…
10:04 AM
If $f$ is gauge integrable and $G$ is a continuous a.e. differentiable $N$-function (for example an indefinite integral of a gauge integrable function), then $\int_{G(a)}^{G(b)} f = \int_a^b (f\circ G)\cdot G'$ if $(f\circ G)\cdot G'$ is gauge integrable
For example, $G$ can be an absolutely continuous function here

3 hours later…
12:37 PM
@Jakobian in your proof, when you use Vitali's covering theorem, how do you get that $\lambda( \bigcup \mathcal{A} \setminus \bigcup_{n=1}^{\infty} I_{n}) = 0$?
oh I guess you just keep approximating with finitely many intervals, is that what you're doing?
also, I don't see why $\bigcup_{n=1}^{\infty} \bigcap_{k=1}^{\infty} E^n_{k} \subset E$, I can see why the other inclusion holds though

1:12 PM
@porridgemathematics This is a corollary to Vitali's covering theorem
For all $\varepsilon > 0$ one can find intervals $I_1, ..., I_n$ such that $A\setminus \bigcup_{i=1}^n I_i\subseteq \bigcup_{i=n+1}^\infty J_i$ and $\sum_{i=n+1}^\infty |J_i|\leq \varepsilon$
The intervals $J_i$ are enlargements of intervals $I_i$
In the proof of Vitali's covering theorem its really that $I_1, I_2, ...$ can be fixed all at once

sure, Im okay with that part
I think the other part may not be correct though, I believe I found a counterexample

I think that's impossible
what's the counter-example?

it might be, im probably confused
basically why is it enough that if $x$ is in the intersection of $[c,d]$'s (length shrinking to zero) for which $\frac{f(d) - f(c)}{d-c} \geq r + \frac{1}{n}$, that the upper derivative at $x$ is $> r$?

If $x\in \bigcup_n \bigcap_k E_n^k$ (because its $E_n^k$ and not $E_k^n$)
Then we can fix some $q$ such that $\frac{f(d)-f(c)}{d-c}\geq r+\frac{1}{q}$ for some $x\in [c, d]$ where $0 < d-c\leq \frac{1}{k}$

right

1:21 PM
Then write $\frac{f(d)-f(c)}{d-c} = \frac{d-x}{d-c}\frac{f(d)-f(x)}{d-x}+\frac{x-c}{d-c}\frac{f(x)-f(c)}{x-c}$
Notice that $\frac{d-x}{d-c}+\frac{x-c}{d-c} = 1$ and both are in $[0, 1]$
If both of these quotients were $< r+\frac{1}{n}$, then so would be $\frac{f(d)-f(c)}{d-c}$
So $\frac{f(x)-f(c)}{x-c}\geq d+\frac{1}{n}$ for some $c$ such that $0 < |x-c| \leq \frac{1}{k}$

ahh, okay, thats the part I was missing
nice proof

I've mentioned that this inclusion is tricky and requires to see that this quotient is a convex combination in here
but you probably didn't see that

the result is still a little surprising to me
because there arent any assumptions on f

I suppose so
but $G(a) = \sup_{x\in [0, a]} f(x)$ is also measurable I guess, no matter what $f$ is
sometimes things just fit so nicely that we don't need any assumptions

thats true

1:42 PM
I find the proof really cool because it means I can talk about this comfortably without some weird arguments about measurability of derivative
I was trying to see that if $f:[a, b]\to\mathbb{R}$ is differentiable on a measurable set $E$, then $|f(E)|\leq \int_E |f'|$
so I can see why Banach-Zarecki theorem is true
(literally differentiable, not almost everywhere)
Then the argument goes that if $f$ is a continuous $N$-function of bounded variation, then $|f(d)-f(c)|\leq \int_c^d |f'|$
Thus being bounded by an absolutely continuous function $F(t) = \int_a^t |f'|$, $f$ must also be absolutely continuous
(this is the tricky implication btw)
This is because of Lebesgue's differentiation theorem applied to $f$ and that $f'$ is Lebesgue integrable
If $A$ is any set then I can talk about $|A|$ comfortably, but integrals can only really be done on measurable functions, thats why my concern

1:57 PM
don't you immediately get that $f$ is absolutely continuous if you know $|f(d) - f(c)| \leq \int_{c}^d |f'|$? to me thats just saying $|\mu_{f}| \leq |f'| d\lambda$, and $|f'| d \lambda$ is absolutely continuous w.r.t $d\lambda$, thus so is $\mu_{f}$
here $\mu_{f}$ is the measure induced by $f$
(assuming $f$ is continuous and $f'$ is in $L^1$)
but $f'$ in $L^1$ is immediate from bounded variation
there's a correspondence (for measures on the real line) between BV right continuous functions and finite (complex) borel measures
so $f$ being BV and right continuous means its induced measure's absolutely continuous part is finite (on $[a,b]$), which means $f'$ is integrable
similarly theres a correspondence between sigma finite signed measures and BV right continuous functions with bounded positive or negative variation (on the real line)

2:22 PM
Hi, I would need some help to a question. I am trying to prove the symmetry group $S_5$ contains a subgroup of cardinality 20. Right now I have able to prove the set of affine transformations $x\to ax+b$ with $a\in F^x_5$ and $b\ in F_5$ is a group of cardinality 20. And I just need to prove there exists an injective homomorphism from this set to $S_5$ such that the image would be the desired subgroup. However I have the difficulty to formula this map.

2:36 PM
@porridgemathematics Yes. That's why you show that
To prove its absolutely continuous

i was just wondering why we need the lebesgue diff. theorem here
or I guess in 99% of what I was saying, the lebesgue diff. theorem is lurking behind in some justifications
like the correspondence between BV functions and radon measures on the line

Lebesgue differentiation theorem says that if $f$ is a function of bounded variation then $f'$ exists a.e.

ah okay

yeah its the different one of the same name

I see

2:40 PM
I've seen the name only in Bartle but I'm following that book so I'll be repeating it

thats the book on the heinstock kurzweil integral right
well, one of them

yeah
Modern theory of integration
The theorem is in chapter 14 and in the appendix
(which I've realized that the proof in the appendix is actually pretty easy)
and by easy I mean digestible by me

lol, i end up reading the appendix of a lot of textbooks out of some sort of OCD, its gotten better now but I used to not be able to keep studying out of a book if I couldn't digest the proof of some result cited in the appendix
I would have to wait till I digest the proof of the result cited in the appendix and then continue with the main narrative
occasionally the results in the appendix , or the proofs of them, turn out to be pretty useful to internalise - a lot of other proofs in the main narrative often build on tricks used in proofs for appendix results

I do the same, I have to know why things are true. That's why I've opened another book just for the proof of Banach-Zarecki
And now I'm doing the same thing for integration by substitution for absolutely continuous functions

fair enough

2:46 PM
I've used to be more stuck with one book, but I'm more open to exploring and reaching out to other sources now
so I think I'm improving... (also in terms of searching through articles etc.)

@porridgemathematics My experience is that the appendices of books summarize results which the author presumes the reader already knows. Hence I tend to skim appendices before I start reading a book, but I don't sweat them too much after that.

Its also where they push theorems that aren't part of the main discussion (which I think, what you're saying above is also a part of)
i.e. things the author doesn't want to focus on
to think of it, thats precisely what appendix is

@Jakobian Right. Typically, these theorems are kind of understood to be prerequisites. They aren't part of the main narrative, you probably learned them somewhere else, don't worry about it.

In the example of Lebesgue differentiation theorem above, it was more that the result needs the definition of outer measure and some trickery that's few pages long
so it was more that the author didn't want to focus on it, but the proof wasn't particularly easy (or assumed to be learned before)

@XanderHenderson yeah thats true, there are times when things in the appendix aren't prerequisites though (e.g. in hatcher)
or if you're reading a book on geometric analysis, and the chapter is on the hodge decomposition theorem, then a bunch of the sobolev space stuff might be relegated to the appendix (sobolev spaces on manifolds)
also not necessarily prereqs there, just not the main focus of the book , i.e. can be blackboxed and the author intended them to be if they weren't prereqs

2:55 PM
sometimes its also a place to put stuff the author thought was cool

@porridgemathematics I mean, to-may-to, to-mah-to. I feel that if an author is "black boxing" something, that means that you are supposed to already know it (and it isn't vital if you don't, but you really should).
@Jakobian I've seen that a few times, but I think that most books tend to use the last three chapters for "here's some cool stuff that I am not an expert in, but it's cool, right?!" :P

@XanderHenderson are you well versed in differential geometry?

yeah, maybe should know of it, I dont think its really necessary to know the exact proofs of certain things, as long as you have a sense of why the result is true this is usually enough

3:01 PM
or as long as you can say well this is a generalization of this well known result to manifolds, basically sketch the proof, theres no need to take a whole detour to pin the proof down

1 hour later…
4:27 PM
@oscarmetalbreak You need a set with 5 elements on which it acts. Is there an obvious such set? (It might help to think projectively, but you do not need to.)

4:41 PM
$F_5$ obviously
or even better $\mathbb{Z}/5\mathbb{Z}$

1 hour later…
5:51 PM
@oscarmetalbreak same thing

6:01 PM
Consider the ODE $$2tx''+x'+2x=0,\quad t>0.$$ I want to use Frobenius' method, i.e. the attempt $x(t)=\sum_0^\infty a_kt^k$. When I plug this into the ODE, however, I don't seem to get a nice recurrence relation. First I get $$\sum 2a_k k(k-1)t^{k-1}+\sum a_kkt^{k-1}+\sum 2a_kt^k=0,$$ where the index start from $0$. We can re-index the sums so that the above reads $$\sum 2a_{k+2}(k+2)(k+1)t^{k+1}+\sum a_{k+1}(k+1)t^k+\sum 2a_k t^k=0.$$ Again all sums going from $0$.
But here I'd expect the coefficient in front of $t^k$ in second sum to cancel with that in third sum, so I get a recurrence relation. Am I going in the right direction?

6:16 PM
Why on earth do you expect cancellation of those terms?
Why did you rewrite the first sum that way? You should make everything have the same $t^k$ and then combine coefficients.

ah, I see what you mean! I will try that, thank you.

7:02 PM
Hi, I have a weird problem with MathJax. When I use MathSE on my PC (Linux) the characters in MathJax (i.e., between the dollar signs) are displayed with the usual LaTeX Computer Modern font. But when I use MathSE on my laptop (macOS) no matter what browser I'm using, the characters between the dollar signs are of a different font like so: ibb.co/FKXhPcs . What is the main root of this problem? I can't seem to find info on this.

@Eric Hard to say, but it looks like you might have some missing fonts, or are using the wrong renderer.
If you right-click on some rendered math, you can try a different renderer.

I was going to ask. Do you compile LaTeX on your laptop (say with TeXShop)?

yes

That doesn't look familiar. Where is that menu?

7:05 PM
1 min ago, by Xander Henderson
If you right-click on some rendered math, you can try a different renderer.

Ah, I've never done that before.

I'm using that renderer.

It looks like missing fonts to me. That's why I asked about running TeXShop.

Oh it worked by choosing MathML. Cheers!

Interesting.

7:07 PM
no, sorry SVG

SVG looks a little different on my screen. Interesting.

I honestly have no clue about the inner workings of MathJax
It does look a bit different now that you mention it.

i don't think the laptop does well over low heat

@copper.hat Yeah, laptops need to be sautéd over high heat.

No simmering allowed?

7:14 PM
you have rendered me speechless

@TedShifrin Oh, I was looking for a more east-Asian cooking technique. If you are going French, I suppose you could simmer it over low.

Stir-frying is not synonymous with sautéeing, although I'm not sure why.

we always had a tub of lard beside the stove.
it was practically deep fat frying

So healthful.

7:18 PM
@TedShifrin Sautéing uses more fat, and a little less heat (I think).
But it is basically the same thing.

Actually, I disagree on the more fat thing. The main difference is shallow frying pan versus wok.

@TedShifrin I don't agree. You can stir-fry in a shallow pan, and sauté in a wok---if that is the only equipment you have. But I think that the heat level is the major difference. Stir-fry is very hot. Sauté is just... hot.
According to this, I am right about the heat, but backwards about the fat.

For me, the heat level on sauté depends on what I'm cooking :) Yeah, Chinese cooking and Thai cooking use more fat than I ever do for French cooking or Italian cooking.
And I definitely use less than typical Asian restaurants.

@Jakobian I have a quick question regarding sequential closed spaces. If a topological space is normed, then is it true that sequential closed implies closed ?

7:34 PM
I'm doing cardinality in my math class and I'm trying to understand cantor's diagonalization argument
Can someone help me with the reasoning? Even just the lemma of an uncountable set is confusing to me
intuitively I can understand why $\mathbb{R}$ is just incomparably large to $\mathbb{N}$

@Obliv Where are you getting hung up?
The basic idea is that if you have any function from $\mathbb{N}$ to $\mathbb{R}$, it must fail to be surjective. Since there is no surjective function, $\mathbb{R}$ must have greater cardinality.

Ohh
you basically define a $b \in B$ such that $f(a) \neq b$
for any $a \in A$

@Obliv I don't understand.

got it, that was simple enough

And, no, you don't "define" such a thing.

7:39 PM
how do you make it fail to be surjective then

Start with a function $f : \mathbb{N} \to \mathbb{R}$. Then demonstrate that there exists at least one real number $x$ such that $x \not\in f(\mathbb{N})$, i.e. there is some $x$ such that no $n \in \mathbb{N}$ satisfies $f(n) = x$.
But this is not "defining" such an object.
You are "finding" or "discovering" or "constructing" such a thing.

@SineoftheTime yes

@TedShifrin This question feels like something which might be within your area of expertise. Most of the answers indicate that $\mathrm{d}x$ is "merely" notation, which I think is wrong (except for, perhaps, the most intro of students).

OH wait is it because $\mathbb{R}$ is infinite in two ways like decimal places and numbers themselves

@Obliv I really think that this is not a helpful way to think of it.
To me, that feels like saying "Well, the rationals are infinite in two ways: the denominator, and the numerator." But the rationals are still countable.

7:44 PM
it's almost like $\mathbb{N}$ is contained in $\mathbb{R}$. You could have 1.2345... all of the natural numbers as one element

Spaces for which sequential closure and closure are equal are called Frechet-Urysohn spaces

@XanderHenderson oh i forgot about $\mathbb{Q}$

@Obliv Yeah, I don't like that. It isn't the way that I would try to think about things, and I don't know how helpful it is.
As it is very dependent on the way in which you are representing a number.

is it kind of like mapping an infinite line to an infinite area?
eh not really because $\mathbb{N}$ isn't continuous like $\mathbb{R}$

$$\text{normed}\implies\text{metrizable}\implies\text{first countable}\implies\text{Frechet-Urysohn}$$

7:46 PM
@Obliv No, because both the line and the area have the same cardinality?
(the term to Google is "space filling curves")

spaces for which sequential closure is closed are called sequential
$$\text{Frechet-Urysohn} \implies\text{sequential}$$

ok thank you. I've never heard about Frechet-Urysohn spaces
are they cover in an undergraduate course in general?

@SineoftheTime No.
Not usually.

like is that a characteristic of all irrational numbers
wait that was terrible

I think it's kind of a matter of perspective. If you're a differential geometer like Ted, the $dx$ is representative of the fact that the type of objects one integrates aren't function, but forms. If you're a measure theorist, it might be representative of the fact that you don't just integrate functions, but integrate them with respect to measures.
but if you just wanna think of integration as a functional on some function space, the $dx$ may as well just be notation that helps making things like change of variables much more convenient to write down (the reason for this of course becmoing

7:51 PM
do irrational numbers have an infinite number of decimal point digits that may or may not repeat at a given time, but can't strictly be repeating the whole sequence

@SineoftheTime I know as much about education as a brick on the wall

@Thorgott Yes, exactly.

I think I'm starting to understand the landscape a bit more.
@XanderHenderson when evaluated, it's still one element (albeit possibly infinitely repeating but not irrational).

I don't have too much hope in modern education, I think it's better to study math on my own

having hope about anything is fallacious
well one way or the other, Frechet-Urysohn spaces might not be crucial to what they are trying to teach you
either way you need to push and educate yourself before they are able to educate you

7:56 PM
yes, I'm taking a course that is considered as an introductory course to functional analysis

I read this one for functional analysis
but only three first chapters
I gave up because of that exercise I couldn't solve

too hard?

might have been
it didn't exactly "stump" me, I was just confused about the hint and didn't really see a way to get out of it

so you dropped

how to define a bijection $f: \mathbb{R}\to (0,1)$ I did $f(x) = \frac{1}{x}$ but this doesn't work for $x=0$

8:02 PM
well I've emailed the prof and he was of no help so I was just like, alright who cares

ooh wait I can do a trig function nvm

and quit reading the book too

like $f(x) = |\cos x|$

you can use the arctangent

is that your fav trig function?

8:04 PM
no, but works for your example

Or $1-\exp(-\exp(x))$

what is that?

$\exp(x) = e^x = \sum \frac{x^n}{n!}$
the exponential

$\arctan$

yeah but what is $1-\text{exp}(-\text{exp}(x))$ like is that $1-e^{x}(-e^x)$

8:06 PM
You can use something like $1-2^{-2^x}$ instead if you want

$1-e^{-e^x}$

ohh

$\arctan$ avoids all that silliness.

$\arctan$ is pretty silly to me

i just did $|\cos x|$ because I'm boring and unimaginative

8:07 PM
this does not work

exponential function is more fundamental I feel like
than the trigonometric ones, and especially than their inverses

why not @SineoftheTime

and intimately related to $\arctan$.

after all, trig functions are combination of exponentials
@Obliv it maps to $[0,1]$

ohh

8:08 PM
@copper.hat if you were to define it from scratch you'd be better off with exponentials

dang so we can't use $\frac{1}{x}$ because that'd be $(0,1]$ but we can't use that because that's $[0,1]$ so it's either arctan or exponential eh

arctan and exp are the standard examples

If you're able to make $\mathbb{R}$ to $(0, \infty)$ somehow then you're free to use $1/x$

integrating ${1 \over 1+x^2}$ if fairly basic.

wait what is $\mathbb{R}\to \sin^{-1}x$

8:12 PM
arcsin

@Obliv draw a little picture of what it would be like. you are fighting this all the way

also what?

doesn't $\sin^{-1}(0)$ have multiple values tho

You're restricting $\sin$ to an interval where its bijective
then it has an inverse so you're good
thats how inverses of all those trigonometric functions work

except for $\arctan$.

8:14 PM
arctan is included

@Obliv The usual result is to send $\mathbb{R}$ to $(-\pi/2, \pi/2)$ using the arctangnet, then send that to $(0,1)$ in the "obvious" way.
over 2.

splitting hairs

@copper.hat Splitting $\pi$s.

more fun

dude, I'm not understanding this. If we're sending $\mathbb{R} \to \frac{\sin^{-1}x}{\cos^{-1}x}$ so both numerator and denominator are functions defined from $[-1,1]$ right?

8:18 PM
Dude?

wait the imaginary component gets cancelled out

What?

dude, I hope you are not thinking that $\tan^{-1} x= \frac{\sin^{-1}x}{\cos^{-1}x}$

like for example $\sin^{-1}(\pi) = 1.570.. - 1.811i$
dude i am
LOL wait

8:20 PM
i need to ask my sister for help I guess, she is taking pre calc and kicking butt and taking names right now
the prof asked if she wanted to volunteer to take a more difficult test that's more "abstract" lol and she aced it. Yeah I can't even remember $\arctan$ it's g'over

you need to find desmos and play a little

dude dude dude. haven't heard anyone say it in ages.
Hi @copper.hat!!

Like, dude! Let's head down to the beach and catch some gnarly waves!

@Obliv did you draw a little diagram of what you are looking for in the shape of a function?
@Koro Hi!!

$\widehat{\text{copper}}$

8:23 PM
@XanderHenderson bro is more common here
Bro! let's go there...
:)

my son says something like bruh

@Koro I prefer "bruh" or "brah".

i get strange looks when i say brah

of course

actually that statement could be generalised considerably
i get strange looks

8:25 PM

im stuck on a stupid compact operator problem and its driving me insane
the fun things we do for lunch

@copper.hat Can you really be driven to a place where you already habitually reside?

physically yes, mentally no
ah, missed the point there

@Obliv this doesn't work because f(1/2)=2.

@copper.hat Heh.

8:27 PM
getting out of my migraine induced fog of a few days ago
all of the ai/ml related convex questions seem to have disappeared
maybe some ai/ml guild that wants to keep the market tight

8:40 PM
@copper.hat No, it's arcsec and arccsc that have disconnected domains.

i agree? unless you are referring to my mental state.

@XanderHenderson Even I will not subject a beginning calculus student to a discussion of differential forms. But I will tell students the notation is important so that when you change variables you must use $du = u'(x)dx$ and write them in the integrals.

indeed, should be required as basic hygeine

I won't take topology next sem.

Cohomology and Poincaré duality is — finally — the good stuff, but maybe too much homological algebra for you.

8:44 PM
I did it this semester.

Oh.
So what is next semester?

@TedShifrin nuh uh

Meaning?

Smooth manifolds, differential forms on manifolds, integration on manifolds,
Stoke’s theorem, computation of cohomology rings of projective spaces, Borsuk-
Ulam theorem.
• Degree, linking number and index of vector fields, the Poincare-Hopf theorem.
• Definition and examples of principal bundles and fibre bundles, clutching construction,
description of classification theorem

Oh, differential topology. This is great stuff.

8:46 PM
isn't it standard to write $du=u'(x)dx$ ?

I'm not quite comfortable with bundles :(

Well, I assume they will teach you to be more comfortable with bundles :)

I'll just write $\int f$ and $\int f\circ\Phi\cdot \Phi'$

You're not a beginning calculus student, Jakobian, so I don't care what you do.

Bundle was there this semester and I understand it and every exercise looks untouchable to me.

8:47 PM
When I taught the Spivak calculus theoretical course, he writes $\int_a^b f$, and I'm fine with that, but when we're going to do substitutitions I insisted on putting in the $dx$ to emphasize that we're integrating $f(x)$ with respect to $x$.

so many problems arise from sloppy notation

like I feel like I don't even know how to start it unless I'm given sufficient no. of hints etc.
I understand solutions when I see them.

I guess $\int f \mathrm{d}\mu$ would be better, just to indicate the measure

But this doesn't feel productive to me.

Stokes's Theorem and degree done with preimage of regular value, intersection theory, etc., are super beautiful, Koro. I don't know that bundles per se are a huge part of the course.

8:48 PM
$\int_a^b f$ sucks

@Jakobian I cannot write $\int_a^b x^2$. I have to write $f(x)=x^2$ and then write $\int_a^b f$. Your $d\mu$ is of course what $dx$ is.
@Sine It doesn't really, but I would never do it with "standard" calculus students.
Lunchtime. Bye.

enjoy

@SineoftheTime $\int f \chi_{(a,b]}(x) \lambda (dx)$
most complicated out there in jungle.
@TedShifrin I'll see. I'll try taking all analysis courses. If that won't be possible then masochist me will take topology.
again