I hope $a$ is either $0$ or $1$.

it isn't

all we're told is that

$a\in [0,\infty]$

@TedShifrin

the hint I just got is that we can find a set $B$ in the borel algebra generated $[0,1]$ such that $f$ coincides with the characteristic function on $B$ almost surely

no idea how to construct such a B

maybe $f^{-1}(a)$

but i'm not able to get it to work

can you paste the problem here verbatim

Sure

Oh, duh. $f$ is the characteristic function of a set of measure $a$.

oh, yes

dur

now sure how that's obvious @TedShifrin, could you elaborate?

Okay so the problem I guess is: Suppose $f:[0,1]\rightarrow \mathbb{R}$ is a positive borel measure function such that $f>0$ and such that $\int_{0}^1f^ndx=a$ for all n, and $a\geq0$ . Show that there exists a a measurable set such that f coincides with the characteristic function on the measurable set almost surely

Yup.

@orientablesurface, what is $\int_0^1 \mathbb{1}_{A}$ for some Borel set $A$?

is that riemann integral @JoeShmo ?

say lebesgue

and then, what is $(\mathbb{1}_A)^n = \ldots ?$ for any $n$

Note that if $a>1$ there is no solution.

mhm

no you had it right

the middle step wasn't necessary but it wasn't wrong

and yes, to Ted's point, observe that $\int_0^1 \mathbb{1}_A \le 1$

okay but i\m not sure how $\int_{0}^1 1_{A}$ relates

to the question

for one thing, your hint says that $f=\mathbb{1}_B$ for some $B$ a.e.

yeah, I suppose my issue is that im not sure how to start constructing such a B and im not sure if my integral means lebesgue measure or riemann

and yes, I see why the integral of characteristic is less than or equal to 1

because $A\subseteq [0,1]$

Thorgott told me here yesterday that a homomorphism $f:S_3\rightarrow C_n$ has a non-trivial kernel. How do I prove that?

youre integrating on $[0,1]$

when I googled it I only found theorems about when the domain of the homomorphism is cyclic

@Sophie, if not, $S_3$ would be a subgroup of an abelian group.

yeah @JoeShmo , what's the relationship between $\int_{0}^1f^n$ and $\int_{0}^1 1_{A}$?

well, start with showing that the exercise is true for $f=\mathbb{1}_A$

@orientablesurface Ask yourself, “Self. How can all the integrals be the same number?”

sorry to interrupt, JoeShmo.

you started helping him first, I interrupted.. :-)

i yield.

wait, sorry I reread your second post of the question - it wasn't a hint, it's what youre asked to prove

yeah, the question I asked is to show a relationship, I emailed the lecturer for a hint, and he wrote that

because I was having trouble proving that that's the unique solution. It's possible there's say, a mollifier out there that concentrates its mass around a point, under exponentiation.

and yeah, it is to show that $f= 1_{A}$

this is an intro to lebesgue measure theory course

OK. Well that's easy. Say $A \subseteq [0,1]$. Then in particular $\mu(A) \le 1$ (again, say $\mu$ is the Lebesgue measure)

then guess what kinda integral that is :)

integral of characterisitc

that I understand

Let $A$ be such that $a = \mu(A)$

such an $A$ exists in $[0,1]$ -- $[0, a]$, for example

why can we assume $a\leq 1$?

OK. Now, question for you: Does $f = \mathbb{1}_A$ fit the bill for $n=1$?

we had shown up top that that's the only possible case. We don't have to assume that

You have to answer my question, I think.

what was your question?

@TedShifrin okay I give up. I don't see why that is either. Why?

You can scroll, JoeShmo.

@Sophie: Is $S_3$ abelian?

no

it's "Ask yourself, “Self. How can all the integrals be the same number?”" @JoeShmo

yeah, I saw

is any subgroup of abelian abelian?

yes, it inherits the groups properties

well, how can they orientable?

@JoeShmo what do you mean?

as in, how can all integrals be the same number

Commas matter, JoeShmo.

exponentiation is a radical operation

So, doesn’t that answer your question, @Sophie?

i'm not sure how to answer Ted's question

which function that you can think of remains invariant under exponentiation?

there's 1 obvious one

I'm so punny today

$1_{A}$, sure

yeah

@TedShifrin I'm figuring it out

even slightly simpler than that - $f(x) = 1$

@orientablesurface What happens if $f$ is different from $0$ and $1$ on a set of positive measure?

are you asking what happens as $n \rightarrow \infty$?

Nah.

not quite sure, @TedShifrin

Play with it.

btw, what I said about the mollifier earlier isn't true, since the support isn't dwindled under exponentiation. Meaning $f$ would either explode or annihilate as $n \rightarrow \infty$

Yeah, I ignored that ...

all I can see is that the integral of $f$ must be>0

Think about how the integral is the same for $n=1,2,3,....$

btw is $\int_{0}^1fdx$ supposed to be interpreted as the riemann or lebesgue integral?

What sort of integral makes sense for Borel sets?

lebesgue

So why do you keep asking?

because in in the notes he keeps using dx notation for riemann

and for lebesgue he uses

$d\mu$

OK, put $d\mu$.

if it was riemann, do they coincide in that case?

Then the a.e. stuff doesn't make sense. You lose Riemann integrability.

Think about the char fn of the rationals.

Ok, so it's lebesgue , it makes intuitive sense that integral must be almost surely characteristic

because of the invariance under exponentiation

So I'm telling you to work that out.

okay i'm gonna try

the last thing I said is the proof that $f = \mathbb{1}_A$ a.e.

I don't think so.

hm, why?

We're talking integrals, not functions.

you guys are speaking through your intuition, and my intuition at the moment isn't as strong as yours, i'm still having trouble figuring it out

"f would either explode or annihilate as $n\rightarrow \infty$ doesn't help me much"

well, Ted, I claim $f \le 1$ a.e. Otherwise, suppose $f|_{E} > 1$ where $\mu(E) > 0$, then since $f > 0$, we have $\int_0^1 f^n \ge \int_E f^n \rightarrow \infty$ as $n \rightarrow \infty$ likewise, $f \not< 1$ a.e.

the last piece I suppose is because already $\int_0^1 f^2 < \int_0^1 f = a$

Is $E$ supposed to be $[0,a]$? and by $f\leq 1$ $a.e.$ , do you mean $f\leq 1_{A}$ almost surely

I mean $f(x) \le 1$ for a.e. $x$

no, $E$ is a hypothetical set of positive measure on which the assertion that $f \le 1$ fails

and I show that given the conditions, such a set cannot exist

anyone got any tips on how to show this

apparently the first equation is supposed to help

I'm using the famous fact that if $\alpha > 1$, then $\alpha ^n \rightarrow \infty$ as $n \rightarrow \infty$

Likewise, if $0 < \beta < 1$, then $\beta^n \rightarrow 0$ as $n \rightarrow \infty$

so if $f(x) > 1$ on a set of positive measure $E$, then in particular the Lebesgue integral over it matters, and we see that $\int_E (f(x))^n \rightarrow \infty$ as $n \rightarrow \infty$, and certainly $\ne a$

if you'd like, you don't even have to take the limit here. the integral over $f^2$ is already going to be $> a$

sorry, there was a test fire alarm in my building, we had to evacuate. I'm back now

The bottom line is that $f = \mathbb{1}_A$ a.e. for some set $A$ of measure $\mu(A) = a$, and that's the only possible solution

i'm having a hard time understanding why $\int_{E}f(x)^n\rightarrow \infty$

because for every $x \in E$, $f(x) > 1$, and therefore $f(x)^n \rightarrow \infty$

Thus the integral explodes, too. Again -- important here is that $\mu(E) > 0$

okay so you want to show $f\leq 1$ $a.e.$, this literally means that we must show that $\mu( x\in [0,1]: f(x)> 1 \}$ is 0

yes

you rephrased of what I showed, but yes, sure

You suppose for a contradiction, and assume it isn't $\leq 1$ $a.e.$ this means you assume $\mu(x\in [0,1]:f(x)>1 \}>0$

mhm

and then the premise of the problem breaks down

btw, it only works because $f > 0$, so there's no negative mass to compensate for the positive mass. If we didn't have that, it wouldn't be true that $f = \mathbb{1}_A$

Okay so I can see why $f\leq 1$ a.e. Because then there exists some $N$, such that the integral is way larger than $a$

yeah. outright diverges eventually, in fact.

now we ought to rule out $f < 1$, as well, so that we are left with $f = \mathbb{1}_{\text{some set}}$

ohhh

:)

okay okay i'm getting there

now, $\text{some set}$ had better be such that $\mu(\text{some set}) = a$, ey?

so just to clarify, because $f\leq 1 a.e$ and $f\geq 1 a.e.$ this mean $f=1_{some set}$ almost everywhere?

yes

use mathbb if you wanna be cool

what would be your advice to get better with these types of questions and to have an easier time conceptualising these arguments? Would it be weird if I asked you how your thought process started?

no, your intuition will come with experience, I promise

tbh it may not seem much to you, but you have motivated me to work so much more and improve

so thank you

really

i'll try to fill in the details right now with the problem I had

:) I'm glad I could help!

also, measure theory isn't all that fancy. Read baby Rudin, work out the exercises and proofs by hand.

I will, there's this question i've been willing to ask... if i get stuck on a problem, how should I in general approach the situation?

ik it's a broad question, but advice from someone experienced will certainly help

anyone got any ideas

OK, I'm not that experienced. And I (and everyone else) get stuck on problems all the time. There's no magical algorithm to get to the answer. But what does help is writing down on a piece of paper AAALLLL the facts you know about the problem

and if you're lucky and you can shake it the right way, the solution is gonna fall right out. The rest is experience. Nothing can replace that.

hey joe

do you think when approaching a proof

you should keep thinking about it

like for hours and hours

or like just read more about it

thanks @JoeShmo. Really

@JosephRock, both. If you feel that you don't understand the problem well enough, you gotta go back to read the relevant mathematics. Otherwise it's gonna naturally bother you everywhere.. when you walk the dog, on the train, in the shower

also, I haven't conducted real mathematical research. So ask the real mathematicians.

@JoeShmo if $f<g$ then, $\int f< \int g$ right?

absolutely

the so called Monotonicity of the Lebesgue integral -- en.wikipedia.org/wiki/…

@JosephRock were you wondering about how to get the second equation from the first, or how to solve the second?

hi rob

i am trying to solve the second

to show that Y(t) solves the below equation

however since you said that , do you think the first and second are related? @robjohn

What I have tried so far is doing Laplace transformation on the equation but it seems that there p and q are not solvable and I would need more information..

The first and second equations are the same equation.

why do u say so

just use the substitution they give $Y(t)=y\left(e^t\right)$

ok

what do u think about the second statement

as p and q are unknown

They are the same $p$ and $q$ in the first and second equations

hmm

You will also need to use $x=e^t$

ok will try it out

by solving do they mean we can get an actual value of t?

i think i am missing something here

how does x become t?

Why is (0) not simple? No nonzero submodule/group exists that is contained in (0)

in measure theory

can a positive function

$f\geq 0$

?

what do you mean?

if a function is strictly positive, then it's $f > 0$ (at least a.e.)

by definition

anyone can provide some insight ?

My lecturer said

let $f:[0,1]\rightarrow [0,\infty)$

be a positive function

but notes in other uni have said

positive function can take on value 0

Did he say nonnegative or positive

positive

positive means $> 0$

if 0 was included, would the argument for the problem above still work? @JoeShmo i don't think so for the $f\geq 1$ almost everywhere

@SayanChattopadhyay

Wake up bro

anyone knows what happens if p = 0 here?

would i still get a defined function

$f$ of course need not be integrable over all of $(0, \infty)$

yes @BalarkaSen

however if p=0

then wouldn't the equation always be zero

What?

I mean if i set p=0

then the equation would no longer be well defined

as any $f$

the place would "send" that f to zero

What equation? $\int_0^\infty f(x) dx = \lim_{b \to \infty} \int_0^b f(x) dx$ is a definition not an equation

yes I am using that defn in general

but my specific eqn is above

because if the general eqn does not allow p = 0

Then mean $p > 0$ in the question above, yes.

but why doesnt

the general case have to $p > 0 $ too

What general case?

What are you talking about

the defn

above

why is p= 0 allowed in the defn

but not the example

can anyone provide some insight to this

The definition is writing down what an improper integral means.

oh..

but the defn didnt specify values of p

If the limit doesn't exist for $p = 0$, that's fine.

$\int_0^\infty g(x) dx = \lim_{b \to \infty} \int_0^b g(x) dx$ is a valid definition for any function $g$, if the limit on the right hand side doesn't exist the improper integral doesn't exist. That's all

erm..

Now plug in $g(x) = e^{px} f(x)$

g(x)?

where does g(x) come into play here

I give up

lol it's ok

thanks for trying

ok i unds what u mean now

but what i mean is F(0) here affects the p, sending every function to zero, does that mean F(0) is well defined or what does it mean to be well defined

help someone

Let $f:[0,1]\rightarrow \mathbb{R}$ be a positive measurable function. If $f\leq 1$ almost everywhere and $f\geq 1$ almost everywhere (with respect to lebesgue measure). Show that $f=1_{A}$ for some $A\in B[0,1]$.

?

what is that

@JosephRock try showing that if $x=e^t$ that $\frac{\mathrm{d}x}{\mathrm{d}t}=x$ and thus, $\frac{\mathrm{d}}{\mathrm{d}t}=x\frac{\mathrm{d}}{\mathrm{d}x}$. Furthermore, $\frac{\mathrm{d}^2}{\mathrm{d}t^2}=x^2\frac{\mathrm{d}^2}{\mathrm{d}x^2}+x\frac{\mathrm{d}}{\mathrm{d}x}$. This also means that $x^2\frac{\mathrm{d}^2}{\mathrm{d}x^2}=\frac{\mathrm{d}^2}{\mathrm{d}t^2}-\frac{\mathrm{d}}{\mathrm{d}t}$. Apply this to $Y(t)=y(x)$.

@Sophie no it does not , so i guess it is not well defined..

@robjohn ok i will try those out

@robjohn so in ur notation i see that we use the differential operator like its a variable?

is there some rule behind this

when we use the differential operator like an interchangable multiple

@JosephRock No, we use it like an operator. But we do have that $\frac{\mathrm{d}}{\mathrm{d}t}z=\frac{\mathrm{d}z}{\mathrm{d}t}=\frac{\frac{\mathrm{d}z}{\mathrm{d}x}}{\frac{\mathrm{d}t}{\mathrm{d}x}}=\frac{\mathrm{d}x}{\mathrm{d}t}\frac{\mathrm{d}}{\mathrm{d}x}z$

hello

hi

how's it going? @robjohn

@skullpatrol kinda hectic right now. how are you doing?

@robjohn fine, thanks

@JoeShmo why $\int_0^1 f^2 \leq \int_{0}^1f $ for $f\geq 1$ $a.e.$

for the proof

That's not true

The equality is reversed

@orientablesurface it's the other way around. If $f < 1$ a.e., then $f(x) \cdot f(x) < f(x) \cdot 1 = f(x)$ so that $\int_0^1 f^2 < \int_0^1 f = a$

Hence $f \not < 1$ a.e. or, $f \ge 1$ a.e.

Getting imaginary numbers as a root means that there is no feasible solution right?

function = 5x^3 - 3x^2 + 10x - 5

There seems to be an inflection point valued at 0.2 but no critical points

Imaginary roots mean the root is not real

Unless it's 0

Shouldn't there be no inflection point when there is no maximum/minimum?

No, inflection is wether the second order derivative changes sign

If you take a function $f$ such that its derivative is $f' = x^2+1$

f' is always strictly positive, so there is no critical point

But the second order derivative has to change sign because the derivative goes to infinity at both $+\infty$ and $-\infty$

So to get the inflection point I just have to equate 2nd derivative to 0?

yes

Oh right was confused. Reviewing my calculus after several years. Hah

@EdwardEvans @Alessandro @Astyx

Do you know about Igorrr

yes

CoB are such trolls

ikr

I like how the comment below says "Oops I took drugs again"

That's Alexi

It reminds me a bit of Kammthaar, by Ultra vomit, which is a song about a guy who loves his truck

thats brilliant

while mixing melodies from 80s french hits

@LeakyNun What do you know about stably isomorphic local rings

Tell me everything

@BalarkaSen what's the definition?

No, he asked you to tell about them, he didn't tell you to ask about them

I would guess he's working with k-algebras so that stabilizing means tensoring with k[x]

Probably k=C

I have two local $k$-algebras $A, B$ such that $A \oplus k^n$ is isomorphic to $B \oplus k^n$ as $k$-algebras. I want to conclude $A$ and $B$ are isomorphic as $k$-algebras. Give me some hypothesis so I can

I don't care about counterexamples

if the residue fields are not $k$ then I think they're isomorphic lol

What an awful notion of stabilizing

$k$ is algebraically closed, residues fields are $k$

big rip

@MikeMiller You saw what I wrote down in garbology, I'm just trying to use that

algebra is awful by definition

no u

TELL ME HOW TO DO IT MF

Any comment on Follands Real analysis

@LeakyNun Oh $A$ and $B$ are finite dimensional $k$-modules.

amazon.com/Real-Analysis-Modern-Techniques-Applications/dp/…

Strong restriction. Does it help?

then don't they look like $k[\varepsilon]/(\varepsilon^m)$?

Something something Artinian feels

No, right? $k[x, y]/(x^2, xy, y^2)$

tell me they don't

aha ok

If they all looked like that singularity theory would be very easy

And boring

Folland is a great book

ur mom is a great book

she doesnt have any exercises

ur mom is a great cook

Lmao

ur mom is a great look

i was waiting for thorgott to respond with that but alas

so how do you do direct sum of $k$-algebras

is it just product with $k$ diagonally embedded

I'd assume so

yes

then calling that a direct sum is awful

it's a product not a coproduct

shutup

u stabilize by taking sums

not products

whats stabilize

thicken something to increase space

two thickened things might look same because you increased space

but the unthickened versions might be different

but geometrically aren't you taking product with $\Bbb A^n$

so it should be tensoring with $k[x_1, \cdots, x_n]$?

oh yeah but this comes from a different context not scheme theoretic

oh you non-scheme-theoretic lot

i have written it down in garbology

tell me

if I localize $A \oplus k^n$ at $\mathfrak m \oplus k^n$

do I get $A$ back

do you?

I'm asking you

too hard man

geometrically this is clear since it's a disjoint union with $n$ dots

ahh

nice man nice picture

geometry

lol

not localize at $\mathfrak{m}\times\{0\}$?

that isn't a prime ideal

hm, thats true

but who cares about prime ideals

lol I can't even define map from each side

aha the map from $A$ is tricky

you send $x$ to $\frac{(x,0)}{(1,1)}$

damn son

and then you need to show that $\frac{(1,0)}{(1,1)}$ is the unit of the localization

in order for the map to be unital

I don't believe that's always injective

its difference with $1$ is $\frac{(0,-1)}{(1,1)}$, and $(0,-1) (1,0) = 0$

this is why the map from $A$ is unital