I was asked this question:

This is my answer (2 images):

I'm wondering if there is some sort of basic conceptual point I'm missing.

Like, if I know$\mu_f$ for all intervals $(a, b]$, do I have a complete description of $\mu_f$?

I'm trying to remember what I was thinking when I wrote this. I think the point of my last paragraph is to say basically, "Look, this is what $\mu_f$ does on intervals $(a, b]$, but if you want to know what it does on an arbitrary set it might get complicated."

Another classmate just gave a case-by-case definition of $\mu_f\big((a, b]\big)$ and got full marks.

(Yes, I know the "We can't read your professor's mind" is coming)

I'm trying to reason my way through this. The question asks me to "compute $\mu_f$", but I suppose there are so many weird sets you could ask to measure that you can't necessarily explicitly describe what their measure would be in all situations. Maybe I was just overthinking it.

Why not write it explicitly in terms of the arbitrary measurable set, intersecting with positives?

I think I get the first part of your question, i.e. ideally we could describe $\mu_f(A)$ for any measurable set $A$, but I don't understand what "intersecting with positives" means.

(Also, thanks for trying to help)

Does anyone ever describe a measure for any measurable set? I will look for some examples in my text.

Something like $\mu(A)-\frac12\mu(A\cap \Bbb R^+)$.

Wow, that looks weird. I've never seen that before.

I'll think a bit about what that means.

There are other ways to write this, but they’re equivalent.

So you're saying the general formula is $\mu_f(A) = \mu_{\mathcal L}(A) - \frac{1}{2}\mu_{\mathcal L}(A \cap \mathbb R^+)$? Did you just eyeball this based on experience?

$-$, not $+$

Typo, sorry.

It follows from your analysis. Split $A$ into its positive and negative parts and see what maps to each piece.

I will think about this a little bit.

So, for arbitrary measurable set $A \in \mathcal B_{\mathbb R}$, we partition $A$ into $A^- = A \cap \mathbb R^-$ and $A^+ = A \cap \mathbb R^+$. Then we have $$\mu_f(A) = \mu_f(A^-) + \mu_f(A^+).$$

(I am trying to figure out the rest.)

By definition $\mu_f(B) = \mu_{\mathcal L}\big(f^{-1}(B)\big)$ for any measurable set $B$.

Since $f$ acts as the identity on $A^-$, we must have $\mu_f(A^-) = \mu_{\mathcal L}\big(f^{-1}(A^-)\big) = \mu_{\mathcal L}(A^-)$.

Now I just need $f^{-1}(A^+)$ and that should take care of it, I think.

I think $f^{-1}(A^+) = \{ x \in \mathbb R \colon x = \frac{1}{2}a \text{ for } a \in A^+\}$.

Not sure if I wrote that right, but the idea is that for example, the preimage of the unit interval would be $[0, 0.5]$.

Even if $A^+$ were some weird Cantor-like set, I figure that description would still work.

If that's right then I just need to figure out how to write the Lebesgue measure of that set.

Yup. You can write the set in a more suggestive manner.

Like $\frac{1}{2}A^+$?

[fixed typo]

And then you used a property of the Lebesgue measure to pull the half outside?

Sure.

What's that called?

I don't think you can do that for just any measure.

You can also deduce it from your analysis, writing $A^\pm$ as unions of intervals.

Of course. It’s true for Lebesgue. Homogeneity, I guess.

How do I know that the positive and negative parts of $A$ can be written as unions of intervals? I know a bit of point-set topology but, an arbitrary measurable set isn't necessarily open (i.e. union of intervals).

Sorry. Use the Borel algebra.

I'm not sure I fully get what you're saying, but I figure that if I reason as above and write $\mu_f(A) = \mu_{\mathcal L}(A^-) + \mu_{\mathcal L}\big(\frac{1}{2}A^+\big)$ then that covers it.

But you had a minus sign in what you wrote above, hmm.

Were you intending to give me the precise answer, or just point me in the right direction?

These are equivalent. Look carefully.

I'm looking now.

Of course.

wonders how he got trapped in measure theory, given that he hates it

I think I see it now.

Who hates it? You?

because the light was on.

$\frac 12 = 1-\frac 12$.

Yup. Me.

Hi, Munchkin’s pet.

hi. i'm away from the keyboard more than usual because my wife is at a conference, and i can't leave munchkin alone for very long without something happening. at the moment, she's drawing on some paper in my office.

My problem with measure theory is that I have trouble seeing or feeling what is going on. Like, I look at look at the Dominated Convergence Theorem or Fatou's Lemma and all I see is limits moving around.

that's also the state of enlightenment. it seems like more for a while but does wrap back to just limits moving around.

Well, integration is more interesting than measures, but now you know why I’m.a geometer.

unless you're ted. then it's limits moving around a big picture of the hopf fibration, or something.

Um …

i was just trying to think of something visual. that's the best i could do.

my daughter's currently drawing the hopf fibration.

She’s way more talented than her dad.

What’s an example of a diff fn $f$ for which the derivative of $\int_0^x f’$ fails to be $f’$ somewhere (assuming $f’$ integrable)?

Thanks for your help, Professor Shifrin. I'll probably go now.

Fare thee well.

beats me. probably the key word is that f should not be 'absolutely continuous.' and then find that in the index of your favorite book and grab an example.

counterexamples to FTC are a cesspool of analysis. we should hate them more than the measure part of measure theory.

I mean Riemann.

hm, do any of the usual examples where f is not C^1 work? you'd think that one of them ought to.

Not obviously.

Let R be a Euclidean domain and let a,b in R be non zero. Does the symbol $a/(a,b)$ make sense?

(a,b) is gcd of a and b.

I ask because (a,b) doesn’t need to have inverse.

The exercise asks to prove that if a divides bc then a/(a,b) divides c.

If I assume that R is $\mathbb R$, then I have no problem.

Huh?

koro i am early/late to this but if x divides a then there's something y with a = xy and in a domain i think the y will be uniquely determined. it would make sense to conceive of "a/x" as a reference to y in this case, even if x isn't a unit.

although it might not be advisable from a notation point of view

maybe there's ambiguity in a general domain because (-x)(-y) = xy or something like that and you might not be able to single out 'the positive one' like you can with integers. i see a reference to ideals in the question posted on SE so maybe i am out of my depth here

just a thought. ted's "huh" may have been with reference to R being $\mathbb R$ and not for example $\mathbb Z$

Leslie, if x divides a then I say that there is a k in R such that a=kx

It’s an exercise problem from Dummit and Foote’s. I don’t understand a/(a,b) in the exercise.

koro if (a,b) has been defined, does it divide a? and if so, wouldn't the "k" implicit in "(a,b) divides a" be a candidate meaning for that notation?

this is more or less what i wrote above

just like in Z, it makes sense to write 6/2 even though 2 is not a unit in Z

you may be thinking a little too rigidly here

@leslietownes yes, I’d like to believe that. So a/(a,b) assumes existence of some k such that a=(a,b)k

Sloppily.

Greatest common DIVISOR. Give me a break.

@TedShifrin Argh

Right!

Thank you so much @leslie et @Ted.

(I deleted the post.)

@Koro $a\,|\,bc\implies a\,|\,(ax+by)c=(a,b)c$

@RandomVariable So it is useful that that function is not continuous at that point. Nice.

math.stackexchange.com/questions/4421296/…

@robjohn yes, I had also shown that in the now deleted post.

@Koro so you'd already shown that $\left.\frac{a}{(a,b)}\,\middle|\,c\right.$

which is equivalent to $a\,|\,(a,b)c$ since $(a,b)\,|\,a$

yes. My confusion was: what does a/(a,b) mean in a general Euclidean domain?

Then it became clear to me.

for a linear map L: V \to W (not necessarily finite dim if that even matters). Why are the cosets in cokernel = W/im(L)? Why are they \bar{w} = wImL instead of w + Im L? I am trying to show ker L^* = cokernel L, and I am stuck at (y,L^*x) = (Ly, x) = 0, this seems to imply the cosets look like xLy = 0, but I don't understand why it's not addition \bar{w} = w + imL.

@Koro Done ;-)

Okay, I give up on trying to understand voting on this site.

@robjohn yay!! Thanks.

I want to compute $u_t(x,t)$ where $u(x,t) = \int_0^{t-x/c}\int_{-cs+ct-x}^{-cs+x+ct} f(y,s) dyds$ where $c$ is a constant $f$ and $f$ is some function. How can I compute this?

I tried to apply Leibniz rule but it's too complicated

@barista writing out the chain rule explicitly may help, i.e., let $g(x,s,t)=\int_{-cs+ct-x}=\int_{-cs+ct-x}^{cs+ct-x}f(y,s)\,dy$, so that your original integral becomes $u(x,t)=\int_0^{t-x/c} g(x,s,t)\,ds$

ignore the first equality, my brain apparently went missing while typing it

differentiating that will give you $u_t(x,t)= g(x,t-x/c,t)+\int_0^{t-x/c} g_t (x,s,t)\,ds$ which seems more tractable

i guess i shouldn't call that "chain rule" exactly---more just writing it in a more manageable way

@Koro why were you interested in that question? I think it was closed as a duplicate wrongly.

@Semiclassical The original problem is much more complicated in fact. I'm trying to show $v(x,t) = h(t-{x\over c}) + {1\over 2c}\int_{0}^{t-{x\over c}}\int_{-cs+ct-x}^{-cs+x+ct}f(y,s)dyds + \int_{t-{x\over c}}^{t}\int_{cs+x-ct}^{-cs+x+ct}f(y,s)dyds$ satisfies $v_{tt} - c^2 v_{xx} = f(x,t)$ for $x>0,t>0$.

I computed that using Leibniz rule many times and get bunch of messy terms.

And I failed anyway.

@barista ew

The reason I did $u_{xx}$ is because $u_{tt}$ is much much more complicated

@Semiclassical You know this is really checking the derived solution of a wave equation is actually a solution.

yeah, d'Alembert or something

one thing you can take advantage is linearity. if you show that $(\partial_{tt}-c^2 \partial_{xx})h(t-x/c)=0$, then you can drop that term and focus on the rest of it. that's not much of a simplification but it does get rid of the first term

@Semiclassical Vanishing $h$ term is not a big deal. Real problem is the double integral part.

yeah, that's the miserable part

lol, i'm trying to check it with mathematica

and for some reason mathematica isn't seeing that $\int c^2 f(x,t,s)\,ds$ is identical to $c^2 \int f(x,t,s)\,ds $

even if i replace $c$ with some value like $2$

How is it supposed to know $c$ is constant?

because i'm not writing it as c[s]

maybe i'm making assumptions about how mathematica treats scalars tho

i think it's something to do with how mathematica treats definite integrals

it correctly knows that $\int c^2 f(x,t,s)\,ds=c^2 \int f(x,t,s)\,ds$

but if it give it bounds of integration it refuses

(and it continues to do that if i take $c=2$ for instance)

@barista the slicker approach to this is to change variables first, namely $\xi=x+ct$ and $\eta=x-ct$

in which case both the wave equation and the integrals become nicer

What is ${\partial \over \partial x} F(x,t-{x\over c})$ where $F(x,s)= \int_{-cs+ct-x}^{-cs+x+ct}f(y,s)\ dy$? Someone pointed out one computation error in my post.

in that case $F(x,t-x/c)=\int_0^{2x} f(y,s)\,dy$, so $\partial_x F(x,t-x/c)=2 f(2x,s)$

@Semiclassical without specific instructions otherwise, undefined quantities, like $c$, are treated as functions. At some point, you might specify $c=s^2$ or something. Mathematica has to keep things general until told otherwise.

@robjohn fine, but then why would it not recognize 2 as a vaild scalar factor?

Have you tried substituting $2$ for $c$, not just setting the variable value?

yes

And does that show equality?

nope

Simplify[Integrate[2^2 f[x, t, s], {s, 0, t}] == 2^2 Integrate[f[x, t, s], {s, 0, t}]] just gives back $\int_0^t 4 f(x,t,s)\,ds == 4\int_0^t f(x,t,s)\,ds$

See what FullSimplify returns

same thing

Maybe it’s not sure the integral converges and is being picky about that.

also, it doesn't seem as if the number of arguments on $f$ is causing it. you get the same behavior with Integrate[ 2 f[s],{s,0,t}]==2Integrate[ f[s],{s,0,t}]

yeah, that seems plausible

annoying but plausible

@robjohn I tried to solve it but got stuck. That’s why :)

@Semiclassical Simplify[Integrate[2^2 f[x, s], {s, 0, 1}] == 2^2 Integrate[f[x, s], {s, 0, 1}]] returns False also. However, Simplify[Integrate[2^2 f[s], {s, 0, 1}] == 2^2 Integrate[f[s], {s, 0, 1}]] returns True. This sounds like a question for Mathematica.

wacky

so something with how mathematica deals with symbolic parameters in definite integrals?

Copying from Mathematica inserted line feeds that kept the link from showing properly.

@Semiclassical it seems so

@robjohn huh, that's weird: my version of Mathematica gives False for the last one too

or, well

it doesn't give True

@Koro It was closed as a duplicate, but it is not a duplicate. Certainly, the other question can be used to answer the question you cited, but it is not a duplicate. There is also a better way to answer it, which gives a better constant.

@robjohn :(

The OP didn’t ask about finding better constant. That may be the reason why.

Greetings @robjohn

Hey, @Ted

@Koro I know that, but it is a method that is directly to that integral rather than having to splice together two applications of the other answer.

I just wonder if there are better answers, yet, because I know my answer does not give the best constant.

gotta run for a few hours. BBL

What exactly is a reflection group of type $A_{n-1}$ ($A_n$?)? Is it just $S_n$?

yes. it's a notation reflective of a classification of coxeter groups. i don't know where it comes from.

Hmm...sounds dumb.

it looks like he just started with the beginning of the alphabet. some classifications are like that - the symbols involved are kinda arbitrary.

see also type I, II, and III von neumann algebras.

Lol true. What about $II_1$ factors? Where does the $1$ come from?

i honestly don't know. they're finite, so maybe you want to tag it with something that suggests that. and you often normalize a finite trace to send 1 to 1. although this doesn't line up with the subscript n in type I_n.

i kinda like that some fields still have clunky notation and nobody spends too much time trying to improve it.