@Jakobian It's helpful because if you don't award the bounty manually it will be awarded automatically, possibly to a different answer to the one you'd choose. See meta.stackexchange.com/a/16067/334566 for details. Section "What is automatic awarding?"
Let $(X,\overline{\mathcal M},\overline{\mu})$ be the completion of $(X,\mathcal M,\mu)$. Let $g$ be $\mathcal M$-measurable, hence also $\overline{\mathcal M}$-measurable. Is it true that $$\int g\,d\mu=\int g\,d\overline{\mu}?$$
@Jakobian well, I haven't read it anywhere so far and I'm almost past the integration chapter. It's strange. I guess it is somewhere in there, but I haven't been able to find it.
There's the following proposition:
> Proposition. Let $(X,\overline{\mathcal M},\overline{\mu})$ be the completion of $(X,\mathcal M,\mu)$. Then if $f$ is an $\overline{\mathcal M}$-measurable function, there is an $\mathcal M$-measurable function $g$ such that $f=g$ $\overline\mu$-a.e.
He has also said that we "identify" the spaces $L^1(\mu)$ and $L^1(\overline{\mu})$ due to this proposition.
I guess it follows from this proposition...
It follows that $$\int f\, d\overline{\mu}=\int g \,d\overline{\mu}.$$ Though I still do not see how $\int g \,d\overline{\mu}=\int g\, d\mu$.
@psie well I guess for this definition the argument is more subtle, but whats matters is both are pointwise limits of increasing sequence of simple functions, which are $\mathcal{M}$-measurable
(since for your definition of $\int gd\mu_0$ we take $A_k$ to be $\overline{\mathcal{M}}$-measurable)
so for your definition we would need to use something like Lebesgue's monotone convergence theorem
Regarding my post, in the link they define a lift of a curve $\alpha: S^1 \to \tilde S$ as a lift of the map $\alpha \circ \pi: \Bbb R \to \tilde S$, where $\pi: \Bbb R\to S^1$ is the usual covering.
so for your definition we would need to use something like Lebesgue's monotone convergence theorem
you can directly use that a non-negative $\mathcal{M}$-measurable function $f$ is an increasing pointwise limit of simple functions $f_n\to f$ with $f_n$ being $\mathcal{M}$-measurable. From Lebesgue's monotone convergence theorem, $$\int f d\overline{\mu} = \lim_n \int f_n d\overline{\mu} = \lim_n \int f_n d\mu = \int f d\mu$$
Do we need the MCT? Let $S$ be the set of $\sum_{k=1}^na_n\mu(A_k)$ where the sets $A_k$ are $\mathcal M$-measurable. Let $\overline S$ be the set of $\sum_{k=1}^na_n\mu(A_k)$ where the sets $A_k$ are $\overline{\mathcal M}$-measurable. Evidently $S\subset\overline S$ and since for every $s=\sum_{k=1}^na_n\mu(A_k)\in\overline S$ we can find sets $B_k\in\mathcal M$ such that $B_k\subset A_k$ and $\mu (B_k)=\overline{\mu} (A_k)$.
Then $s=\sum_{k=1}^na_n\mu(A_k)=\sum_{k=1}^na_n\mu(B_k)\in S$. So also $\overline S\subset S$
Can this matrix problem be formulated as an ILP? Given an n by n binary matrix, I want to find the smallest number of bits that need to be flipped to reduce the rank of the matrix over the field of integers mod 2. I don't think there is a fast algorithm so I was hoping it could be formated as an ILP problem. But I am not sure if the rank restriction allows that.
I'm trying to think of examples of spaces that are path connected but not locally path connected. If we take a circle in the plane and remove a point, this is an example, right?
@SineoftheTime But for this "Criteria of exactitude for differential forms C⁰" its mean to write the characterization theorem of exact forms? Is that all or is there something else to write?
it's worth being suspicious of any academic or technologist making broad claims about "science" vs. "bureaucracy." i think my default position is that serious people do not do things like that. at a high level i would certainly agree that science is often at odds with bureaucracy. the sky is also blue
@ZacU. Honestly, I got maybe three or four paragraphs in before it felt like b*llsh*t. There seems to be a certain uncritical fawning over rich technologists (Musk and Wolfram, in particular) which I find really off-putting.
> Exercise: Let $(X\times Y,\mathcal L,\lambda)$ be a complete measure space. If $f$ is $\mathcal{L}$-measurable and $f=0$ $\lambda$-a.e., then $f_x$ and $f^y$ are integrable for a.e. $x$ and $y$, and $\int f_x\,d\nu=\int f^y\,d\mu=0$ for a.e. $x$ and $y$.
I have worked towards a solution to this exercise, but the solution doesn't use the fact that $f$ is $\mathcal L$-measurable. Why is this needed?
Here $\lambda$ is the completion of $\mu\times\nu$.
psie just in the background, while this kind of hypothesis chasing is necessary in the textbook, in many "real life" applications of measure theory you will stay in the world of measurable functions if you start in it, so it is common for people to tack that kind of hypothesis on just as a security blanket because that's what they're going to be confronting even if it isn't crucial to really anything at all in the argument
so it wouldn't upend my understanding of what an author is trying to do if they added in a measurability hypothesis that seemed superfluous or even was provably superfluous
with lebesgue measure in particular, once you deal with functions having any kind of structure there are a lot of 'automatic measurability' results where, if you do your homework, anything that satisfies whatever you are talking about is going to be measurable, but the proof and argument style used for the proof is just so far from whatever else you might be doing that people ignore it and throw in the superfluous assumption instead
you should be sensitive to completion issues whenever you deal with product measures and i would be generally concerned with measurability concerns if you are doing anything resembling an 'infinite product' construction, people get lazy and sometimes f up in that setting
but not in just like doing calculus or PDE or whatever on R^n
Consider the above proposition. I'm trying work a proof of another theorem where this proposition is needed. I'd like to make two assumptions, one of which I feel confident about, the other which I'm doubting.
1. If $f\geq0$ in the above proposition, I can assume without loss of generality that $g\geq0$. I understand this, because we could just consider $h=|g|$, which will be nonnegative and still measurable if $g$ is. Also, we still have $h=f$ $\overline{\mu}$-a.e. if $f\geq0$.
2. The second assumption I'd like to make is that if $f\geq0$, we can assume without loss of generality that $g\leq f$. Is this possible? In other words, if we have $g>f$ on the $\overline{\mu}$-null set, can I redefine $g$ so that it still satisfies all the conditions in the proposition and we get $g\leq f$?
I thought about simply defining $h=g\chi_{E}$, where $E^c$ is the $\overline{\mu}$-null set, though I'm not sure if this function is $\mathcal M$-measurable.
Anyone know what happens to the point $(0,-3)$ when $f(x)=-3\cdot\cos(45(x))$ becomes $f(x)=-3\cdot\cos(45(x-2^{\circ}))$? My thinking is that since it deals with trig functions that $-2^{\circ}$ is the same as $\frac{2}{2\pi}$ or just $\frac{1}{\pi}$ & that this may be useful supposed to be dealing in degrees the whole time.
would the points on the graph just be equal to radians in a sense? 3.14... is denoted as $\pi$ so wouldn't subtracting 2 degrees just be a shift to the right by 0.0349066 radians?
@VulpesInculta Using the same letter to mean multiple things is generally a bad idea. You have a function $f$, defined by the formula $f(x) = -3\cos(45^\circ x)$. You then define a new function, say $g$, by the formula $g(x) = -3x\cos(45^\circ(x-2^\circ))$.
Observe that $g(x) = f(x-2^\circ)$. So the graph of $g$ is identical to the graph of $f$, but translated to the right by $2^\circ$. This implies that the point $(0,-3)$ (corresponding to $f(0)=-3$) is translated $2$ units to the right, to the point $(2,-3)$ (corresponding to $g(2)=-3$).
@XanderHenderson it seems pretty obvious now the graph was represented using fractions of $\pi$ so I guess I just assumed I should do everything else in radians lol
@XanderHenderson The joke is that they figured out how to divide by 0, finally, and then nobody cares (as far as I'm aware)
not being able to divide by 0 is a critical problem in mathematics, even to the point that mathematics as a whole is a scam, y'know (according to... you know the kind)
Folland says the Lebesgue measure $m^n$ on $\mathbb R^n$ is the completion of $m\times \cdots\times m$ on $\mathcal B(\mathbb R^n)$, or equivalently, the completion of $m\times \cdots\times m$ on $\mathcal L\otimes \cdots\otimes\mathcal L$. Why is this an equivalence? Are both these measure spaces incomplete and their completion is $\mathcal L(\mathbb R^n)$?
I think these spaces are both not complete. I feel quite shaky by all the surprises one finds when examining product measures on product spaces, so I'm even a bit unsure if $\mathcal B(\mathbb R^n)\subset \mathcal L\otimes \cdots\otimes\mathcal L$.