@Jakobian If $X$ is locally compact metric, and $X=U\cup V$ where $U,V$ are open, then we can refine $\{U,V\}$ to open covers with closures that are compact?
In Folland's text, he talks about the standard representation of a simple function; one that has finite and distinct range, i.e. $\mathrm{range}(f)=\{z_1,...,z_n\}$ where $z_i$ is distinct from $z_j$, and where the sets of the indicator functions partition the domain. I don't see how a standard representation is unique, yet in the next section on integration of simple functions, he talks about "the" standard representation of a simple function. This confuses me.
Example; in proposition 2.13, I quote: "...let $\sum a_j \chi_{E_j}$ and $\sum b_k \chi_{F_k}$ be the standard representations of $\phi$ and $\psi$." (emphasis mine)
@SineoftheTime I saw a video where he explained a trick on integrals for example: $\int x \tan^-1 (x) dx$ , then I would do $\frac{x^2}{2} \tan^-1(x) - \int \frac{x^2}{2} \cdot \frac{1}{x^2+1} dx$
In this case, i can choose instead of for example $\frac{x^2}{2}$ , i can choose $\frac{x^2+1}{2}$
so : $\frac{x^2+1}{2} \tan^-1(x) - \frac{x}{2} + C$
@Jakobian just to be clear, refinement here does not mean the following: $\{U_i\}_i$ is an open cover, then there exists an open cover $(V_i)_i$ such that $\overline{V}_i\subseteq U_i$ for each $i$, $\overline{V}_i$ is compact. Right?
in the first integral, you find the primitive of $e^x$ :$\int e^x \sin x dx= e^x\sin x -\int e^x \cos x dx$. In the second integral, you find the primitive of $\sin x$: $\int e^x \sin x dx = -e^x \cos x +\int e^x \cos x dx$
What about the following: In a locally compact metric space, every open cover can be shrunk to one where the boundaries of the elements of the cover are compact
Even though the standard representation is unique, it would not be satisfactory to define the integral in terms of it, unless every other representation gave the $same$ result. Hence, the proofs in Cohn, et al- — MatematletaSep 20, 2016 at 20:59
@monoidaltransform I think that if $X$ is an infinite-dimensional Banach space, $U = \{x : \|x\| < 2\}$, $V = \{x : \|x\| > 1\}$ then it has no such shrinking
Let $X$ be a locally compact metric space, $\mathcal{U}$ a cover of $X$, does there exists a shrinking such that boundaries of its elements are compact
What goes wrong with this argument (it goes back to a question before the one now): Let $X$ be locally compact metric space. Let $\{U_i\}_i$ be an open cover. Let $\mathcal{A}= \{A\subseteq X$ $:$ $A$ is open, $\overline{A}$ is compact, $\overline{A}\subseteq U_i$ for some $i$ $\}$. Then $\mathcal{A}$ is an open cover for $X$ and we can shrink to open cover $(V_i)_i$ such that $\overline{V}_i$ is compact, and $\overline{V}_i\subseteq U_i$
Ah maybe I have a better example. Consider an infinite genus torus and cover it with two open sets $U$ and $V$ "cutting every hole in half", so that $U\cap V$ is countably many pairwise disjoint annuli. Can we argue that open sets in the shrinking must have some boundary in every component of $U\cap V$?
By paracompactness, we have a refinement $\{B_k\}_{k\in I}$. Observe, $\overline{B_k}$ is compact in some $A$ in $\mathcal{A}$ So for every $k$ we can pick $f(k)$ such that $\overline{B}_k\subseteq U_{f(k)}$. So put $V_i= \bigcup_{f(k)=i}B_k$
@SineoftheTime wanna know my galaxy brain move? I once saw that promoting to a queen would cover a diagonal and that would cause a stalemate, so I promoted to a bishop.
@Semiclassical I don't feel like I've even come across an answer with major formatting problems in a long time
on the other hand, somebody brought a case here where a question had 3 highly upvoted answers, of which one claimed that the other two are straight up wrong
none of the answers had any downvotes
that's what I mean by proofreading
the issue is certainly not that the answers weren't getting enough attention
