Hi everyone, wondering if someone can point me in the right direction before I resort to asking a question. I'm analysing numerical stability of an operation and can incur in an error that is (number of elements of the vector smaller than machine precision)×ε, which obvs. ≤ (size of the vector)×ε, but this seems super loose?
i hav a 1D manifold which is the real line, R. i define two atlases : Atlas 1 : ψ(x)=x. Atlas 2: ψ(x)=2x,x≥0;ψ(x)=−3x,x<0 why arent these two atlases two different smooth structures?
the first one is a smooth structure becuz it's a homeomorphism onto $R^1$ and there's only one chart so the compatibility of charts holds. the second one is also smooth for the same reason
sorry i meant $\psi (x)=2x ; x\geq0$ and $=3x , x<0$ for the second atlas
@RyderRude what are you objecting. Which part of what I said
@RyderRude there's multiple smooth structure on $\mathbb{R}$ but no matter which one you equip $\mathbb{R}$ with, those will be, up to diffeomorphism, your standard $\mathbb{R}$
I'm reading a proof of the fact that the space $C_c(\mathbb R^n)$ of compactly supported continuous functions on $\mathbb R^n$ is dense in $L^1(\mathbb R^n)$. The proof starts by a reducing the proof to simpler special cases. We can assume $f$ has compact support by DCT and moreover that it's positive. I understand that. However, then it is claimed, since $f$ is measurable, there is an increasing sequence of simple functions of, note, compact support, that converge to $f$.
I understand why there's such a sequence, but I don't understand why the sequence can be compactly supported. Why?
@YourLordJoyBoy Tags can be given descriptions. You generally shouldn't touch that since if you add a useless tag and give it a description then it will stay forever until it gets moderated
Unless you're trying to give description to an already existing tag
> Theorem 2.25. A subset $A \subset \mathbb{R}^{n}$ is Lebesgue measurable if and only if for every $\epsilon>0$ there is an open set $G$ and a closed set $F$ such that $G \supset A \supset F$ and $$\mu(G \backslash F)<\epsilon.$$ If $\mu(A)<\infty$, then $F$ may be chosen to be compact.
I'm reading a proof of the fact that the space $C_c(\mathbb R^n)$ of compactly supported continuous functions on $\mathbb R^n$ is dense in $L^1(\mathbb R^n)$. Let $A$ be a bounded set. Somewhat out of the blue, it is claimed that from the Borel regularity of Lebesgue measure, there exists a bounded open set $G$ and a compact set $F$ such that $G \supset A \supset F$ and $\mu(G \setminus F)<\epsilon$.
I know the theorem above, but it doesn't say that we can choose $G$ to be bounded. Does anyone have any clue why $G$ can be bounded?
the simpler question might be, what is the quickest path in your document to seeing that G can be bounded. i might look at the proof of "theorem 2.25" to see where the existence of its "G" comes from. it might be as simple as tracing through that argument
as a background vibe, you could think about covering of A by "simpler" open sets (e.g. dyadic rectangles, or other sets whose diameter you can control), and convince yourself that if A is bounded and the sets in your cover have controlled enough diameters, the union of the cover will be bounded too
but that might not be the simplest route in your reference
I think I found an answer. If $A\subset\mathbb R^n$, then $\mu^\ast(A)=\inf\{\mu(G):A\subset G, G \text{ open }\}$. If $A$ is measurable, we can find an open set $G$ such that $\mu(G)<\mu(A)+\epsilon$ for some $\epsilon>0$, and if $\mu(A)<\infty$, we'll have $\mu(G)<\infty$.
@AlessandroCodenotti I figured there exist spaces which are Dieudonne complete but not Cech complete, and there are spaces which are Cech complete but not Dieudonne complete
