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6h
comment Irrational number problem
It's so-called Dirichlet's approximation theorem.
7h
comment A sebset of $\Bbb C^2$
Well, you can try $m(f^{-1}(I)\cap F)$ is small if $m(I)$ is small enough, for any compact set $F$, then $m(f^{-1}(K)\cap F)$ should be zero. I forgot to restrict the domain on a finite measure space.
14h
comment Show that a proper continuous map from $X$ to locally compact $Y$ is closed
In fact, locally compact spaces are compactly generated, and a continuous from a topological space to a compactly generated Hausdorff space is proper if and only if it's closed and preimages of singletons are compact.
15h
comment The sum of reciprocal squares: estimating the remainder
For the asymptotic of the remainder term, try Euler-Maclaurin formula.
1d
comment Moving the branch cut of the complex logarithm
It works if we rewrite $f(z)=\log(z-\sqrt[3]2)+\log(z-\sqrt[3]2\omega)+\log(z-\sqrt[3]2\omega^2)$, but I want to obtain good insight on how Riemann surfaces work. For example, what's the exact meaning of $\log(fg)=\log f+\log g$ w.r.t. Riemann surfaces when $f,g$ are multi-valued functions.
1d
comment Moving the branch cut of the complex logarithm
Taking a single-valued branch is always annoying to me. +1
1d
comment A sebset of $\Bbb C^2$
Suppose $f\colon\mathbb C^2\to\mathbb R,(z,w)\mapsto\lvert zw\rvert$. Note that if $I$ is an interval such that $m(I)$ is small, then $f^{-1}(I)$ is measurable and $m(f^{-1}(I))$ is also small.
1d
comment How to prove that $\max\{f,g\}$ is Riemann integrable?
Note that $\max(f,g)=(f+g+\lvert f-g\rvert)/2$.
1d
comment Outer measure and Caratheodory's criterion
@GiuseppeNegro Generally, if $\mathcal A$ is a collection of subsets closed under countable intersection, then there's $A_0\in\mathcal A$ such that $m^*(A_0)=\alpha=\inf_{A\in\mathcal A} m(A)$. We again choose $A_n$ such that $m^*(A_n)\ge\alpha+1/n$, then let $A_0=\bigcap_n A_n\in\mathcal A$ and by def $m^*(A_0)\ge\alpha$, but $m^*(A_0)\le m^*(A_n)\le\alpha+1/n$.
1d
comment Outer measure and Caratheodory's criterion
@GiuseppeNegro $\inf$ in my expression is really $\min$, i.e. if my expression of $m^*(A)$ is right, then it should be a regular measure, since suppose $m(E_n)\le m^*(A)+1/n$ and $E_n\supseteq A$, then consider $E=\bigcap_n E_n$. Since $E\supseteq A$, we have $m(E)\ge m^*(A)$, but $m(E)\le m(E_n)\le m^*(A)+1/n$, thus $m(E)=m^*(A)$.
2d
comment Proof about Number Fields
Note (elementarily) that if $0\neq f\in\mathbb Z[X]$ and $m/n\in\mathbb Q$ is irreducible, then $m$ divides $f(0)$ and $n$ divides the leading coefficient of $f$.
2d
comment Outer measure and Caratheodory's criterion
@GiuseppeNegro Good. I took the wrong book so I didn't reach the same content. Now I borrowed the right book and found that the measure satisfying the condition I listed in the original post is just so-called regular outer measure in Munroe's book. Please post an answer filling the details and I'll accept your answer.
Apr
13
comment Proving that a function has a removable singularity at infinity
Comment on the last paragraph: as hinted in the preceding exercise of Ahlfors, when $f$ is entire, $f(z)=o(\lvert z\rvert)$ follows from Shwarz integral formula.
Apr
13
comment Outer measure and Caratheodory's criterion
@GiuseppeNegro Did you mean Munroe's GTM89 Introduction to Measure Theory and Integration? I've looked up and found nothing about this, but a discussion on the agreement of premeasure and the corresponding outer-measure.
Apr
12
comment Is there a simple proof for Fundamental theorem of finitely generated abelian group?
Try Serge Lang's Algebra or Michael Artin's Algebra.
Apr
11
comment Formula for the following sum?
Such an algorithm exists for hypergeometric version of anti-difference. Gosper-Zeilberger's algorithm. See, for example, here.
Apr
11
comment I'm not able to solve the following indefinite integral
They are so-called elliptic integrals.
Apr
11
comment Bounded entire function constant
And additionally, $g(z)\to0$ as $z\to\infty$ by condition, thus $g(z)=0$.
Apr
11
comment Function coincides with a function of bounded variation almost everywhere
Sorry, my network is very bad, so I postponed to accept your answer.
Apr
7
comment Function coincides with a function of bounded variation almost everywhere
It seems sound to me. I'll discuss it with my classmates. There are two key steps. The first is reduction to Lebesgue points. The second is using approx. to id as test functions.