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Oct
31
comment Could give me an informal, but detailed explanation of what Cauchy sequences are?
An informal layman's explanation is that the tail end or those terms near the end of the sequence are very close together.
Oct
31
comment Tao's Analysis Proposition 2.1.16
N to N indicates the domain and codomain respectively. In layman's terms it means $f_n$ takes in a natural number and outputs a natural number. Note that codomain is different from range.
Oct
30
comment $G$ is an abelian group, $H \triangleleft G$, show that $G/H$ is abelian.
You may want to let the two elements in G/H be aH and bH, where a, b are in G
Oct
29
accepted Measure on $\mathbb{N}$ which has measure $1/n^2$ at each point $n$.
Oct
29
comment Measure on $\mathbb{N}$ which has measure $1/n^2$ at each point $n$.
I see! The lower bound was what was bothering me. Thanks
Oct
29
asked Measure on $\mathbb{N}$ which has measure $1/n^2$ at each point $n$.
Oct
29
comment Cauchy sequence in $L^p$, existence of a set with finite measure, and integral is less than epsilon over the complement
May I ask what is the main purpose of introducing a new measure $\mu$? Also, I am a bit lost on the part of formally applying Markov's inequality to $|f|^p$.
Oct
29
comment Cauchy sequence in $L^p$, existence of a set with finite measure, and integral is less than epsilon over the complement
Thanks, I upvoted your answer. However, I don't really understand your solution completely...
Oct
29
revised Cauchy sequence in $L^p$, existence of a set with finite measure, and integral is less than epsilon over the complement
added 148 characters in body
Oct
29
accepted lim sup inequality (with infinite sum)
Oct
29
comment Is $ac \equiv bc\:(mod\: n) \iff a \equiv b\:(mod\: n)$ valid?
if $n$ is a prime, both directions hold
Oct
28
comment Outer measure of symmetric difference zero implies measurable
@DanielFischer Eureka!! $E\cap F=E\setminus (E\setminus F)=E\cap (E\setminus F)^c$, thus $E\cap F$ is measurable as it is the intersection of two measurable sets. Thus $F$ is the union of $E\cap F$ and $F\setminus E$, two measurable sets and thus is measurable. Thanks for your guidance!
Oct
28
comment Outer measure of symmetric difference zero implies measurable
@DanielFischer is the difference of two measurable sets measurable? I only learnt so far that the union of two measurable sets is measurable.
Oct
28
comment Outer measure of symmetric difference zero implies measurable
Hmm... $E\setminus (E\setminus F)=E\cap F$? Thanks for your patient help.. I think I am close to the solution but having a mental block
Oct
28
comment Outer measure of symmetric difference zero implies measurable
@DanielFischer If $m^*(A)=0$, then $A$ is measurable.
Oct
28
comment Outer measure of symmetric difference zero implies measurable
@DanielFischer Hmm.. I know $m^*(E\setminus F)$ and $m^*(F\setminus E)$ are both zero. $m^*(E\cap F)=m^*(E\cup F)$. How do I proceed from there? Thanks!
Oct
28
asked Outer measure of symmetric difference zero implies measurable
Oct
28
answered Finding $N$ in the relation $de\equiv 1\bmod N$, given $d$ and $e$
Oct
27
comment Show (using Vitali Covering Lemma) that the union of any collection of closed, bounded non-generate intervals is measurable.
Thanks for your hints.. However I am still kind of lost. Any more explicit hints / sketch of proof?
Oct
27
comment Show (using Vitali Covering Lemma) that the union of any collection of closed, bounded non-generate intervals is measurable.
@DanielFischer I managed to get $m^*(E_n)\leq m(\cup_{k=1}^m J_k)$. Am I supposed to show that each $E_n$ is measurable, and thus $E$ is the countable union of such $E_n$ and thus measurable?