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Feb
23
comment Does continuous convergence imply uniform convergence?
@DavidMitra You're right. Could you pleased post an answer and I'll accept yours?
Feb
23
accepted Unnecessary simple function in the proof of Lebesgue's convergence theorem in Baby Rudin?
Feb
23
asked Does continuous convergence imply uniform convergence?
Feb
23
comment The limit of the measures of monotone decreasing sets
@Thomas If $\{A_n\}$ are measurable, it seems that $\mu(A_n)\to0$ since $A_n$ are bounded. We can write $A_1=\bigcup_n(A_n\backslash A_{n+1})$, since $\forall a\in A_1,\exists N: a\in A_N\land a\not\in A_{N+!}$, and $\mu(A_1)=\sum_n\mu(A_n\backslash A_{n+1})$. Since both sides are finite, we have $\mu(A_m)=\sum_{n\ge m}\mu(A_n\backslash A_{n+1})\to0$ as $m\to\infty$.
Feb
23
comment The limit of the measures of monotone decreasing sets
@Thomas The extension process is referred to Baby Rudin 11.7 to 11.10, just like the Lebesgue measure extended from the length function of elementary sets.
Feb
23
comment The limit of the measures of monotone decreasing sets
@Thomas From Baby Rudin: A nonnegative additive set function $\phi$ defined on $\mathcal E$ is said to be regular if the following is true: To every $A\in\mathcal E$ and to every $\epsilon>0$ there exists $F\in\mathcal E$, $G\in\mathcal E$ such that $F$ is closed, $G$ is open, $F\subset A\subset G$ and $\phi(G)-\epsilon\le\phi(A)\le\phi(F)+\epsilon$.
Feb
22
answered If $f_n(x_n) \to f(x)$ whenever $x_n \to x$, show that $f$ is continuous
Feb
22
comment A Limit of a sum related to the exponential series.
@SeanEberhard Yeah, for example, $k/(n+k)\le\ln(1+k/n)\le k/n$.
Feb
22
answered Asymptotic behaviour of a sequence
Feb
22
comment if $(n+1)a_{n+1}=(n-2)a_n+\frac{(n+1)^2(3n+2)}{4}(n\geq 2),~a_1=a_2=0 $ then $a_n=?$
@rigordonma I haven't gone into the calculation of $\sum_k (k+1)^2k(k-1)(3k+2)/4$.
Feb
22
answered A simultaneous system of equations
Feb
22
revised The limit of the measures of monotone decreasing sets
edited body
Feb
22
revised The limit of the measures of monotone decreasing sets
edited title
Feb
22
comment How to show this is compact?
Let $f_0=f$. Fix $\{f_{n_k}(p_k)\}_{k>0}$ to be an arbitrary sequence in $S$. WLOG, $p_k\to p$, since $K$ is compact. If there's some $m$, such that there're infinite $k$'s such that $n_k=m$, then there's a subsequence converging to $f_m(p)$. Otherwise, $f_{n_k}(p_k)\to f(p)$.
Feb
22
asked The limit of the measures of monotone decreasing sets
Feb
22
comment Unnecessary simple function in the proof of Lebesgue's convergence theorem in Baby Rudin?
In Theorem 11.24, it's proved that $\phi(A)=\int_Afd\mu$ is nonnegative and countably additive on $\mathfrak M$, and $E_1\subset E_2\subset\dotsb$ and $E=\bigcup_nE_n$ so $\phi(E_n)\to\phi(E)$, therefore $\lim\int_{E_n}fd\mu=\int_Efd\mu$. But what your said is somewhat right. I took out Rudin's Real and Complex Analysis and found that it proved $\lim\int_{E_n}fd\mu=\int_Efd\mu$ by Lebesgue's monotone convergence theorem, without proving $\phi$ is countably additive first.
Feb
22
comment Unnecessary simple function in the proof of Lebesgue's convergence theorem in Baby Rudin?
$\forall x\in E_n$, $f_n(x)\ge cf(x)\implies f_{n+1}\ge f_n(x)\ge cf(x)\implies x\in E_{n+1}$, so $E_n\subset E_{n+1}$. Why descending?
Feb
22
comment if $(n+1)a_{n+1}=(n-2)a_n+\frac{(n+1)^2(3n+2)}{4}(n\geq 2),~a_1=a_2=0 $ then $a_n=?$
Multiply $n(n-1)$ on both sides and put $b_n=n(n-1)(n-2)a_n$, we have $b_{n+1}=b_n+(n+1)^2n(n-1)(3n+2)/4$ for $n\ge2$.
Feb
22
asked Unnecessary simple function in the proof of Lebesgue's convergence theorem in Baby Rudin?
Feb
21
comment Does it exist a function for which the derivative changes sign more than countably many times?
@mrf As Barbara said, all we need is $f\in\mathcal C^1$, and according to my hint, it's unrelated to the fact that $f^\prime$ is a derivative (which only depends on the fact that it's continuous), so I think the answer is imperfect, and unless I get an idea to solve in the general case, I won't post it.