Let $(\mu_n)_{n\in\Bbb{N}}$ be a sequence of probability measure on $\Bbb{R}$ with characteristic functions $(\phi_n)_{n\in\Bbb{N}}$. Assume that $\lim_{n\rightarrow\infty}\phi_n(t)=1$ for all $t\in[-\delta,\delta]$ for some $\delta&gt;0.$ Prove that $\mu_n$ converges weakly to $\delta_0$, and hence $\phi_n$ converges uniformly to 1 on any bounded interval.  My thought: I try to prove the property of tightness of $(\mu_n)$. Then if subsequence of $(\mu_n)$ can be proved to weakly converge to $\delta_0$. By the uniqueness, we are done.
120568
Prove the following statement:   $ X_n \Rightarrow 0 $ (convergence in distribution) if and only if $ (\exists\; \epsilon&gt;0: |t|&lt;\epsilon) \;\; \phi_n(t) \rightarrow 1  $, where $\phi_n(t)$ is the characteristic function of the random variable $X_n$.  We could write $X_n \rightarrow 0 $ in probability since convergence in probability to a constant and convergence in distribution to a constant are the same concept.  I guess Lévy's continuity theorem implies the 'only if' part.  Thank you very much for your help in advance!
2723
Question: Suppose $f: [0,1] \rightarrow \mathbb{R}$ is continuous on $(0,1)$. Prove there is a sequence of step functions $\left\{f_{n}\right\}$ which converges pointwise to $f$ on $[0,1]$.    I saw the same question was asked  but I was still confused regarding the solution to the question.   I had attempted a solution but I now realize this solution is wrong:  Proof: Since $f$ is continuous on $(0,1)$, then for every $x \in (0,1)$ and for every $\epsilon&gt;0$, there exists a $\delta(x)$ such that $|f(x)-f(y)|&lt;\epsilon$ whenever $y \in (0,1)$ and $|y-x|&lt;\delta(x)$. In particular, let $\epsilon_{n} = 1/2^{n}$ and let $\delta_{n}(x)$ be the corresponding $\delta$ which is guaranteed by the continuity of $f$. Fix a value of $n$ and choose a partition $x_{0}=0 &lt; x_{1} &lt; x_{2} &lt; \ldots &lt; x_{m} = 1$ such that $x_{i}-x_{i-1} &lt; \delta_{n}(x)$. Let $f_{n}(x):= f(x_{i-1})$ for all $x \in (x_{i-1}, x_{i})$. Clearly $f_{n}(x)$ is a step function and hence $\left\{f_{n}\right\}$ is a sequence of step functions. Observe that $f_{n}(x)=f_{n}(x_{i-1})=f(x_{i-1})$. Finally, by the continuity of $f$ and by the fact that $x_{i}-x_{i-1} &lt; \delta_{n}(x)$, we may conclude that \begin{align*} |f(x)-f_{n}(x)| \leq |f(x)-f(x_{i-1})| &lt; \epsilon_{n}  \end{align*} and so the sequence converges pointwise to $f$.  End of proof.  I was hoping someone can shed some light on the correct solution to this problem. Any hints would be appreciated. Thanks!
15358
Let $(X,\mathcal{M},\mu)$ be a measure space. We say that $\{f_n\}$ converges to $f$ in measure if, for any $\epsilon&gt;0$   $$ \lim\limits_{n \to \infty} \mu\Big(\{x\in X: |f_n(x)-f(x)| \ge \epsilon  \}\Big)=0.$$         Show that $\{f_n \}$ converges to $f$ in measure if $$\lim\limits_{n\to \infty} \int_X \frac{|f_n(x)-f(x)|}{1+|f_n(x)-f(x)|}d\mu(x)=0.$$   Suppose $\mu(X) &lt; \infty$. Show that    $$\lim\limits_{n\to \infty} \int_X \frac{|f_n(x)-f(x)|}{1+|f_n(x)-f(x)|}d\mu(x)=0$$   if $\{ f_n \}$ converges to $f$ in measure.
57008
Suppose that $(u_n)$ is unbounded above. Then, we pick any arbitrary monotone increasing subsequence $(v_n)$ of $(u_n)$. But by hypothesis, we can find a subsequence of $(v_n)$ that converges to $0$. Hence, $\lim v_n =0$. Therefore, $(u_n)$ must be bounded above.  Again, let $(u_n)$ be unbounded below. Then we pick any monotone decreasing subsequence $(w_n)$. By similar arguments, $(u_n)$ must be bounded below.   Being bounded, it has a finite $\limsup =l$ and $\liminf =m$.  Hence, there must be convergent subsequences $(a_n)$ and $(b_n)$ that converge to $l$ and $m$ respectively. But every subsequences of the above two must converge to $l$ and $m$ respectively (treating the two as convergent sequences). Again, by hypothesis, they both have subsequences that converge to $0$. Therefore, both $(a_n)$ and $(b_n)$ converge to $0$.   In conclusion, $\limsup u_n = \liminf u_n = 0 =\lim u_n$  Is this alright?
47764
How can I prove that if a sequence of functions $\{f_n\}$ that converges to $f$ in measure on a space of finite measure, then there exists a subsequence of $\{f_n\}$ that converges to $f$ almost everywhere?
116702
Is  $$\bigcup_{p&lt;\infty}\ell_p=c_0 ?$$  At least one inclusion obvious: every $p$-summable sequence converges to zero.
92501
Let ($x_n$) be a monotone sequence and contain a convergent subsequence. Prove that ($x_n$) is convergent.  I know that by the Bolzano-Weierstrass Theorem, every bounded sequence has a convergent subsequence. But I need some hints as to how to prove the question above. Thanks for your help!
102050
Let $(X,S,\mu)$ be a measure space. Suppose $\mu(X)&lt;\infty$. Put $E_n=\{$ $x\in X:$ $n-1\leq |f(x)|&lt;n \}$ Then, $f\in \mathcal{L}^p$ if and only if $\sum_{n=1}^{\infty}n^p\mu(E_n)&lt;\infty$. My attempt: Notice, $X=\bigcup_{n=1}^{\infty}E_n$. Since $(E_n)_n$ are pairwise disjoint, we deduce by a standard property of measures that $\mu(E_n)&lt;\infty$ for each $n$. As a consequence, $\int |f|^p d\mu=\sum_{n=1}^{\infty}\int_{E_n}|f|^p d\mu$ Now, observe that for each $n$, $n^p\chi_{E_n} \geq |f|^p\chi_{E_n}\geq (n-1)^p\chi_{E_n}$ Upon taking integrals, we obtain: $ \sum_{n=1}^{\infty}(n-1)^p \mu(E_n)\leq\int |f|^p d\mu \leq \sum_{n=1}^{\infty} n^p\mu(E_n)$ So, if $\sum_{n=1}^{\infty}n^p\mu(E_n)&lt;\infty$ then $f\in \mathcal{L}^p$. Now for the reverse, not that if $\mu(E_n)=0$ for each $n$ then we are done. If there is some $n$ such that $\mu(E_n)=0$  if $f\in \mathcal{L}^p$, then $\sum_{n=1}^{\infty}(n-1)^p \mu(E_n)&lt;\infty$. Now the question is how do we determine whether a series coneverges given another one converges? The answer is the $\textbf{limit comparison test}$: Theorem: Let $\sum_{n=1}^{\infty}a_n$ and $\sum_{n=1}^{\infty}b_n$ be two series such that $a_n\geq 0$ and $b_n&gt;0 0$ for all $n$.  Put $c= \lim_{n\rightarrow \infty}\frac{a_n}{b_n}$.  If $0&lt;c&lt;\infty$ then either both series converge or both series diverge.  To invoke this theorem notice that if for all $n$, $\mu(E_n)=0$ then we are immediately done. So, we may assume that for some n , $\mu(E_n)&gt;0$. If for all $n$, this is the case, then we are immediately done. If only finitely many satisfy this property, then we may disregard the ones with 0 measure. Thus, we may assume for each $n$, $\mu(E_n)&gt;0$. In which case, the limit comparison test above allows us to deduce that $c=1$. Since $\sum_{n=1}^{\infty}(n-1)^p \mu(E_n)$ converges, so must $\sum_{n=1}^{\infty}n^p \mu(E_n)$.
3370
Can anyone prove this question? I tried but I didn't get any I idea, so I hope someone can solve it.     Let $X_n\leq Y_n$ for each $n\in \Bbb N$. Show that $\liminf X_n \leq \liminf Y_n$ and $\limsup X_n \leq \limsup Y_n$.   Please prove this question - thanks.  The definition I have:     Let $X_n$ be a sequence in real number and let $$E=\{x\in \Bbb R^\sharp:(X_{n_k}) \rightarrow x \text{ for some subsequence }(X_{n_k})\text{ of }(X_n)\}$$ for all $n \in \Bbb N$ and $k$ from $1$ to $\infty$. Then by definition $\lim\sup X_n = \sup E$ and $\lim\inf X_n = \inf E$.
97532
Define $(p_n)_n$ recursively by $p_0(x) = 0$ and $p_{n+1}(x) = p_n(x) + \frac{1}{2}(x - p_n(x)^2)$.  Prove that $p_n(x) \rightarrow \sqrt{x}$ uniformly for $0 \leq x \leq 1$ as $n \rightarrow \infty$.  I'm not sure where to go with this.  I can show inductively that $\forall n,\  0 \leq p_n(x) \leq 1$ for $0 \leq x \leq 1$, and I think that $p_n(x)$ is increasing with $n$, but I have no idea whether this helps me.  I know that I need to show that $\forall \epsilon&gt;0: \exists N \in \mathbb{N}: \forall n \geq N: \forall x\in [0,1]: \big|p_n(x) - \sqrt{x}\big| &lt; \epsilon$ Edit:  I just had a thought.  I'm given a version of Dini's theorem that reads  Let $X$ be a compact metric space, $(f_n)_n$ a sequence of real-valued continuous functions on $X$, $f$ a continuous function on $X$ such that: (i) $f_n \rightarrow f$ pointwise on $X$, and (ii) $f_n(x) \geq f_{n+1}(x) $ for all $x \in X, n \in \mathbb{N}$. Then: $f_n \rightarrow f$ uniformly on $X$.  Obviously, (ii) is not true for my sequence, but what if I took $f_n(x) = -p_n(x)$ and $f(x) = - \sqrt{x}$?  Then I could prove (ii) to be true, I know that $[0,1]$ is compact, and $f_n(x)$ is real-valued.  The only part I would need to prove is (i), which is where I get confused.  Correct me if I'm wrong, but after that I could say that since $(-p_n)_n \rightarrow -\sqrt{x}$, then $(p_n)_n \rightarrow \sqrt{x}$.  Any advice?
19896
I am trying to show the following     Let $X_{n}$ be a sequence of random variables which converge in probability to $X$. Show if $f$ is a continuous function on $\Bbb R$, then $f(X_{n}) \to f(X)$ in probability.   I think if we know $f$ is uniform continuous, then this will be quite trivial. I try to proceed from this point by using compact exhaustion, but it didn't work. Could someone give me help on this?  Thanks in advance!
1736
Let $X$ be a metric space and let $\{ f_n \}_{n \geq 1}$ be a sequence in $C(X)$. Show that if $\{ f_n \}_{n \geq 1}$ converges uniformly on $X$ then $\{ f_n \}_{n \geq 1}$ is both uniformly bounded and equicontinuous.  Could I get a hint?
23668
Let $\mu$ be a $\sigma$-finite measure on a $\sigma$-algebra $\mathcal{A}$ of subsets of $\Omega$. Show that there exists probability measures $\mu_n$ on $\mathcal{A}~(n \in \mathbb{N})$ and $a_n&gt;0$ such that for every set $A \in \mathcal{A}:$ $$\mu (A)= \sum_{n=1}^{\infty} a_n \mu_n (A).$$  Is there anyone can give me hints or show how to start proving this question?
80485
$(a_n)$ is a bounded sequence with the following condition  $a_{n+1}\geq a_n-\frac{1}{2^n}$   The sequence converges, but how do we show it?
83662
Let $\{f_k\}_{k=1}^\infty\in L^{loc}_1(\mathbb R^n)$ be a sequence of real value functions such that  $\mbox{supp}(f_k)\subseteq\{|x|\leq k^{-1}\}$, $\quad \int f_k(x)dx=1$ $\quad for\quad all \quad k=1,2........$  show that the sequence $\{f_k^2\}_{k=1}^\infty$ does not converge in $\mathscr{D}^\prime(\mathbb R^n)$ $as \quad k\rightarrow \infty$
36284
Let $(X,\mathcal{A})$ be a measurable space, with measure $\mu$. Let $\{E_n\}_{n \in \mathbb{N}} \subseteq \mathcal{A}$ be a sequence of measurable sets, with $E_{n+1} \subseteq E_n, \ \forall n \in \mathbb{N}$, that is a decreasing sequence, and $\mu(E_n)=+ \infty, \ \forall n \in \mathbb{N}$. Let $E=\bigcap_{n \in \mathbb{N}} E_n$ the limit set of the sequence, $E$ is measurable for definition. Is true that $\mu(E)=+\infty$? If not in all cases, is true with $X=\mathbb{R}^{N}$, $\mathcal{A}$ the collection of Lebesgue-measurable sets, and $\mu$ the Lebesgue measure on $\mathbb{R}^N$?
44461
I would appreciate your help!  How can we show that if a sequence of positive real numbers $a_n$ converges to $a\in\mathbb{R}$ with $a&gt;0$, then $(a_0 a_1\cdots a_n)^{\frac{1}{n}}$ converges to $a$.