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14

No. The cardinality of the Borel $\sigma$-algebra on $\mathbb R$ (or on $[0,1]$) is the same as that of $\mathbb R$. This is a non-trivial fact that can be proved by transfinite induction; see, for example, Billingsley (1995). On the other hand, the cardinality of the set of all subsets of the Cantor set is the same as that of $2^{\mathbb R}$, the power set ...


4

HINT: Consider the change of variables $u=x^2-y^2$, $v=xy$. Elaborating: By symmetry, we get $4$ times the integral over the region $S=\{(x,y): x,y\ge 0,\ -1\le x^2-y^2\le 1,\ xy\le 1/2\}$. Consider $R=\{(u,v): -1\le u\le 1,\ 0\le v\le 1/2\}$ and $g\colon R\to S$, where $g^{-1}(x,y) = (x^2-y^2,xy) = (u,v)$. Note that $\det(Dg^{-1})(x,y) = 2(x^2+y^2)$. ...


3

There is something which is sometimes called the subsequence principle. It states that $x_n \to x$ holds if and only if every subsequence $(x_{n_k})_k$ admits a further subsequence $(x_{n_{k_l}})_l$, which converges to $x$. Note that the limit $x$ has to be fixed. This holds as soon as the notion of convergence is induced by a topology. This should help ...


3

Let $f_k(x) = \exp(-kx \sin^2 x)$. We show that $f_k$ is not integrable on $[0,+\infty)$. We have $$ \begin{align*} \int_{\pi/2}^{+\infty} f_k(x) \, dx &= \sum_{n \geq 1} \int_{n\pi-\pi/2}^{n\pi +\pi/2} \exp(-kx \sin^2 x) \, dx \\ &= \sum_{n \geq 1} \int_{-\pi/2}^{\pi/2} \exp[-k(x + n\pi)\sin^2 x] \, dx \\ &=\int_{-\pi/2}^{\pi/2} \left( \sum_{n ...


2

If $X:\Omega\rightarrow\mathbb R$, i.e. if $X$ takes values in $\mathbb R$, then the statement is always true. We have $\{X\in\mathbb R\}=\bigcup_{n\in\mathbb N}\{|X|<n\}$ where $\{|X|<1\}\subseteq\{|X|<2\}\subseteq\cdots$. Based on that it can be shown that $P(\{|X|<n\})$ converges to $P(\{X\in\mathbb R\})=1$. Edit: For a proof of that ...


2

Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space and $X:\Omega\to\mathbb R$ a random variable. Let $S_n=\{\omega\in\Omega : |X(\omega)|<n\}$. Then $S_n\subset S_{n+1}$ and $$\bigcup_{n=1}^\infty S_n = \Omega, $$ so $$\mathbb P\left(\bigcup_{n=1}^\infty S_n\right)=\lim_{n\to\infty} \mathbb P(S_n)=\mathbb P(\Omega) = 1. $$ $\mathbb P(S_n)$ is an ...


2

As a first approach you can do this component wise. The trouble with this is that usually you want to have a definition which does not depend on the choice of a basis, so this is what you would have to check in this case. A (very) general definition can be found in several places, e.g. in Rudin's functional analysis, chapter 3, section 'Vector valued ...


2

The point of this exercise is to demonstrate the counterintuitive fact that there exist strictly increasing singular-continuous functions, i.e. strictly increasing continuous functions whose derivative vanishes almost everywhere. The Cantor function $F$ is continuous, increasing (but not strictly) and $F^\prime=0$ a.e. (see an earlier chapter). Let ...


2

A probability measure is defined on a $\sigma$-algebra $\mathcal{A}$, not just an algebra. Also, you don't have to show anything, it's part of the definition of a probability measure $m$ that $m\bigl(\cup_{n=1}^\infty A_n\bigr) = \sum_{n=1}^\infty m(A_n)$ for disjoint $A_1, A_2, \dots$.


2

Hint: Any (nonempty) union or intersection of elements of this $S$ contains both positive and negative reals. This is true of the generating "double intervals" which are symmetrically chosen unions of two half intervals. So when these are combined to fill out the generated sigma algebra, one only gets sets having both positive and negative reals in them (or ...


2

Let $A$ be the $\sigma$-algebra generated by the intervals $(a,b]$ for $a,b \in (0,1]$ with $a < b$, and let $B$ be the Borel $\sigma$-algebra. To show $A \subseteq B$, it is enough to show that each interval $(a,b]$ as above is a Borel set. If $b=1$, this is clear, because $(a,1]$ is open. If $b < 1$, then $(a,b]$ is the intersection of the countable ...


2

By the Monotone Convergence Theorem, $$\int X_n\,d\mu\to \int Y\,d\mu\ge \int_{\{Y=\infty\}} Y\,d\mu=\infty\cdot \mu(Y=\infty)=\infty, $$ since $\mu(Y=\infty)>0$. An alternate proof: Let $\mu(Y=\infty)=a$. For any $M$, let $E_{M,n}=\{X_n\ge M\}$. Since $X_n\to Y$, it follows $E_{M,n}\nearrow \{Y_N\ge M\}\supseteq\{Y_n=\infty\}$, so ...


2

Since $(|f_n-f|)_{n\geqslant 1}$ is a sequence of non-negative functions we have $$ \int \sum_{n\geqslant 1} |f_n-f| \,\mathrm d\mu=\sum_{n\geqslant 1}\int |f_n-f|\,\mathrm d\mu<\infty. $$ Therefore, $\sum_{n\geq 1}|f_n-f|<\infty$ almost surely and hence also $\lim_{n\to\infty}|f_n-f|\to 0$ almost surely.


2

You need to fix $x>0$. Then there is $K$ such that $\frac{1}{2^k}<x$, for $k>K$. Then $f_k(x)=0$, for $k>K$. Then $f(x)=0$. For $x\leq0$, $f_k(x)=0$, for all $k$. Then $f(x)=0$. Now, for $L^1$ we can integrate. ...


2

You have that $$\mu(E) \leq \sum_{n \geq 1}\mu(E_n),$$ where the $E_n$ cover $E$. Since this inequality is true for every collection $E_n$ covering $E$, you have that $\mu(E)$ is a lower bound for the set $$\left\{\sum^\infty_{n=1}\mu(E_n)|E_n\in R, E\subset\cup^\infty_{n=1}E_n\right\},$$ so that $\mu^*(E) \geq \mu(E)$ follows from the definition of infimum. ...


2

Let we set $x=\rho\cos\theta$ and $y=\rho\sin\theta$. The first condition translates into $|\cos 2\theta|\leq\frac{1}{\rho^2}$, the second one into $|\sin(2\theta)|\leq\frac{1}{\rho^2}$. If we fix $\rho>0$, the Lebesgue measure of the set: $$ E_{\rho}=\left\{\theta\in[0,2\pi): \max(|\cos(2\theta)|,|\sin(2\theta)|)\leq\frac{1}{\rho^2}\right\} $$ is $2\pi$ ...


2

take $X := \{a,b,c\}$ and $A_1 := \{ \{a\}, \{b,c\}, \emptyset, X\}$, $A_2 := \{ \{b\}, \{a,c\}, \emptyset, X\}$ and show that $A_1 \cup A_2$ is not a $\sigma$-algebra


2

The set $A = \{x \in \mathbb{Q}^{\omega} : \{ x(n) : n < \omega \} \text{ is reverse well order as a suborder of rationals} \}$ is not Borel. Luzin and Sierpinski showed that it is in fact $\Pi_1^1$-complete - see Kechris, Classical descriptive set theory, page 213. Now let $f_n:\mathbb{Q}^{\omega} \to \mathbb{R}$ be $f_n(x) = x(n)$.


2

I'm not sure what you mean by $C_c([0,1])$, but you can use $C[0,1]$. Let $f_j$ be a dense sequence in $C[0,1]$ (this being a separable Banach space). The sequence $\frac{1}{n} \sum_{i=1}^n f_1(x_i)$ is bounded, so for some increasing sequence $N(1,1), N(1,2), N(1,3), \ldots$ of positive integers, $\frac{1}{N(1,n)} \sum_{i=1}^{N(1,n)} f_1(x_i)$ converges. ...


1

$$f \mapsto \mu_n(f) = \frac{1}{n}\sum_{i=1}^n f(x_i)$$ is a probability measure. So you have a sequence of probability measures $(\mu_n)$ and we want to show that is has a convergent subsequence ( convergence in the weak topology). Take $(f_l)$ a dense family in $C[0,1]$. Start with $l=1$. The sequence $(\mu_n(f_1))_{n\in \mathbb{N}}$ is bounded in ...


1

Let it be that $X=A\cup B=U\cup V$ with $A\cap B=\emptyset=U\cap V$. Then $\mathcal{A}=\left\{ \emptyset,A,B,X\right\} $ and $\mathcal{V}=\left\{ \emptyset,U,V,X\right\} $ are both $\sigma$-algebras. Is $\mathcal{A}\cup\mathcal{V}$ a $\sigma$-algebra? Not if $A\cup U\notin\mathcal{A}$ and $A\cup U\notin\mathcal{V}$.


1

$\{\emptyset,[0,3/4),[3/4,1),[0,1)\}$ and $\{\emptyset,[0,1/2),[1/2,1),[0,1)\}$ are $\sigma-$algebras but their union $\{\emptyset,[0,3/4),[3/4,1),[0,1/2),[1/2,1),[0,1)\}$ is not. $[0,3/4)\ minus \ [1/2,1)=[1/2,3/4)$ is not in the union.


1

Note that $d(A,B) = \| 1_A-1_B\|_1$, so we can use properties of $L^1(\mathbb{R})$ to show completeness. As Michael noted in the comments above, one needs to be careful with infinities. Suppose we have a $d$-Cauchy sequence of sets $A_n$. Without loss of generality we can presume that $d(A_n,A_1) < 1$ for all $n$. Let $f_n = 1_{A_n} - 1_{A_1}$. We have ...


1

Note: The following is not an answer, but merely some thoughts which might or might not be helpful to you. First note that you confused(?) your inequality signs. I think you want $$ \gamma_{n}\left(\left\{ x\in\mathbb{R}^{n}\,\mid\,\left\Vert x\right\Vert ^{2}\geq\frac{n}{1-\varepsilon}\right\} \right){\color{red}\leq}e^{-\varepsilon n/4} $$ and $$ ...


1

Your sequence is going to zero pointwise. On the one hand, $f_k(0) \equiv 0$. On the other hand, if $x>0$ and $k>-\log_2(x)$, then $f_k(x)=0$. Additionally, your sequence is dominated by $f(x) = \frac{1}{\sqrt{x}} \chi_{[0,1]}(x)$, which is in $L^p$ for $p \in [1,2)$. So it is definitely going to zero in $L^p$ for $p \in [1,2)$. It does not go to zero ...


1

It seems the following. The answer is negative provided $n>2$. Let the space $X$ consists of $n+2$ elements of equal measure, $\{A_i\}$ be the family of all two element subsets of $X$. Then $\mu(A_i)=2/(n+2)>1/n$ for each $i$. But for any $n$-element subset $X’$ of $X$ we have $X\setminus X’=A_i$ for some $i$.


1

I don't think your solution is correct. In particular If $X_i \in \mathcal M$, then $\bigcup_{i = 1}^n X_i, \bigcap_{i = 1}^n X_i \in \mathcal M$ is going to be true (assuming $M = \mathcal P(X)$ is the power set). Yes the unions are countable, but all you need to show to get that the set is a $\sigma$-algebra is that countable intersections, unions are in ...


1

If $X \colon \mathbb{R} \longrightarrow \mathbb{R^{+}}\bigcup\left\{0\right\}$ is defined as $ X(x) = x^2$, then observe that, for $b\ge a\ge 0$, $$ \begin{eqnarray*} ...


1

Yes. Lebesgue $n$-dimensional measure on $\mathbb R^n$ is $n$-dimensional Hausdorff measure, and for any nonexpansive map $f$ of metric spaces and any $d \ge 0$, it is easy to show from the definition that $\mathscr H^d(f(A)) \le \mathscr H^d(A)$ where $\mathscr H^d$ is $d$-dimensional Hausdorff measure.


1

Hint: There are actually a few proofs; here's one. Positive measures preserve null sets: $$\lambda(N)=0:\quad\lambda(E)=0\quad(E\subseteq N)\implies\mu(E)=0\quad(E\subseteq N)$$ So the Radon-Nikodym must vanish here: $$E\subseteq N\quad0=\mu(E)=\int_E u\mathrm{d}|\mu|\implies u1_N=0\mod{|\mu|}$$ That in turn gives absolute continuity: ...



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