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4

Homeomorphisms, in general, preserve neither completeness nor total boundedness. Consider $\Bbb R$ and $(0,1)$.


3

Asaf's answer is quite fine, but perhaps too specific to Lebesgue measure. More basically, you can use just the regularity of Lebesgue measure. This tells you that if your measurable set $X$ is not null, then it contains a closed set $C$ which is not null either. This $C$ is certainly uncountable, and so (as for any uncountable Polish space) one can embed ...


3

Consider $(\mu_n)_{n\geqslant 1}$ a Cauchy sequence for the metric $\rho$. Then for each $f$ (measurable) and bounded by $1$, the sequence $\left( \int_X f(x)\mathrm d\mu_n(x)\right)_{n\geqslant 1} $ is Cauchy. In particular, for each measurable subset $A$ of $X$, the sequence $(\mu_n(A))_{n\geqslant 1}$ is convergent. By the Vitali–Hahn–Saks theorem, we ...


3

Let $\varepsilon>0$, then since $\int_{\Omega}|X|dP<\infty$, there exists an $n>0$ such that: $$nP(|X| > n) = \int_{\{\omega\in\Omega : |X(\omega)|>n\}}ndP \leq \int_{\{\omega\in\Omega : |X(\omega)|>n\}}|X(\omega)|dP < \varepsilon.$$ Therefore we have $$nP(|X| > n) < \varepsilon.$$


3

Let $A \in \mathcal{F}$ be a measurable set. Fix $\epsilon>0$. By assumption, there exists an open set $B$ such that $K \backslash A \subseteq B$ and $$\mu(B) \leq \mu(K \backslash A) + \epsilon. \tag{1}$$ The set $\tilde{K} := K \backslash B = B^c \subseteq A$ is compact and satisfies $$\begin{align*} A \backslash \tilde{K} = A \backslash B^c &= A ...


3

Let $C \subset \Bbb{R}$ denote the Cantor set, and $V \subset \Bbb{R}$ denote the Vitali set. It is well known that $C$ is Borel of measure zero, $V$ is not measurable, and that they both have the cardinality of continuum. Let $f:C \longrightarrow V$ be any bijection. Extend $f$ to a function $f': \Bbb{R} \longrightarrow \Bbb{R}$ setting $f'(x)=0$ if $x ...


3

As Did remarks, $\alpha-1/n\in\mathbb{R}$, so $$A_{\alpha-1/n}=\{x\in X:f(x)>\alpha-1/n\}\in X.$$ Everything is well-defined. Kind of strange that they make this into an explicit claim.


3

If I am not mistaking, the statement Sphere is the only closed surface in $\Bbb R^3$ that minimizes the surface area to volume ratio. is equivalent to either of the following statements: Sphere is the only closed surface in $\Bbb R^3$ that minimizes surface area, enclosed volume fixed. Sphere is the only closed surface in $\Bbb R^3$ that maximizes ...


2

Let me expand upon njguliyev's comment. Note that $$E_1\cup E_2 = E_1\cup(E_2\setminus E_1) = E_1 \cup (E_2\cap E_1^c)$$ so \begin{align*} T\cap(E_1\cup E_2) &= T\cap(E_1\cup(E_2\cap E_1^c))\\ &= (T\cap E_1)\cup(T\cap(E_2\cap E_1^c))\\ &= (T\cap E_1)\cup(T\cap E_2\cap E_1^c) \end{align*} and therefore \begin{align*} m_e(T\cap(E_1\cup ...


2

The problem of to show that the surface with minimal area that enclose a given volume is a sphere is not at all simple. Using the calculus of variation we can show that the surface must have constant curvature. I sketch the proof: Given a smooth closed surface $S$ in $\mathbb{R}^3$, let $A$ its area and $V$ the volume enclosed. As varied surface consider ...


2

If $Y,Z:\Omega\rightarrow\mathbb R$ are both $\mathcal G$-measurable then so is the function $X:\Omega\rightarrow\mathbb R^2$ prescribed by $\omega\mapsto\langle Y(\omega),Z(\omega)\rangle$. Function $f:\mathbb R^2\rightarrow\mathbb R$ prescribed by $\langle y,z\rangle\mapsto y-z$ is continuous hence measurable if domain and codomain are both equipped with ...


2

Note. This answer is based on the assumption that your random variables have codomain $\Bbb R$, i.e., $Y, Z : \Omega \to \Bbb R$. Do you remember this definition for a random variable $X$ to be $G$-measurable? It is measurable if for every $\alpha \in \Bbb R$, $\{ \omega \mid X(\omega) > \alpha \} \in G$. Now, do you believe that $X$ is $G$-measurable ...


2

Fix $\epsilon>0$. Since $X$ is integrable, there exists $R>0$ such that $$\int_{|X| \geq R} |X| \, d\mu < \epsilon.$$ Thus, $$\begin{align*} \left| \int_{A_n} X \, d\mu - \int_A X \, d\mu \right| &= \left| \int (1_{A_n}-1_A) X \, d\mu \right| \\ &\leq 2 \int_{|X| \geq R} |X| \, d\mathbb{P} + \int_{|X|<R} |1_{A_n}-1_A| \cdot |X| \, d\mu ...


2

Another approach: suppose $x_n$ increases to $x$. Then $A \cap (-\infty,x_n]$ is an increasing sequence of sets, whose measure converges to $m(A \cap (-\infty,x))=m(A \cap (-\infty,x])$ by continuity of measure from below, along with the fact that the measure of a singleton is zero. Here we do not need that $A$ has finite measure. If you suppose $x_n$ ...


2

It is not a trivial theorem, and it certainly needs the axiom of choice (well, after a certain number of unions and complements have been applied anyway). So the proof is not very obvious. First we show that given a Polish space $X$ (separable and completely metrizable space), then every uncountable closed set has size continuum, this is done by essentially ...


2

Just use Fubini and integrate, $$ 0 = \int_Y 0 d\lambda(y) = \int_Y \mu(E^y)d\lambda(y) = \int_Y\int_X 1_{E} d\mu(x)d\lambda(y) $$ $$ = \int_X\int_Y 1_{E} d\lambda(y)d\mu(x) = \int_X \lambda(E_x) d\mu(x) $$ Since $\lambda(E_x)$ is positive and its integral over $X$ w.r.t $\mu$ is zero, it must be that $\lambda(E_x) = 0$ for $\mu$-a.e. $x$.


2

On 1): Let $A\in \mathfrak{K}_{\infty}$ and $B\in\mathfrak{T}$. Then $A\in \mathfrak{X}_n$ for some $n$ and since $\mathfrak{X}_n$ and $\mathfrak{T}$ are independent $\sigma$-algebra's we are allowed to conclude that $P(A\cap B)=P(A)\times P(B)$. This proves that $\mathfrak{K}_{\infty}$ and $\mathfrak{T}$ are independent. Here $\mathfrak{K}_{\infty}$ is ...


2

Let $(\Omega,\mathcal A,P)$ be a probability space and let $\mathcal B$ denote the Borel $\sigma$-algebra on $\mathbb R$. Any random variable $X:\Omega\rightarrow\mathbb R$ induces a probability $P_X$ on measurable space $(\mathbb R,\mathcal B)$. This by: $$P_X(B)=P(\{X\in B\})$$ for $B\in\mathcal B$. This $P_X$ is the distribution of $X$ and is ...


2

If $Y$ is a nonnegative random variable then $\mathbb{E}Y=\int_{0}^{\infty}P\left\{ Y\geq y\right\} dy$ (Are you familiar with that?) From $P\left\{ Y\geq\lceil y\rceil\right\} \leq P\left\{ Y\geq y\right\} \leq P\left\{ Y\geq\lfloor y\rfloor\right\} $ it follows that $$\sum_{n=1}^{\infty}P\left\{ Y\geq n\right\} =\int_{0}^{\infty}P\left\{ Y\geq\lceil ...


2

$$\int_\Omega |X|\, dP < \infty \implies \int_{\{|X| > n\}} |X|\, dP \to 0$$ by the dominated convergence theorem. Since $nP(|X|>n)$ is bounded above by the last integral, we have it.


2

By Fubini's theorem, we have $$\begin{align*} \int_0^\infty t^k\mathbb P(X\geqslant t)\mathsf dt&= \int_0^\infty t^k \mathbb E\left[1_{\{\omega:X(\omega)\geqslant t\}} \right]\mathsf dt\\ &=\int_0^\infty t^k\int_{\Omega} 1_{\{\omega:X(\omega)\geqslant t\}}\mathsf d\mathbb P\;\mathsf dt\\ &=\int_{\Omega}\int_0^{X(\omega)}t^k \mathsf dt\; \mathsf ...


2

When you calculate $\Lambda_ng $, you are you are adding over more boxes, i.e, you are subdividing the boxes from $\Omega_n $. But $g $ is constant over those additional boxes. Concretely, a fixed $x\in P_N $ corresponds to some box $Q\in\Omega_N $; then, as $g $ is constant in $Q $, $$ 2^{-nk}\sum_{y\in P_n\cap Q}g (y)=2^{-nk}g (x)\sum_{y\in P_n\cap Q}1 ...


2

Hints: Fix $\epsilon>0$. Fix $k \in \mathbb{N}$. Using Markov's inequality, show that $$A_N^k := \left\{x; \exists n \geq N: |f_n(x)-f(x)| \geq \frac{1}{k} \right\}$$ satisfies $$m(A_N^k) \leq k \sum_{n=N}^{\infty} \|f_n-f\|_{L^1}.$$ Conclude from the first step that there exists $N=N(k)$ such that $m(A_{N(k)}^k) \leq \epsilon 2^{-k}$. Set $$A := ...


2

Basic is that $\nu\left(\bigcup_{n=1}^{\infty}A_{n}\right)=\sum_{n=1}^{\infty}vA_{n}$ if the $A_{n}$ are measurable and disjoint. On this it can be proved that $A_{n}\uparrow A$ implies $\nu A_{n}\uparrow vA$. This by setting $B_{1}:=A_{1}$ and $B_{n}:=A_{n}-A_{n-1}$ if $n>1$. Then: ...


2

No, pick $n=1$, $f(x)=\xi(x)e^{-|x|}$ and $g(x)=e^{x^2}$, where $\xi \equiv 0$ near $x=0$. But even $f(x)=e^{-x^2}$, as Daniel Fischer points out in a comment. Here the point is that any continuous function is locally integrable, and $f$ cannot compete with functions that blow up at infinity faster than polynomials.


1

Check here or here for the definition of a Radon measure. A measure $\nu: \mathcal{F} \to \mathbb{R}$ is concentrated on a set $A \in \mathcal{F}$ if $\nu(A^c) = 0$. Thus if $\nu$ is concentrated at a point (say $\omega$), then for any $A \in \mathcal{F}$ we have $$ \nu(A) = \begin{cases} \nu(\Omega) &\text{ if $\omega \in A$} \\ 0 &\text{ if ...


1

This is a nice problem. $$\int_0^a g(s)ds=\int_0^a\int_s^a\displaystyle\frac{f(t)}{t}dtds $$ Following your idea $$\int_0^a\int_s^a\displaystyle\frac{f(t)}{t}dtds=\int_0^a\int_0^a\displaystyle\frac{f(t)}{t}\chi_{[s,a]}(t)dtds$$ Using Fubini we can change the order of integration: ...


1

Both $F$ and $[0,1]\setminus C$ must be dense in $[0,1]$ (for different reasons: the former because it has measure 1 and the latter because $C$ is nowhere dense). Moreover, since $m(F) = 1$ and $m(C) > 0$, we know that $F \cap C \not= \emptyset$. Take $x \in F \cap C$. Any interval around $x$ contains points of $F \setminus C$, which shows the function in ...


1

No, the proof in the book is correct. By definition, $$[0,T_A[ = \{(\omega,t); 0 \leq t < T_A(\omega)\}.$$ Since $T_A(\omega) = \infty$ for $\omega \notin A$ and $T_A(\omega)=T(\omega)$ for $\omega \in A$, we get $$[0,T_A[ = \{(\omega,t); \omega \in A, 0 \leq t< T(\omega)\} \cup \{(\omega,t); \omega \notin A, t \geq 0\}.$$ Since $$\{(\omega,t); ...


1

This is an application of Fatou's Lemma. Let $E \subset X$ be $\mu$-measurable. By using Fatou's Lemma twice we get $$\begin{align*} \int_E f \, \mathrm d\mu &= \int_X f \, \mathrm d\mu - \int_{E^{\mathrm c}} f \, \mathrm d\mu \\ &\geq \lim_{n \to \infty} \int_X f_n \, \mathrm d\mu - \liminf_{n \to \infty} \int_{E^{\mathrm c}} f_n \, \mathrm d\mu ...



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