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0

The following version of the open mapping theorem holds: If $T:Y\to X$ is a bounded operator between Banach spaces with $B_X \subseteq c \,\overline{T(B_Y)}$ for some constant $c>0$ (and the unit balls $B_X$ and $B_Y$) then $T$ is surjective (and open). Sometimes such a $T$ is called almost open. (The proof is in every book on functional analysis, the ...


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 $x_n$ ...


1

The first statement is written incorrectly. Here is the correct result, which follows from the inclusion-exclusion principle; $B = A_1\cup A_2\cup A_3$, so $$\begin{align*} 1_B &= 1_{A_1\cup A_2\cup A_3} \\ &= 1_{A_1} + 1_{A_2} + 1_{A_3} - (1_{A_1\cap A_2} + 1_{A_1\cap A_3} + 1_{A_2\cap A_3}) + 1_{A_1\cap A_2\cap A_3}\\ &= \sum_{i=1}^3 1_{A_i} - ...


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 ...


0

$\newcommand{\E}{\operatorname{E}}$If $\E(|X|)<\infty$ then, since $\lceil |X|/c\rceil\le |X|/c+1$, you have $\E\lceil |X|/c\rceil <\infty$. Let $W=\lceil |X|/c\rceil$. Then use the fact that for a positive integer-valued random variable $W$, $$ \E W = \sum_{w=1}^\infty \Pr(W\ge w). $$ A somewhat similar argument handles the "only if" case.


2

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 $$nP(|X| > n) < \varepsilon.$$


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.


1

Let: $E\in S$. For every $F\in \mathcal R$, we know that $E\cap F \in \mathcal R$. So, $F-(E\cap F) \in \mathcal R$. And it's easy to see that, $E^c \cap F=F-(E\cap F)$. Thus, $E^c \cap F \in \mathcal R$. So, $E^c\in S$.


1

Some steps: The intersections of two algebras $\mathcal F_1$ and $\mathcal F_2$ still is an algebra. Indeed, the whole set belongs to the intersection of the algebras, as well as the complement of an element of $\mathcal F_1\cap \mathcal F_2$. Stability by finite intersections also holds. Therefore, the question is actually equivalent to the following ...


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.


1

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 ...


1

Let $(Y^m)_{m \in \mathbb{N}}$ be a sequence of bounded continuous processes converging in $L^2$ to $Y$. By assumption, there exists a sequence $(Y^{m,n})_{n \in \mathbb{N}}$ of simple processes such that $$Y^{m,n} \stackrel{n \to \infty}{\to} Y^m$$ for each fixed $m \in \mathbb{N}$. Taking a further subsequence (if necessary), we may assume that ...


0

The even $p$ norms corresponds directly to the $p$th moment of the random variable. $$E[X^p] = ||X||_p^p$$ The association for odd $p$ is not as clear because the definition of the norm requires taking the absolute value of the outcomes. At best you can note: $$E[|X|^p] = ||X||_p^p$$ In this way, for a symmetric distribution, you can ...


0

Following the hint from Exercise 24 here consider $$ F_N := \bigcup_{n \geq N} E_n \Delta E.$$ By monotone convergence for sets $$\lim_N m(F_N) = m(\bigcap_N F_N) = 0,$$ the last equality coming from $\chi_{E_N} \rightarrow \chi_E$ a.e.. Therefore $$ 0 \leftarrow m(F_N) \geq m(E_N \Delta E) \geq m(E_N \setminus E) = m(E_N) - m(E),$$ thus the conclusion.


1

As $X = \{0,1\}$, there is not much freedom one has for $\mu^*$. Because $\mu^*$ is an outer measure, we are forced by subadditivity to satisfy \begin{equation} \mu^*(A) \leq \mu^*(X) \leq \mu^*(\{0\}) + \mu^*(\{1\}). \end{equation} where $A = \{0\}$ or $\{1\}$. Suppose we choose $\mu^*(X), \mu^*(\{0\})$, and $\mu^*(\{1\})$ in such a way so that we have ...


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 ...


1

example Without assuming some "joint" measurability of $X_t(\omega)$ in $t$ and $\omega$ you are out of luck. We use the Continuum Hypothesis. There is a set $A \subseteq [0,1] \times [0,1]$ such that: $\qquad$For each $t \in [0,1]$,$\qquad \{\omega \in [0,1] : (t,\omega) \in A\}$ is countable, $\qquad$For each $\omega \in [0,1]$,$\qquad \{t \in ...


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 ...


0

Each $x\in (0,1)$ has a unique dyadic expansion $.x_1 x_2 \dots $ that doesn't end in all $1$'s. Using only these expansions, the map $f_k(.x_1x_2\dots ) = x_k$ is well defined on $(0,1).$ Now it's easy to see that if $F$ is the floor function and $g:(0,1) \to \mathbb {R}$ is measurable (measurable = Borel measurable), then $F\circ g$ is measurable. Thus ...


1

Fix any $k\in\mathbb N$ and define $$T_k(x)=\mathord.\,x_{n_1}\ldots\, x_{n_k}\quad\forall x\in(0,1].$$ I leave it to you to check that this function is measurable. Then, check that for each $x\in(0,1]$, $\lim_{k\to\infty} T_k(x)=T(x)$ pointwise and use the fact that the pointwise limit of measurable functions defines a measurable function. Hint: $T_k$ is ...


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

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 := ...


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 ...


1

The approximation of non-negative measurable functions works as follows: Sombrero lemma: Let $(E,\mathcal{A})$ be measurable space and $f: E \to \mathbb{R}$ be a non-negative mesasurable function. Then there exists a sequence of simple functions $(f_n)_n$ such that $0 \leq f_n \leq f$ and $f_n \uparrow f$ (i.e. $f_1(x) \leq f_2(x) \leq \dots$ and $\sup_n ...


0

Let $g\in L^1(|v|). For \;$j=r,i$,\; we \:have, v_{j} ^ {\pm} \leq v_j^{+}+v_j^{-}=|v_j| \leq |v|$, then $g\in L^1(v_r^{+})\cap L^1(v_i^{+})\cap L^1(v_r^{-})\cap L^1(v_i^{-})$ , then $g\in L^1(v)$


0

Actually you have answered what sets are of measure $0$. They are exactely subsets of union of all $B \in \mathcal{A}$ such that $x \not \in B$. Asume that there is mesurable set $A$, than: $$(1) \: \: \: \:\delta_x(A)+\delta_x(A^c)=1$$ and what is more measure of every set is $0$ or $1$ so if $A$ wasn't of measure $0$ it was of measure $1$ and thanks to ...


11

Yes. This is a trivial consequence of a theorem by Steinhaus: Suppose that $X$ has a positive measure, then $X-X=\{x-y\mid x,y\in X\}$ contains an interval around $0$. It is not hard to prove that if $X$ is infinite, then $X$ and $X-X$ are equipotent (there is a surjection from $X^2$ onto $X-X$, and there is an obvious injection from $X$ into $X-X$). ...


1

Take $f \in L^2$, $f \geq 0$. Define $A_\varepsilon = \{ x \in \Bbb{R} : f(x) \geq \varepsilon\}$ for every $\varepsilon >0$. Now $$ \mu(A_\varepsilon) < \infty\,, $$ hence $\mu( \Bbb{R} \setminus A_\varepsilon) = \infty$. Now $$ \mu( \Bbb{R} \setminus A_\varepsilon) = \mu \left( \bigcup_{q \in \Bbb{Q}} B(q, \varepsilon) \cap (\Bbb{R} \setminus ...


2

As pointed out in comments, you can't just set one value of $g$ to be negative and set $g = f$ everywhere else, because in order to define $L^p$ you make equivalence classes of functions that agree everywhere except a measure-zero set. So then you would essentially have $g = f$ under this paradigm, so this would not show the interior of your set is empty. ...


0

Hints: Denote by $\pi_j: B(n) \to Y, x \mapsto \pi_j(x) := x(j)$ the projection onto the $j$-th coordinate for each $j \in \mathbb{Z}$. Recall the following statement: Let $(X,\mathcal{A})$ be a measurable space such that $\mathcal{A}$ is generated by some family of sets $\mathcal{G}$, i.e. $\sigma(\mathcal{G}) = \mathcal{A}$. Then a mapping $\sigma: ...


0

For example if you assume that $$\liminf_{n\to\infty} f_n =0$$ almost everywhere and $$\int_A f_n d\mu >c>0$$ for almost all $n$.


1

If $A_n=A$ for each $n\in\mathbb N$ then evidently $\limsup A_n=A$. For any function $f:\mathbb N\rightarrow\mathbb N$ we have $A_{f(n)}=A$ for each $n\in\mathbb N$ so that also $\limsup A_{f(n)}=A$. Then $$\limsup A_{f(n)}=A\subseteq A=\limsup A_n$$ but no conditions like $f(n)\rightarrow\infty$ on function $f$ are necessary.


0

See Theorem 1.5.2 in Deitmar and Echterhoff's Prinicples of Harmonic Analysis.


3

It seems the following. I am not an expert in this question, but I think you may encounter some problems defining Haar measure on such a group. One of ways to define a (Haar?) measure $\mu$ on a non-locally compact group $G$ may be to consider a completion $\hat G$ of the group $G$. If the group $\hat G$ is locally compact and has a Haar measure ...


1

The Radon-Nikodym “derivative” is an a.e. define concept. Suppose $(X,S)$ is a measure space and $\mu,\nu$ are finite measures on $(X,S)$ with $\mu\ll\nu$, then the theorem is: Theorem. There exists $f\in L^{1}(X,\nu)$ a non-negative real-valued function, with $\mu(A)=\int_{x\in A} f(x)~\nu(dx)$ for all $A\in S$. There are all sorts of generalisations (to ...


1

The Radon-Nikodym derivative is a thing which re-weights the probabilities, i.e. it is a ratio of two probability densities or masses. It is used when moving from one measure to another, for whatever reason you have to do so. So, say $X$ is a random variable and you want to work out $\mathbb{E}_\lambda[X]$ - i.e.the expectation of $X$ in ...


1

Let $R$ be so large that $\Omega\subset B(x,R)$ (where $B(x,R)$ is the ball centered at $x$ and with radius $R$). Then, since the function is positive, you have, by domain-monotonicity, $$ \int_\Omega|x-y|^{1-n}\lambda^n(y)\leq \int_{B(x,R)}|x-y|^{1-n}\lambda^n(y). $$ Now, let $u=x-y$, and you get that the integral to the right equals $$ ...


0

In general, no. It depends on which $\sigma$-algebra equips the measurable space. Rudin's definition relies on the fact that if $Y$ is a topological space, we usually equip it with the Borel $\sigma$-algebra, which is generated by open sets.


0

It seems the following. The space $X_0$ of arbitrary functions $f : \mathbb{R} \to \mathbb{R}$ endowed with the topology of pointwise convergence is just the Tychonov product $\mathbb{R}^\frak c$. Thus in order to construct a required counterexample $X\subset X_0$ it suffices to construct a Tychonov (that is completely regular) counterexample $X$ of weight ...


0

Let $f, g: \mathbb{R} \to X$ be Borel measurable and $h(s) = f(s) + g(s)$. Consider mappings $$ \begin{align} \mathbb{R} \ni s & \stackrel{h_1}{\mapsto} (s, s) \in \mathbb{R}^2 \\[1ex] \mathbb{R}^2 \ni (s, t) & \stackrel{h_2}{\mapsto} (f(s), g(t)) \in X^2 \\[1ex] X^2 \ni (x, y) & \stackrel{h_3}{\mapsto} x+y \in X. \end{align} $$ Then $$s ...


2

In general $\mathbb{E}[X \cdot Y] = \mathbb{E}[X] \cdot \mathbb{E}[Y]$, which you are trying to do, applies only if $X,Y$ are independent.


2

Not without further assumptions, I guess (your last equation holds in general only if $X$ and $\mathbf{1}_A$ are independent). Take $X$ the random variable equal to $0$ with probability $9/10$, and $10$ with probability $1/10$; and $A=\{X=0\}$. Then $$\mathbb{E}[X] = 1$$ but $$\mathbb{E}[X\mathbf{1}_A] = 0$$ while $\mathbb{P}(A) = 9/10$.


1

Suppose $\alpha := \mu(\{e\}) > 0$, then $\mu(\{x\}) = \alpha$ for all $x\in G$. Now if $\epsilon := \alpha/2 > 0, \exists U$ open such that $e\in U$ and $$ \mu(U) < \mu(\{e\}) + \epsilon $$ Conclude that $U = \{e\}$ must hold. Suppose $G$ is not compact, then by local compactness, choose a neighbourhood $U$ of $e$ such that $\overline{U}$ is ...


1

Yes, this is correct. Let $(\Omega, \beta, \mathbb P)$ be a probability space and $\beta_1\subset\beta$ a $\sigma$-algebra. If $X:\Omega\to\mathbb R$ is $\beta_1$-measurable, then for each Borel set $B$, $X^{-1}(B)\in\beta_1\subset\beta$, so $X$ is $\beta$ measurable. Clearly the converse is not true (just take any non-degenerate random variable and ...


1

Here's a detailed sketch that (I hope) doesn't spoil all the fun: Define the arc length function by $$ s(t) = \int_{0}^{t} \|\gamma'(\tau)\|\, d\tau. $$ Because $\gamma$ is an injective parametrization and piecewise $C^{1}$, the function $s:[0, 1] \to [0, L]$ is strictly increasing and continuously differentiable. (This probably looks obvious to a ...


2

Let $\Omega = \{a,b\}$, $\mathcal F=2^{\Omega}$, $\mathbb P(\{a\})=0$, $\mathbb P(\{b\})=1$. Define $X(a)=0$, $X(b)=1$. Then $\mathbb P(X=1)=1$ so $X$ is almost surely constant, but $$X^{-1}(\{1\})=\{b\}\notin\{\varnothing,\Omega\},$$ so $X$ is not trivial $\sigma$-algebra measurable. The converse is true. Suppose $\sigma(X)=\{\varnothing,\Omega\}$ and ...


1

If a r.v. is constant, you have $[X\in B]$ is empty if the constant is not in $B$ and it is $\Omega$ otherwise. This is measurable in the trivial $\sigma$-algebra. It if is a.s. constant, then $[X\in B]$ is either a set of zero measure or measure 1. The completion of the trivial $\sigma$-algebra is the $\sigma$ algebra consisting of sets of full or zero ...


2

You arrived at:$$\sum_{r=1}^nrP_r=\sum_{r=1}^nn{n-1\choose r-1}p^r(1-p)^{n-r}$$ Now realize that the RHS equals: $$np\sum_{r=1}^{n}{n-1 \choose r-1}p^{r-1}(1-p)^{n-r}=np(p+(1-p))^{n-1}=np$$


0

You are starting this proof at the right way, but you are missing one important thing here: $(1) : (x+y)^n=\sum_{k=0}^n x^{n-k}y^k $. I will show the proof, cause it is now pretty straightforward. $\sum_{r=0}^n r Pr = \sum_{r=0}^n r \binom{n}{r} p^r(1-p)^{n-r} = \sum_{r=1}^n r \frac{n!}{r!(n-r)!} p^r(1-p)^{n-r}$ where we used that the first term is $0$. ...



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