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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 ...

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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 $x_n$ ...

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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 ...

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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 ...

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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 ...

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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.

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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 $$\mu^*(A) \leq \mu^*(X) \leq \mu^*(\{0\}) + \mu^*(\{1\}).$$ where $A = \{0\}$ or $\{1\}$. Suppose we choose $\mu^*(X), \mu^*(\{0\})$, and $\mu^*(\{1\})$ in such a way so that we have ...

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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 ...

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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 ...

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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 $g\in L^1(v)$

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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 ...

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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 $X-X$). ...

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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). $$Now, let u=x-y, and you get that the integral to the right equals$$ ...

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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.

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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 ...

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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

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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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