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7

Put $$g_n(x) = \frac{x}{n} f(x) \chi_{[0,n]}(x)$$ Then $$\frac{1}{n}\int_0^n xf(x) dx = \int_{0}^\infty g_n(x) dx$$ Also, $|g_n(x)| \leq |f(x)| = f(x)$ for all $x$, and $g_n(x) \rightarrow 0$ pointwise. Therefore the dominated convergence theorem applies, and \begin{align} \lim_{n \rightarrow \infty} \frac{1}{n}\int_0^n xf(x) dx &= \lim_{n ... 4 For X=\mathbb R, let, for each n\in\mathbb N, \begin{align*} f_n(x)=\begin{cases}1&\text{if x\in(-\infty,0],}\\n^2&\text{if x\in(0,1/n],}\\0&\text{if x\in(1/n,\infty).}\end{cases} \end{align*} Let \begin{align*} f(x)=\begin{cases}1&\text{if x\in(-\infty,0],}\\0&\text{if x\in(0,\infty).}\end{cases} \end{align*} Consider ... 4 First, P(E_n)=1/n simply by symmetry, because E_n happens when \max\{X_1,X_2,\ldots,X_n\} is realized with X_n. Since the common distribution is continuous there is almost surely no ex aequo and \max\{X_1,X_2,\ldots,X_n\} has as much chances to be realized with each X_k with 1\leqslant k\leqslant n, hence the result. Likewise, the ... 4 (1) Yes, your argument is correct. The fact that composition with T^{-1} preserves Sobolev classes also needs to be proved, but the proof is immediate from consideration of what this composition does to Cauchy sequences (wrt W^{1,p} norm) of smooth functions. (2) Yes, and this generalization is one of fundamental results for the theory of Sobolev ... 4 The idea is that, since the series \sum\limits_nP(X_n\ne-1) converges, Borel-Cantelli lemma (the simple one) indicates that, almost surely, there exists some finite N such that X_n=-1 for every n\geqslant N. Every such sequence (X_n) is such that S_n/n\to-1 hence P(S_n/n\to-1)=1. The strong law of large numbers does not apply, although (X_n) ... 3 We have\displaystyle \sum_{n=1}^\infty \mu\{x \in X : |f_n(x)| \geq \epsilon\} = \mu(\displaystyle \sum_{n=1}^\infty1_{ |f_n(x)| \geq \epsilon}) < \infty$$which means \displaystyle \sum_{n=1}^\infty1_{ |f_n(x)| \geq \epsilon} is finite \mu-a.e. That is to say for any \epsilon, there are only finitely many n such that |f_n(x)| \geq \epsilon ... 3 The family of all subsets of \mathbb{N} that are either finite or have a finite complement is an algebra, but not a \sigma-algebra. 3 Consider the map$$ \Phi : L^1 \to L^1, f \mapsto \varphi f. $$By assumption, this is a well-defined linear map. Use the closed graph theorem to show that it is bounded (there are some details to fill in here). This shows that the functional$$ \psi : L^1 \to \Bbb{C}, f \mapsto \int f \varphi d\mu $$is bounded (why)? Now use the characterisation of ... 3 No, such an E need not be measurable: Let A be a non-measurable set. Then E=A\cup(\Bbb Q\setminus A) is non-measurable (otherwise, A=\bigl( A\cup (\Bbb Q\setminus A)\bigr) \cap( \Bbb Q\setminus A)^C would be measurable). But, as \Bbb Q\subset E, every point x in E is a limit point of E \setminus\{x\}. 3 The notation P(Y\mid X=x) probably means the distribution \mu_x where the family (\mu_x)_x is a conditioning of the distribution of Y conditionally on X. Informally, \mu_x(B)=P(Y\in B\mid X=x) for every Borel subset B and every x. Formally, each function G_B:x\mapsto\mu_x(B) is such that P(Y\in B\mid X)=G_B(X) almost surely. 3 By "real random variable" you mean the values are real? That is, we have P(X=\infty)=0 ? Then sets$$ A_n = \{ X < n \} $$have finite measure, and$$ \Omega = \bigcup_{n} A_n . $$2 Consider X_n\equiv 1+\log n everywhere, and \lambda=1. Then every element belongs to LHS but RHS is empty... so this is a counterexample. 2 First, we can show that \overline{\mathcal B(\Bbb R)\times\mathcal B(\Bbb R)} = \mathcal B(\Bbb R^2), as every open disk -hence every open set in \Bbb R^2- can be covered by open rectangles (with sides parallel to the axes). Second, f(x) can also be considered as a function in two variables (but constant in y). More precisely we consider ... 2 Let (X_n)_{n=1}^\infty be a sequence of disjoint sets of strictly positive finite measure. Set g_n = \mathbf{1}_{X_n}. For each non-empty set A\subset \mathbb{N} let$$f_A(x) = \sum_{n\in A}g_n(x).$$Then \|f_A - f_B\| = 1 for distinct subsets A,B\subseteq \mathbb{N}. The power-set of \mathbb{N} has cardinality continuum so L_\infty(\mu) is ... 2 As commenters explained, the boundary of a compact set can have positive measure. How to proceed with the proof of$$\lim_{j \rightarrow \infty} \int_{K_j} f\, dm = \sup \left\{ \int_K\, f dm: K \subset \Omega, \ K \mbox{compact} \right\} $$then? The inequality \le holds because every K_j is also eligible for the supremum on the right. To prove ... 2 In general$$ \eqalign{P(X + Y \le z) &= \iint_{\{(x,y): x+y \le z\}} dx\; dy\; f_{XY}(x,y)\cr &= \int_{-\infty}^\infty dx \int_{-\infty}^{z-x} dy \; f_{XY}(x,y) }$$If X and Y are independent, f_{XY}(x,y) = f_X(x) f_Y(y) so this becomes$$ \int_{-\infty}^\infty dx \int_{-\infty}^{z-x} dy\;f_X(x) f_Y(y) = \int_{-\infty}^\infty dx\; ...

2

A non principal ultrafilter on the set of natural numbers is a tail set but it is far from Borel. What you are missing is that you need to make use of the fact that your set is Borel (or Lebesgue measurable) so that you can apply something like Lebesgue density theorem.

2

The issue is that you are not taking into consideration that convergence in measure guarantees a subsequence which converges pointwise almost everywhere to $f$ - not everywhere pointwise convergence. Everywhere pointwise convergence is necessary to guarantee that $f_n \to f$ pointwise if every subsequence has a further subsequence which converges to $f$ ...

2

For fixed $x \in \mathbb{R}$, define a mapping $$\tau_x: \mathbb{R} \to \mathbb{R}, y \mapsto y-x.$$ Then $\tau_x$ is continuous (hence measurable) and $\tau_x^{-1} = \tau_{-x}$. By definition, $$\int_{\mathbb{R}} |g(y-x)| \, m(dy) = \int_{\mathbb{R}} |g \circ \tau_x(y)| \, m(dy).$$ We can rewrite the right-hand side using image measures: $$... 2 Lemma: Let E \subset \Bbb{R} be uncountable. There is a countable set F\subset E such that E\setminus F satisfies your assumption. Proof: Let$$F := \{x \in E \mid \exists U_x \text{ neighbourhood of } x \text{ s.t. } U_x \cap E \text{ is countable}\}.$$The open(!) covering (U_x)_{x \in F} of F has a countable subcover (U_{x_n})_n, (because ... 1 Uncountable additivity are considered especially in set theory. For example, an uncountable cardinal \kappa is a real-valued measurable cardinal if and only if there exists a nontrivial \kappa-additive probability measure on \kappa. Although in some cases \sigma-additivity is enough. For example, the least cardinal which has a \sigma-additive ... 1 If you write out the equation in components and take the real and imaginary components of those constituent equations, you see that the set cuts out a proper real variety in \mathbb{R}^{2n} \times \mathbb{R}^{2n}, which thus has measure zero. Edit: In fact, you don't need to expand the equation at all: it's enough to see that (1) the equation is ... 1 Your set S is a level set of a smooth function f: \mathbb C^{n\times n }\to \mathbb C, with$$ f(x,y)=x^Ty-x^Tx, $$and hence is has measure zero. In fact S can be viewed as a hypersurface in \mathbb R^{2n\times 2n} of co-dimension 2. 1 I see what you intended now by that web link. Suppose f_n converges to f in measure, and there is a g such that |f_n|\leq g for all n, and \int g < \infty. I think you mean this: Your original sequence is \{\int f_n\}_{n=1}^{\infty}. Consider an infinite subsequence of this with indices in \mathcal{N}. We want to show that there ... 1 Since X is positive, E(X) \geq +\infty P(X = +\infty) = +\infty. Otherwise, you can use the definition of \int XdP, it is defined as the sup of integration of simple functions that approaches X from below. Remark integration of these simple functions on \{X = +\infty\} tends to +\infty 1 Define random variable Y by \omega\mapsto 0 if X(\omega)<\infty and \omega\mapsto +\infty otherwise. Then 0\leq Y\leq X hence 0\leq \mathbb E(Y)\leq\mathbb E(X). Y only take the values in \{0,+\infty\} and P(Y=+\infty)=P(X=+\infty)>0. This leads directly to \mathbb E(Y)=(+\infty)\times P(Y=+\infty)=+\infty 1 There is nothing wrong with the statement if f_n\to 0 in measure, then so does every subsequence f_{n_k}. It is wrong, however, to state that if a subsequence \{f_{n_k}\} converges pointwise, then so must \{f_n\}. No. (and 4) See above. 1 Let \{A_m\}_{m \in \mathbb{N}} be a sequence of sets in X of finite measure such that A_m \nearrow X, otherwise arbitrary. We shall assume \lim_{n \to \infty}\int_X|f_n-f|d\mu \neq 0, and reach a contradiction. By our assumption there is a positive number \hat{\varepsilon} and a subsequence \{f_{n_k}\} so that 2 \hat{\varepsilon} ... 1 For every x \in X, we have three possibilities: f(x) > 0, so f^+(x) is nonzero and f^-(x) is zero, hence x \in B and x \not\in A f(x) < 0, so f^-(x) is nonzero and f^+(x) is zero, hence x \in A and x \not\in B f(x) = 0, so both f^-(x) and f^+(x) are zero, hence x \in A and x \in B In all three cases, x is in either ... 1 Observe |f| + |f_n| - |f-f_n|\geq 0, with Fatou's lemma$$\liminf \int |f| + |f_n| - |f-f_n| \geq 2 \int |f| $$by the assumption \lim \int |f_n| = \int |f|, the left-hand-side of above inequality also equals$$\liminf \int |f| + |f_n| - |f-f_n| = 2\int |f| -\limsup \int |f-f_n|,$$combine them we get$$2\int |f| -\limsup \int |f-f_n|\geq 2 \int |f|  ...

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