# Tag Info

## New answers tagged measure-theory

0

If $M(s)$ is supposed to be the $\sigma$-Algebra generated by $s$, then it should suffice to use the definition of a measure and the general description of an element of $M(s)$. If $M(s)$ is not said algebra, please clarify.

1

It remains to show that the set $\{w\mid f(w)\ne0\}$ has measure $0$. Consider the nonzero rational $a$'s to complete the proof.

1

Indeed, your proof shows that $\{f > 0\}$ has measure zero. Similarly, you show $\{f < 0\}$ has measure zero, and hence $f=0$ a.s. Yes it really is this easy.

1

To answer my own question: We can construct the singletons as follows: $$\{i\} = A_i \backslash \cup_{t = i+1}^{\infty}A_t, \quad \forall i \in \mathbb{N}.$$ Now that we know $\sigma(\mathcal{H})$ contains all the singletons and therefore we know that this sigma-algebra is equal to the powerset of $\mathbb{N}$, because we can construct it by taking ...

0

Define $Y_n:=\left|X_n\right|^r$ and $Y:=\left|X\right|^r$. If $\limsup_n\mathbb E[Y_n]\leqslant E[Y]$, it may be not true that $\mathbb E[Y_n]$ is finite for each $n$. Even if it is the case, the best we can deduce is that $\sup_n\mathbb E\left[Y_n\right]$ is finite (see grand_chat's answer). But this is not enough for the uniform integrability, for example ...

0

Taking $f=1$, the statement would imply that for each compact subset $K$ of $M$, $\mathbb P_n(K)\to\mathbb P(K)$. There are counter-examples even for $M$ compact (say $[0,1]$ with the usual metric): take $\mathbb P_n:=\delta_{1/n}$.

0

Fix a sequence $(A_n \mid n \in \mathbb N)$ of sets. By definition $\liminf A_n = \bigcup_{m \in \mathbb N} \bigcap_{l \ge m} A_l$. In other words: $x \in \liminf A_n$ if and only if there is an $m \in \mathbb N$ such that $x \in A_l$ for all $l \ge m$. Thus $\chi_{\liminf_{n \in \mathbb N} A_n}(x) = 1$ if and only if there is some $m \in \mathbb N$ such ...

1

Yes, that's enough. Remember that $\limsup_n a_n = \lim_n\{\sup_{k\ge n} a_k\}$ for any sequence $\{a_n\}$ of real numbers. If the limsup is bounded above by some $A$, then there exists $N$ such that $$\sup_{k\ge N} a_k \le A +1.$$ But this implies that $a_k\le A +1$ for all $k\ge N$ ( since the sup is an upper bound on every term ), so $A+1$ is a uniform ...

0

"Integrability" of sequences corresponds to absolute convergence of series and not to the existence of the limit of partial sums one.You could check, to find a similar situation in "continuous" integration, that there exists (finite) the limit , for A tending to infinite, of the (Lebesgue) integral over [0,A] of sin(x)/x, while this function is not Lebesgue ...

0

He is probably referring to the Bochner integral. Edit: The Wikipedia page does a good job of presenting the definition. To answer your question in the comments, there is a natural theory of vector-valued measures, but when the vector space is $\mathbb{R}^d$ (and other finite products of measure spaces) these are just tuples $\mu = (\mu_1,\ldots,\mu_d)$ of ...

0

Method 1: Abbreviate $Y:=E[X|\mathcal F]$. Let $g(x)$ denote the right-hand derivative of $\varphi$ at $x$. Because $\varphi$ is strictly convex, we have $\varphi(x)>g(m)(x-m)+\varphi(m)$ for all $x\not=m$. Thus, $$\varphi(X)\ge g(Y)(X-Y)+\varphi(Y)$$ with strict inequality off $\{X=Y\}$ (almost surely). Taking conditional expectations in the ...

0

You can use the change of variables theorem and the fact that given J Jacobian matrix of the composition of a rotation and a translation , |det(J)|=1

2

It isn't true. The standard counterexample is to look at $L^2((0,1))$ with Lebesgue measure and take $f_n(x) = \sqrt{2} \sin(n \pi x)$. The functions $f_n$ are orthonormal in $L^2$, so by Bessel's inequality they converge weakly to 0. But pointwise, the sequence $\{f_n(x)\}$ diverges for every $x \in (0,1)$.

0

Note that it is sufficient to consider sets $A$ of the form $C\times D$, where $C$ and $D$ are open intervals (because these generate the $\sigma$-algebra). For these sets you only have to look at $f^{-1}C\cap g^{-1}D$.

2

Consider a Vitali set $A \subset [0,1]$, which is non measurable. The aim is to find measurable functions $f_a$ for each $a \in A$, such that $\sup\limits_{a \in A} f_a = \mathbb 1_A$. Hint :

1

Note that if $f=\chi_I$ almost everywhere and $f$ is continuous, you must have for example $$f\lvert_{(-\infty,0)}=0$$ since if for some $x \in (-\infty,0)$ you have $f(x)\neq0$, then from continuity you must have that $f(x)\neq0$ on some ball $B_\epsilon(x)$. But then $f$ differs from $\chi_I$ on a set that is not a measure zero set (namely $B_\epsilon(x) ... 0 First of all, you first define a function$X$, then, if you can demonstrate that its preimage is in the$\sigma$-algebra$F$, it is a random variable. As an example let's say you have a deck of 52 card well shuffled, so$\Omega=\{\text{cards in the deck}\}$. Let's say that you want to know the probability of extracting an even card (i.e. a card with an even ... 1 (i) if$\phi_1(x)=\phi(x)$except on a set of measure zero$A$, then$g(x,\phi(x))=g(x,\phi_1(x))$except on that same set of measure zero. Thus two functions which are members of the same a.e. equivalence class in$L^r$give two functions equivalent in$L^s$. (ii) holds because$x\mapsto g(x,\phi(x))$is a real function of$x$defined for$x\in \Omega$... 1 Fix$\varepsilon$. First us tightness to find a compact subset$K=K(\varepsilon)$of$C[0,+\infty)$such that$\mathbb P_n(K)\gt 1-\varepsilon$. Use the uniform convergence of$(f_n)_{n\geqslant 1}$to$f$in order to handle the integral of$f_n$over$K$. Use the uniform bound to handle the integral of$f_n$over the complement of$K$(which has a measure ... 1 On$\{1,2,3,4\}$let$\mu$be uniform on$\{2,4\}$and$\nu$be uniform on$\{1,3\}$. Let$A$consist of just the two sets$E_1:=\{1,2\}$and$E_2:=\{2,3\}$(note$A$is not a$\sigma$-algebra). Then$\mu(E_1)=\nu(E_1)=\frac12$and$\mu(E_2)=\nu(E_2)= \frac12$but$\mu(E_1\cup E_2)=\frac12 <1=\nu(E_1\cup E_2). 3 Let \begin{align}\left( \int_{-\infty}^{\infty} \sqrt{p}\sqrt{q} \;d\mu \right)^2 &= \left( \int_{-\infty}^{\infty} \min(\sqrt{p},\sqrt{q})\cdot \max(\sqrt{p},\sqrt{q}) \;d\mu \right)^2 \\ &\leq \int_{-\infty}^{\infty} \min(\sqrt{p},\sqrt{q})^2 \;d\mu \cdot \int_{-\infty}^{\infty} \max(\sqrt{p},\sqrt{q})^2 \;d\mu \\ &\leq ... 2 We have to assume\mathbb{E}(|X|^r)<\infty$; otherwise the expession$\|X_n-X\|_{L^r}$might not even be finite. Note that $$\|X_n-X\|_{L^r}^r = \int_{|X_n-X| \leq \epsilon} |X_n-X|^r \, d\mathbb{P} + \int_{|X_n-X| > \epsilon} |X_n-X|^r \, d\mathbb{P} \tag{1}$$ for any$\epsilon>0$and$n \in \mathbb{N}$. Obviously, $$\int_{|X_n-X| \leq ... 1 "The collection \mathcal{F} of all unions of sets A_j" should really mean the following: for any subset S\subseteq\{A_1,\dots,A_N\}, the union of all the elements of S is in \mathcal{F}, and every element of \mathcal{F} is obtained in this way. To get that \emptyset\in\mathcal{F}, you then just take S=\emptyset: the union of the empty ... 1 The measure theoretic part was answered, so let me complement it by answering the choice related question. The axiom of countable choice is needed on a far more fundamental level when you talk about measure theory. It is consistent that the real numbers are a countable union of countable sets. In that case there is no \sigma-additive Borel measure ... 4 Yes. Suppose not, look at X^c which is of positive measure. By Lebesgue Density Theorem, there exists \sigma, \tau\in 2^{\omega} (WLOG might assume they have the same length) such that X and X^c has measure >\frac{1}{2} above \tau, \sigma respectively. By hypothesis, X has measure >\frac{1}{2} above \sigma too. But then above ... -1 (a) For any fixed t, we know from the condition that X\mid_{[0,t-\frac{1}{N}]\times \Omega} \to (\mathbb R, \mathcal B(\mathbb R)) is \mathcal B([0 ,\ \ t-\frac{1}{N}]) \times \mathcal F_t measurable, therefore, for any A\in \mathcal B(\mathbb R), (X\mid_{[0,t)\times \Omega})^{-1}(A)= \mathop{\cup}_{N}\left ((X\mid_{[0,t-\frac{1}{N}]\times ... 1 Yes, the surface area can be bounded by a constant only depending on n. To start with, I'll use the Minkowski content rather than the Hausdorff measure. Then, we can give a simple bound. For a set S\subseteq\mathbb{R}^n and r > 0, I will use S_r to denote the set of points x\in\mathbb{R}^n within a distance r of S. i.e., such that \lVert ... 0 \Omega \in \mathcal{A}_i follows from the fact that \Omega \in \mathcal{F}_{i,j} for all i,j. Regarding your second question: Let A \in \bigcup_{j=1}^{m(i)} \mathcal{F}_{i,j} for some fixed i. Then there exists j_0 \in \{1,\ldots,m(i)\} such that A \in \mathcal{F}_{i,j_0} and therefore we can write$$A = \bigcap_{j=1}^{m(i)} A_{i,j}$$where ... 4 The integral diverges. To see this, we can write$$\int_0^n \left(1-\frac{3x}n\right)^ne^{x/2}\,dx=\int_0^{n/3} \left(1-\frac{3x}n\right)^ne^{x/2}\,dx+\int_{n/3}^n \left(1-\frac{3x}n\right)^ne^{x/2}\,dx \tag 1$$We will present two parts. In Part 1, we will show that the first integral on the right-hand side of (1) converges. In Part 2, we will ... 0 Since F_n \in F^* for all n \in \mathbb{N}, it follows directly from the definition of \alpha that$$\mu(F_n) \leq \sup_{F \in F^*} \mu(F) = \alpha<\infty.$$Since this holds for all n \in \mathbb{N} and F_n \uparrow F_{\infty} := \bigcup_{j \in \mathbb{N}} E_j, the continuity of the measure \mu implies$$\mu \left( \bigcup_{j \in ... 3 Alternative Solution: Take any$\epsilon>0$. We will show that$\mu\{x: \limsup_n (f_n(x))^{\frac{1}{n}}>1+\epsilon\}=0$. Define$A_n=\{x: f_n(x)> (1+\epsilon)^n\}$, and$A=\{x: \limsup_n (f_n(x))^{\frac{1}{n}}>1+\epsilon\}$. Then,$A=\limsup_n A_n$. Now, ... 1 This is my try. Let:$f_n (y) = \frac{e^y}{n^2y^4+1} \mathbb I_{[0,1]}$, where$\mathbb I$is indicator function. We have:$f_n(y) \to 0$, as$n \to \infty$. For integrability and domination condition, for every$n \in \mathbb N$,$|f_n(y)| \leq \frac{e^y}{y^4+1} \mathbb I_{[0,1]} \leq e^y \mathbb I_{[0,1]}$. This function is integrable on$\mathbb R$, ... 3 The difference$\gamma = \alpha - \beta$is a signed measure. The corresponding bounded linear functional$\phi$on$C[0,1]$must satisfy$\phi(1) = 2$,$\phi(x) = 1$,$\phi(x^2) = 0$. Since$1$,$x$and$x^2$are linearly independent, this defines an affine subset of$M[0,1]$of codimension$3$. A typical$3$-dimensional subspace of$M[0,1]$will ... 4 Applying Cauchy's Mean Value Theorem twice says that there are$0\lt h_1,h_2\lt hso that \begin{align} \left|\frac1h\left(\frac{\cos(x+h)-\cos(x)}h+\sin(x)\right)\right| &=\left|\frac{\cos(h)-1}{h^2}\cos(x)+\frac{h-\sin(h)}{h^2}\sin(x)\right|\\ &=\left|-\frac{\cos(h_1)}2\,\cos(x)+\frac{\sin(h_2)}2\sin(x)\right|\\[4pt] &\le1 \end{align} ... 0 No, you can't. Look at the following example: IfA$is the unit disc in the$(x_1,x_2)$-plane then Fubini's theorem says that $$\int_A g_1(x_1) g_2(x_2)\>{\rm d}(x_1,x_2)=\int_{-1}^1 \int_{-\sqrt{1-x_1^2}}^{\sqrt{1-x_1^2}} g_1(x_1)\> g_2(x_2)\>dx_2\>dx_1\ ,$$ as you have learned in Calculus 102. (You are allowed to take the factor$g_1(x_1)$... 3 Let$f_n(x)=\left(1-\frac{3x}{n}\right)^ne^{x/2}. Then $$\lim_{n\to\infty }f_n(x)\lambda_{[0,n]}(x)=\lim_{n\to\infty }f_n(x)\cdot \underbrace{\lim_{n\to\infty }\lambda_{[0,n]}(x)}_{=\lambda_{[0,\infty [}(x)}=\lambda_{[0,\infty [}(x)\lim_{n\to\infty }f_n(x).$$ Therefore \int \lim_{n\to\infty }f_n(x)\lambda_{[0,n]}(x)\mathrm d x=\int \lambda_{[0,\infty ... 0 It turns out that \int_{X} |f_{n}|^{2} \, d\mu \leq C implies (f_{n}) is uniformly integrable. Convergence then follows from Vitali's Theorem. To see that (f_{n}) is uniformly integrable, suppose A is a measurable set. For each n, we have by Holder's inequality, \begin{align*} \int_{A} |f_{n}| \, d\mu &= \int_{X} |f_{n}| \chi_{A} \, d\mu \\ ... 0 Alternative Solution: We need to show that\int_X |f_n-f| \ d\mu \to 0$$as n\to\infty. We know that \mu(X)<\infty. Fix \epsilon>0. Let us write the above integral as sum of two components as follows:$$\int_X |f_n-f|\mathbb{I}\{|f_n-f|\le M\} \ d\mu+|f_n-f|\mathbb{I}\{|f_n-f|>M\} \ d\mu \ \ \ \ (1)$$. Now, consider the 2nd part: By ... 0 Let X = \bigcup E be the set of which all members of E are subsets. For good measure, assume that X\in E (if it isn't, use E'=E\cup {X} in what follows). Note that R(E) is closed under intersection: if A,B\in R(E), then (X\setminus A)\cup (X\setminus B) = X\setminus(A\cap B) \in R(E), so A\cap B = X\setminus (X\setminus (A\cap B)) \in R(E). ... 2 Hint: Show that$$ \int_0^1 \frac{\sin x}{x^{3/2}}dx \quad\text{exists } $$And$$ \lim_{n \rightarrow \infty} n \int_{\frac{1}{n}}^{1} \frac{\cos(x+\frac{1}{n})-\cos(x)}{x^{\frac{3}{2}}}dx=-\int_0^1 \frac{\sin x}{x^{3/2}}dx $$1 We can write the foloowing: take a test function \psi and integrate it against \nabla \phi_s:$$\int_{\Bbb R^n}\nabla\phi_s(x)\psi(x)dx = \int_{supp\,\nabla\phi_s}\nabla\phi_s(x)\psi(x)dx=\int_{supp\,\nabla\phi_s}\nabla(\phi_s(x)\psi(x))dx-\int_{supp\,\nabla\phi_s} \phi_s(x)\nabla\psi(x)dx=\int_{\partial \{supp\,\nabla\phi_s\}} \phi_s(x)\psi(x)\cdot ... 0 Following @robjohn's excellent answer, I thought of bounding to find a dominating function for the integrand directly as \begin{align}\left|\frac{\cos(x+1/n ) - \cos x}{1/n}x^{-3/2}\right| &=\left|\frac{\cos x \cos(1/n) - \sin x \sin(1/n) - \cos x}{1/n}x^{-3/2}\right| \\ &= \left|\cos x \frac{\cos(1/n) - 1}{x/n}- \frac{\sin x}{x}\frac{\sin ... 0 For every \epsilon>0,\delta>0, there exist N,such that P(|X_n-X|>\epsilon)<\delta, for all n\ge N. Actually, you can verify it's the same as for every \epsilon>0, there exist N,such that P(|X_n-X|>\epsilon)<\epsilon, for all n\ge N. 1 From what you have you can do the following. Let B be the set where f_n\to f uniformly on B, and \mu(B^c) <\epsilon. You have \int_X |f-f_n| = \int_B|f-f_n| + \int_{B^c}|f-f_n|. $$The first integral limits to 0 by the uniform convergence, hence can be made smaller than \epsilon for large enough n. For the second integral, apply the ... 0 For (a) the sequence f_m(u) = 1/(1+u)^m will work. 1 Try f(x):=I_{[0,1]}(x), i.e. f is a box of height 1 over the interval [0,1]. Then$$f(nx)=I_{[0,1]}(nx)=I_{[0,1/n]}(x)$$is a box of height 1 over the interval [0,1/n]. Then the sum \sum f(nx) converges for every x\ne0 (since the sum terminates for every x\ne0), while the sum of the areas is 1 + \frac12 + \frac13 +\frac14 +\cdots . 0 The set of finite subsets of X is the smallest ring of sets (in the sense of measure theory) containing the singletons. Indeed every finite set is a finite union of singletons. Moreover the difference of two finite sets is also a finite set. The fact that X is uncountable is irrelevant. 0 almost there. Let F = f^s, G = g^s, apply your result so ||FG||_{1} <= ||F||_{p/s}||G||_{q/s} from here, you should get your result by subsituting F and G with f^s and g^s 1 If f \in L^p, then |f|^s \in L^{p/s}. Using \frac{1}{p/s} + \frac{1}{q/s} = 1 and Hölder's inequality yields the result:$$\|fg\|_s = \||f|^s|g|^s\|_1^{1/s} \le \||f|^s\|_{p/s}^{1/s} \cdot \||g|^s\|_{q/s}^{1/s} = \|f\|_p \|g\|_q$$You can easily verify the two equalities I've used by plugging in the definition of the occuring L^r-norms. 4 Your idea that if the conclusion fails, then \sum f_n = \infty on a set of positive measure is almost what we need to conclude. Indeed, consider the modified sequence g_n := f_n / n^2. Then$$ \int \sum_n g_n \, d\mu = \sum_n \int g_n \, d\mu = \sum_n 1/n^2 < \infty,$so that we get$\sum_n g_n < \infty\$ almost everywhere. But by the root test, ...

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