New answers tagged

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 $0\lt h_1,h_2\lt h$ so 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: If $A$ 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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