5

It's true that the definition of absolute continuity is the same either we state it in the form of the finite and disjoint intervals or we state it in the form of countable and disjoint intervals. In detail, suppose we have the following definitions of absolute continuity: Definition 1 Let $f:\mathbb{R}\to \mathbb{R}$ be a function. We say that $f$ is ...


4

Let me discuss the case $n=1$ only. Let me denote by $f_x$ the weak derivative of $f$ with respect to $x$. Let $\phi,\psi$ be two smooth test functions. Then $$ -\iint f_x(x,y) \phi(x)\psi(y)dx\ dy=\iint f(x,y) \phi_x(x)\psi(y)dx\ dy, $$ using Fubini on both sides gives $$ \int \left( \int -f_x(x,y) \phi(x) + f(x,y) \phi_x(x)\ dx\right) \psi(y) \ dy = 0 $$ ...


3

$$\int_0^\infty f_{Y-X}(z) \, dz = \int_0^1 f_{Y-X}(z) \, dz + \int_1^2 f_{Y-X}(z) \, dz + \int_2^\infty f_{Y-X}(z) \, dz.$$ Plugging in $1/2$ for the integrand in the first integral yields $\int_0^1 \frac{1}{2} \, dz = \frac{1}{2}$. The second integral yields the second term $\int_1^2 \frac{1}{2} (2-z) \, dz$. The third integral is zero. By the way, I ...


3

Recall that the outer measure $m^*$ is defined on the entire power set of $\Bbb{R}^n$, while $m:= m^*|_{\mathcal{L}(\Bbb{R}^n)}$; i.e we define $m$ to be the restriction of the outer measure $m^*$ to the Lebesgue sigma-algebra. Clearly, from the very definition itself, it's clear that $m\neq m^*$ as functions. Afterall, how can they be equal, they have ...


3

Suggestion: If the $A_n$ are pairwise disjoint and If $\bigcup_nA_n$ is countable, then all the $A_n$ are countable and $\mu(A_n)=0$ for all $n$. So no problem here If the $A_n$ are pairwise disjoint and $\bigcup_nA_n$ is uncountable, then and at least one of the sets $A_n$ must be uncountable. The question is, can there be more than one uncountable ...


3

Suppose that $l(I)>m^*(I)$ and let $\varepsilon=\frac{l(I)-m^*(I)}2$. Then $l(I)>m^*(I)+\varepsilon$ , but we are assuming that $(\forall\varepsilon>0):l(I)\leqslant m^*(I)+\varepsilon$.


2

The only thing missing is to check that $$ \rho(x,y)=\frac{|x-y|}{1+|x-y|} $$ is a metric on $\mathbb{R}$. Here is a simple proof: Consider the function $$ f(t)=\frac{t}{1+t}, \qquad t\geq0$$ notice that $f(t)=0$ iff $t=0$, $f$ is monotone non decreasing on $[0,\infty)$, and that $\rho(x,y)=f(|x-y|)$. That $\rho$ satisfies the triangle inequality is a ...


2

You cannot write $\Bbb R$ as a union of two nowhere dense sets, because the union of nowhere dense sets is still nowhere dense. And without interior points (after the question change): take $A$ negative rationals plus positive irrationals), and the other set its complement.


2

The structure on $[0, \infty]$ is that of a monoid, so you only have addition and an identity element $0$. When saying $\infty + 5 = \infty + 3$ implies $5 = 3$, we add $-\infty$ on both sides, but we do not assume we have inverses, so that is not a counter example here. You can check the axioms of a monoid to see that $([0, \infty], +, 0)$ is well defined ...


2

Hint: if $X\geq 0$ then $\int_0^\infty\textbf{1}_{\{-X>x\}}dx=0$ and $$ \int_0^\infty\textbf{1}_{\{X>x\}}dx = \int_0^{X}\,dx =X $$ if $X<0$ then $\int_0^\infty\textbf{1}_{\{X>x\}}dx=0$ and $$ \int_0^\infty\textbf{1}_{\{-X>x\}}dx = \int_0^{-X}dx =-X. $$


2

Let $\mu$ and $\nu$ any complex measures in $\mathbb{R}$, then $\mu\times \nu$ is a complex measure in $\mathbb{R}^2$, and you have that $$ \begin{align*} \mu\times \nu([a,b]\times [a,b])&=\int \mathbf{1}_{[a,b]\times [a,b]}(x,y)(\mu\times \nu)(dx,dy)\\ &=\iint \mathbf{1}_{[a,b]}(x)\mathbf{1}_{[a,b]}(y)\mu(dx)\nu(dy)\\ &=\iint \mathbf{1}_{[a,b]}(...


2

The convolution of $f$ and $\chi_{\{1\}}$, applied to $x$, is given by: $$ (f\star \chi _1)(x) = \int_G f(y)\chi_{\{1\}}(y^{-1}x)\,dy. \tag 1 $$ You cannot replace $y$ with $x$ in this integral, because $x$ and $y$ have very distinct roles. That would be the same as replacing $i$ by $n$ in the formula $$ \sum_{i=1}^n i = \frac{n(n+1)}2, $$ ...


2

Note that $X = \bigcup_{N = 1}^{\infty} \{|m| \leq N\}$. Suppose $f \in L^{p}(\mu)$. Note that $|f \chi_{\{|m| \leq N\}} - f|^{p} \leq 2^{p} |f|^{p}$ so Lebesgue's dominated convergence theorem gives \begin{equation*} \lim_{N \to \infty} \int_{X} |f \chi_{\{|m| \leq N\}} - f|^{p} \, d \mu = 0. \end{equation*} At the same time, since $m \chi_{\{|m| \leq N\}}$...


2

The pair $(X,Y)$ is uniformly distributed on the rectangle $[0,1]\times[0,2],$ whose area is $2.$ The set $\{(x,y)\in[0,1]\times[0,2] : y>x\}$ is the interior of a trapezoid whose are is $1.5.$ Thus the probability is $1.5/2= 3/4.$


2

Given a set $\Omega$ with some subsets $A_i \subset \Omega$, for some indices $i\in I$. We define a generated $\sigma$-algebra $\sigma(A_i)$ to be the smallest $\sigma$-algebra that contains $A_i$. Is it always the case that $ \sigma\left( \bigcup_{i\in I} \sigma(A_i) \right) = \sigma\left( \bigcup_{i\in I} A_i \right) $? For the question as stated, the ...


2

The converse is true. Notice that since $\text{dist}(A,B)>0$, we have $\overline{A}\cap\overline{B}=\emptyset$. Since $\mathbb{R}^n$ is a metric space, it is normal (in fact $T_6$); and so we can find an open set $U$ such that $A\subseteq U$ and $A\cap B=\emptyset$. Then, as $\mu$ is a Borel measure, \begin{align*} \mu(A\cup B) &= \mu((A\cup B)\...


2

is there an example of non-measurable $\{A_k\}$ such that $m^{*}(\cap_{k=1}^{\infty} A_k) \ne \lim_{k \to \infty} m^{*}(A_k)$ even if for each $k$, $m^{*}(A_k) < \infty$? The answer is YES. Here is a simple example. Let $(\Bbb N, \Sigma, m)$ be a measure space where $\Sigma =\{\emptyset, \Bbb N\}$ and $m$ is a measure defined as $m(\emptyset)=0$ and $m(\...


2

Given the set $\mathcal{C} = \{a\}$ of $X$, where $a \in X$, the $\sigma$-algebra generated by $\mathcal{C}$, $\sigma \mathcal{C}$ will be equal to $\{X, \{a\}, \{a\}^C, \emptyset\}$. So you were quite close; you just needed the complement of $\{a\}$, $\{a\}^C$. If you didn't have $\{a\}^C$ then $\sigma \mathcal{C}$ would not be closed under complements, ...


2

It is the indicator function of an event (set). Here, $$ 1_{\{X\leq t\leq X+Y\}} = \begin{cases} 1 & \text{ if } X\leq t\leq X+Y\\ 0 &\text{ otherwise} \end{cases} $$


2

If you define $\operatorname P_x$ to be the image of $\operatorname P_0$ under the translation $w\mapsto w+x:=\{w(t)+x, t\ge 0\}$ on $C[0,\infty)$, then $x\mapsto \operatorname P_x$ is Borel measurable, in the sense that $x\mapsto\int F(w)\operatorname P_x(dw)$ is Borel measurable for each bounded $\mathcal B(C[0,\infty))$-measurable function $F$. Indeed, $x\...


2

Let $E$ be a Lebesgue measurable subset of $[0, 1]$, and define $B={\{x \in [0,1] : \lambda(E \cap (x-\varepsilon,x+\varepsilon)) > 0 \ \text{for all} \ \varepsilon>0 }\}$. Show that $B$ is perfect. A perfect set is a closed set with no isolate points (see here). According to this definition, the empty set is a perfect set. Let us prove the result. ...


2

Suggestions: $\{Y_n\}$ is uniformly integrable. (Why?) Because of 1, $\{X_n\}$ is uniformly integrable. (Why?) Show that $M_n:=\sup_{m\ge n}|X_m-X|$ converges to $0$, as $n\to\infty$, in $L^1$ (not just a.s.). $E\left[|X_n-X|\mid\mathcal F_n\right]\le \sup_{m\ge n} E\left[|X_m-X|\mid\mathcal F_n\right]\le E[M_n\mid\mathcal F_n]$. By 3., the middle term ...


2

The proposed inequality is equivalent to \begin{align*} [\mathbb{P}(A\cap B)]^{2} \leq \mathbb{P}(A)\mathbb{P}(B) \end{align*} which is true indeed. Since $A\cap B\subseteq A$ and $A\cap B\subseteq B$, the monotonicity of the probability measure implies that \begin{align*} \begin{cases} \mathbb{P}(A\cap B)\leq \mathbb{P}(A)\\\\ \mathbb{P}(A\cap B)\leq \...


2

The statement is false, by example take the Dirac measure concentrated at zero, then $$ \lim_{n\to\infty}\delta (\{0\}-1/n)=0\neq \delta (\{0\}-0)=1 $$


2

One element of confusion is coming from a somewhat imprecise notation. Writing $\partial_1 f$ is asking for trouble when you invoke a change of variables. If instead you write $\frac{\partial f}{\partial y_1}$ or $\partial_{y_1} f$ this removes (in principle) any ambiguity. For example if $f(y_1,y_2)=y_1^2 y_2$ then (omitting domains) $$\int \partial_{y_1} f ...


2

Hints: Where does the issue of convergence arise? Is it near the origin or near $\infty$? Think first about $N=1$. This is a standard result (called integral p-test...whatever it is called it is a simple matter to evaluate the integral $\int_1^{\infty}\frac{1}{x^s}\,dx$ and decide for what values of $s$ the integral converges). Use spherical coordinates ...


1

In general not. All $\sigma$-algebras in the following should be taken to be the Borel $\sigma$-algebra. Suppose that $\Omega = [0,1]$ and $\mathbb P$ is a Dirac at $\{0\}$ so that $\mathbb P((0,1]) = 0$. Then, let $X(\omega) = 0$ and $Y(\omega) = \omega$. Clearly the laws of $X$ and $Y$ are both Diracs at zero. Now choose $\phi : [0,1] \to [0,1]$ be a non-...


1

Denoting by $\mu_1$ the distribution of $X_1$ and $\mu_2$ that of $X_2$ we have $$ \mathbb E[F_2(X_1)]=\int_{\mathbb R}F_2(x_1)\,\mu_1(dx_1)=\int_{\mathbb R}\int_{\mathbb R}1_{\{x_2\le x_1\}}\,\mu_1(dx_1)\,\mu_2(dx_2). $$ We have a similar expression for $\mathbb E[F_1(X_2)]$. The desired equality is then just the observation that $1_{\{x_2\le x_1\}}+1_{\{...


1

Your argument needs the following change: Since only one term in $\sum f_n$ is non-zero at any point $x$ it folows that $|\sum f_n(x)|=\sum |f_n(x)|$ for all $x$. Now you can replace the inequality in your proof by an equality. The proof is now complete.


1

Sets of measure $+\infty$ are not at all unusual: even in $\Bbb R$ in the standard Lebesgue measure we must allow this for sets as $\{x \in \Bbb R\mid x > 1\}$ e.g. We have to keep in mind we can do addition and have an order, but we have no subtraction in $[0,+\infty]$. Measures are usually taken with values in $[0,+\infty]$, as in your definition. ...


Only top voted, non community-wiki answers of a minimum length are eligible