New answers tagged

3

Since it is true for characteristic functions, you can do a simple linearity argument to show it is true for all simple functions. Then do an approximation argument to finish.


0

To improve readability, we will first fixed our space to $X:=[-n,n]$, any subset of $X$ that is measurable is also measurable in $\Bbb R$. For any $A\subset X$, $A^c$ denotes its complement in $X$. Any arbitrary $E$ satisfying $$ m^*(E)+m^*(E^c)=m^*(X), $$ we'll show that there exist $F,G\subset \Bbb R$ such that $F\subset E \subset G$ and $m(G\backslash F)=...


0

This seems close: $$ I(A(a)) = \frac{n^2}{||A(a)||^2_F}$$, where $||A(a)||_F$ is the matrix entrywise 2-norm. For the example in the question: $ I(A(0)) = 3$ and $ I(A(1)) = 9/5 = 1.8$


0

The appeal you speak of, which I believe is/was shared by many, is that the fuzzy variants of the traditional mathematical objects are, intuitively, more realistic. After all, in real life we never know anything with certainty, so why not replace all of logic and set theory by their fuzzy cousins - surely the mathematics we'll get will be better. Well, ...


2

Not in general. First, note that if $XY,Z$ are independent: $$ \mathbb{E}[XY\mid Z] = \mathbb{E}[XY] $$ which is "just a number." Now, it's tempting to try and say something in the case that if $XY,Z$ are not independent. But... If $X=f(Z)$ for some function $f$, then: $$ \mathbb{E}[XY\mid Z] = \mathbb{E}[f(Z)Y\mid Z] = f(Z)\mathbb{E}[Y\mid Z] = X\mathbb{...


1

Yes, we can insure the desired nestedness. First: If $A,B$ are closed disjoint subsets of $ \mathbb R,$ then there exists a continuous $\varphi $ on $\mathbb R$ such that $\varphi = 1$ on $A,$ $\varphi = 0$ on $B.$ So in the problem at hand, there is a closed $F_1\subset E$ with $m(E\setminus F_1) < 1$ and a continuous $g_1$ such that $g_1=f$ on $F_1.$ ...


1

$|f_n + g_n - (f+g)| \le |f_n - f| + |g_n - g|$, so wherever $|f_n - f| < \epsilon/2$ and $|g_n-g| < \epsilon/2$ we'll have $|f_n + g_n - (f+g)| < \epsilon$. Thus $$ \eqalign{\{x: |f_n(x) + g_n(x) &- (f(x) + g(x))| \ge \epsilon\}\cr& \subseteq \{x: |f_n(x) - f(x)| \ge \epsilon/2\} \cup \{x: |g_n(x) - g(x)| \ge \epsilon/2\}}$$ so if each ...


1

As $\int |f_n - f| \to 0$, we have $\int_E |f_n - f| \to 0$ for any measurable $E$. Now: $$\left| \left| \int_E f_n \right| - \left| \int_E f \right| \right| \le \int_E| f_n - f| $$ So the LHS converges to $0$, i.e. $\int_E f_n \to \int_E f$. Now we also have: $$\int ||f_n| - |f|| \le \int |f_n - f|$$ So the LHS converges to $0$, hence by above, $\int |...


0

A lower dimensional surface always has Lebesgue measure zero. Indeed, if $j < k$, and $\mathcal H^j(E) < \infty$, $\mathcal H^k(E)=0$. This is a straightforward consequence of the definition of the Hausdorff measure; if this isn't obvious, you should try proving it yourself. On the other hand $\mathcal H^n = \mathcal L$. It's not so hard to see that $\...


0

Let $\mu^{\ast}$ be the outer measure on $R$. A collection $\left\{A_i\right\}$ is a partition of $R$ if $A_i \cap A_j=\phi$ if $i\neq j$ and $\bigcup^\infty_{i=1} A_i=R$. Prove that all sets on the partition $\left\{A_i\right\}$ are measurable if and only if $\mu^\ast (B)=\sum^\infty_{i=1} \mu^\ast (B\cap A_i)$ for all subset $B$ of $R$. Proof ($\...


0

Forward implication: First, $$\mu^* (B) = \mu^*\left( B \cap \bigcup_{i=1}^{\infty} A_i \right) = \mu^* \left( \bigcup_{i=1}^{\infty} (B \cap A_i) \right) \le \sum_{i=1}^{\infty} \mu^*(B \cap A_i)$$ Now, by the definition of measurability, we can show by induction that for all $n$, $$\mu^*(B) = \sum_{i=1}^n \mu^*(B \cap A_i) + \mu^* \left( \bigcup_{i=n+1}...


0

(Focusing on the case $X=Y=[0,1]$ with Borel sets.) 1. A probability measure on $[0,1]\times[0,1]$ admits a disintegration $$ \mu(A)=\int_{[0,1]}\mu_1(dx)K(x,A),\qquad A\in\mathcal B([0,1]^2), $$ with $K$ a kernel such that $x\mapsto K(x,A)$ is Borel measurable for each $A\in \mathcal B([0,1]^2$ and $A\mapsto K(x,A)$ is a probability measure on $[0,1]^2$ ...


3

Let $\{E_n\}$ be a partition of a measurable set $E$ (i.e. a countable mutually disjoint collection whose union is $E$). then for $m\in\mathbb{N}$, the collection $\{E_1,\ldots,E_m,\bigcup_{n=m+1}^\infty E_n\}$ is a finite partition of $E$, and thus $$\sum_{n=1}^m|\mu(E_n)|+|\mu\left(\bigcup_{n=m+1}^\infty E_n\right)|\leq \mu_A(E) $$ Since $m$ was ...


1

Recall that a σ-algebra for $X$ is a collection of subsets $\Sigma$ of $X$ such that: The empty set and the whole set $X$ belong to $\Sigma$. $\Sigma$ is closed under all countable unions. $\Sigma$ is closed under all countable intersections. $\Sigma$ is closed under complementation. Recall that a monotone class for $X$ is a collection of subsets $\...


1

Theorem 3.27 a.) If $F\in BV$ if and only if $Re F \in BV$ and $Im F \in BV$. b.) If $F:\mathbb{R}\rightarrow \mathbb{R}$, then $F\in BV$ if and only if $F$ is the difference of two bounded increasing functions; for $F\in BV$ these functions may be taken to be $(\frac{1}{2}(T_F + F)$ and $\frac{1}{2}(T_F - F)$. c.) If $F\in BV$, then $F(x^+) = ...


2

Let $\epsilon>0$ and $\mathbb{R}=\cup_n (-n,n)=\cup_n I_n$. Now, by Egorov, for each $k$ exist $E_k^n$ such that $|I_n-E_k^n|<\epsilon/k$ Note that for each n, $E_k^n$ can be increasing and then $I_n-E_k^n$ is decreasing in k $|\mathbb{R}-\cup_{n,k}E_k^n|=|(\cup_n(\cap_k( I_n-E_k^n))|=|\cup_n(\cap_k(I_n-E_k^n))|\leq \sum_n(\lim_k |I_n-E_k^n|)=0$.


0

I think there is a simpler way to do this by introducing some more machinery. Let $f_n=\sum_{m=1}^\infty\frac{m}{2^n}\chi_{E_{n,m}}$. Show that $f_n\rightarrow f$ pointwise and then that $f_n$ is monotonically increasing. Then apply the Lebesgue monotone convergence theorem.


3

I don't know how familiar with Banach spaces you are. It's a standard fact that "absolutely summable series converge" in a Banach space (in the same way this is true for series of numbers). More precisely, if $(v_{n})_{n \in \mathbb{N}}$ is a sequence in a Banach space $B$ and $\sum_{n = 1}^{\infty} \|v_{n}\|_{B} < \infty$, then $\sum_{n =1}^{N} v_{n}$ ...


1

Suppose that $\{r_j\}_j$ is a sequence of positive real numbers, and $\{x_j\}_j$ is a sequence in $\mathbb{R}^n$. Suppose also that there are $r \geq 0$ and $x \in \mathbb{R}^n$, such that $$\lim_{j\rightarrow \infty}(r_j,x_j) = (r,x)$$ then is it true that $$\lim_{j\rightarrow \infty}\chi_{B(r_j,x_j)} = \chi_{B(r,x)} \textrm{ pointwise?} $$ where $...


7

Let me try to explain the Vitali construction by breaking everything into small pieces. (Usually I find this presented quite rapidly in books, so that everything seems to come all at once and it's not clear what all the pieces even were.) You begin by quotienting $[0,1]$ by this equivalence relation that you stated. You get the full collection of ...


1

I know this question was asked a long time ago but I just came across this question and I think it is a good one that deserves an attempt at an answer. I'm kind of new to measure theory but I think I can give some degree of insight on this question. When it comes to the Lebesgue measure, we want the smallest $\sigma$-algebra, say $\mathcal{B}$ generated by a ...


0

Consider (in $\mathbb{R}^2$ for example/simplicity) $x_n=((-1)^n\frac{1}{n},0)$ and $r=1$ fixed and the point $x^\ast:=(1,0)$. What do you conclude? EDIT: I show for example, that for a point $x^\ast\notin B_{x,r}\cup S(r,x)$ the limit is pointwise zero. The same idea applies for points inside the ball. So, by our assumption on $x^{\ast}$, there exists an $...


0

We have $$f(x) = \chi_{[0,1)}(x) + x^2(1+x)\chi_{[1,2)}(x) + 17\chi_{[2,\infty)}(x). $$ The Lebesgue-Stieltjes measure associated with $f$ is defined by $$\mu(E) = \int_E \mathsf df, \ E\in\mathcal B(\mathbb R)$$ where the integral is interpreted as a Stieltjes integral. Note that $f$ has three jump discontinuities: \begin{align} \Delta_0 := f(0)-f(0-) &...


0

I think you get a complete picture of the Prokhorov metric $\pi$ by combining what Dirk has pointed out and what we already know about the total variation metric. Essentially, $\pi$ is a measure-theoretic analogue of the Hausdorff metric, but loosened up modulo the total variation metric. I will explain what I mean by this. Suppose we have a probability ...


1

If $F\subset\Bbb Z$ is finite then $A^F$ is a finite set, so there is a uniform probability measure $\mu_F$ on $A^F$; that's the measure that assigns the same measure to each point. And there is a natural projection $\pi_F:A^{\Bbb Z}\to A^F$. The uniform measure $\mu_u$ is defined by $$\mu_u(\pi_F^{-1}(S))=\mu_F(S)\quad(S\subset A^F).$$ Asking whether ...


2

Note that a sequence of real numbers converges to a limit $L$ iff every subsequence has a subsubsequence that converges to $L$. Since your hypotheses are preserved by passing to subsequences, it suffices to show that $\int|f-f_{n_k}|\to 0$ for some subsequence $(f_{n_k})$. In particular, since $\int g_n\to 0$, there is a subsequence $(g_{n_k})$ of $(g_n)$ ...


0

No, this is not true in general. For instance, let $S$ be the set of all subsets of $\Omega$ with at most one element, and let $\mu$ and $\lambda$ be counting measure. Then for any topology that makes $\mu$ continuous, $\mu^{-1}((1,\infty])=\sigma(S)\setminus S$ must be open. So unless $S=\sigma(S)$ (which happens only if $\Omega$ has at most one element),...


13

A standard argument to show that $\mathcal{L}(\mathbb R)$ has the same size as $\mathcal{P}(\mathbb R)$ is to note that the Cantor subset of $[0,1]$ has the same size as $\mathbb R$ and measure 0, so any of its subsets also has measure 0. You can use the same idea to find the size of $\mathcal P(\mathbb R)\setminus\mathcal L(\mathbb R)$: Fix a nonmeasurable ...


1

As Prahlad Vaidyanathan mentioned in the comments, consider the delta measure. Fix $x \in X$ and let \begin{align*} m_x(A) = \begin{cases} 1 & x \in A \\ 0 & x\not\in A \end{cases}. \end{align*} Then $m_x$ is a finite measure on $X$. This can be generalized to $m = \sum_{x \in X} p_x \mu_x$ with $p_x \in [0, \infty]$. Another possible measure would ...


0

If $f$ is additive then $f(x)=cx$ for every rational $x$, this correct. Now for the second part you don't need to ask for derivability : if $f$ is continuous then $f(x)=cx$ for every real number, because $\mathbf Q$ is dense in $\mathbf R$. But if the function $f$ is not continuous it's not necessarily of the form $f(x)=cx$. In fact one can show that ...


0

Those other solutions on the reals are not differentiable (in fact, not continuous). So your argument is in a conditional form: IF it is differentiable, THEN it is of the form $c\cdot x$. Note that for complex numbers $x$, the functional equation $f(x+y) = f(x)+f(y)$ has, in addition to continuous solutions $f(x) = c\cdot x$, also continuous solutions $f(x)...


5

Another idea: Put $g=\sum |f_n|$. We see that (as all the functions are $\geq 0$) we have $$\int_0^1g(t)dt=\sum _{n\geq 1}\int_0^1|f_n(t)|dt\leq \sum_{n\geq 1}\frac{1}{n^2}<+\infty$$ Hence $g$ is in $L^1$, hence finite $a.e$, and the serie $|f_n|$ is a.e convergent, and $f_n\to 0$ a.e.


0

If $B_n = \{x: |f_n(x)| > 1/\sqrt{n}\}$, we have $$m(B_n) \le\sqrt{n} \int |f_n|\; dx \le n^{-3/2}$$ so that $$ \sum_{n=1}^\infty m(B_n) < \infty $$ For any $M$, $$m \{x: |f_n(x)| \ge 1/\sqrt{n} \ \text{infinitely often}\} \le m(\bigcup_{n\ge M} B_n) $$ which goes to $0$ as $M \to \infty$. Thus for almost all $x$, $|f_n(x)| < 1/\sqrt{n}$ for all ...


1

For the first part of your question, define $$ h_n=|f|+|f_n|-|f-f_n| $$ Then $h_n\geq 0$ for all $n$, so we can apply Fatou's lemma to obtain $$2\int |f|=\int \liminf_{n\to\infty}h_n\leq \liminf_{n\to\infty}\int h_n=2\int|f|-\limsup_{n\to\infty}\int|f-f_n|$$ and since $\int|f|$ is finite this shows that $$\limsup_{n\to\infty}\int |f-f_n|=0 $$ For the ...


0

To remember the proof, maybe it is best to keep a particular example in mind. Let $E = [0,1]$, the closed unit interval on the line. Let $f_n(x) = x^n$, which is bounded by $M=1$. Then $f_n \rightarrow 0$ almost everywhere on $E$ but not uniformly. But we can exclude the bits where uniform convergence fails (this is Egorov's theorem). In this particular ...


1

For (a), use the fact that for any complex number $z$, we have the inequality $\max\{|\text{Re}z|,|\text{Im}z|\} \leq |z| \leq |\text{Re}z| + |\text{Im}z|$. For the converse direction of (b), write $F = I - D$ where $I$ and $D$ are bounded and increasing. Then if $x_0 < x_1 < \ldots x_n$, we have $$\begin{aligned} \sum_{k=1}^{n}|F(x_k)-F(x_{k-1})|&...


0

$\mathfrak{U}(C)$ is by definition the minimal $\sigma$-Algebra such that the functions $p_t: x\mapsto x_t$ are $\mathfrak{U}(C)$-$\mathscr{B}(\mathbb{R})$-measurable for $t \in [0, \infty)$. Now, it can easily be shown that the system of sets \begin{align*} \mathscr{A}:=\left\{ A\in \mathscr{B}(\mathbb{R}): p_t^{-1}(A) \in \sigma(\left\{ p_t^{-1}(B):B\in ...


0

If $0\le x \le n,$ then $(1-x/n)^n \le e^{-x}.$ Proof: This is certainly true at the end points. For $0<x<n,$ $$\tag 1\ln (1-x/n)^n = n \ln (1-x/n) < n(-x/n) = -x.$$ Here we have used the inequality $\ln (1-u) < - u,$ which is valid for $ 0 < u < 1.$ Exponentiate back in $(1)$ to get the desired inequality. To finish the problem, let $...


0

There is the following formulation: For a system of sets $\mathscr{E}$ which generates the $\sigma$-Algebra $\mathscr{A}$ on $\Omega$ and is closed under $\cap$ and contains a sequence $(E_n)_{n\in \mathbb{N}}$ such that $\Omega = \bigcup_n^\infty E_n$. Then two measures $\mu, \nu$ which satisfy $$\mu(E) = \nu(E) \qquad E\in \mathscr{E}$$ and $$\mu(E_n) = \...


3

A similar limit is used to prove Euler's limit product formula for the $\Gamma$ function: $$ t! = \lim_{n\to +\infty}\frac{n! n^t}{(t+1)(t+2)\cdot\ldots\cdot(t+n)}\tag{1}.$$ For any $x\in\mathbb{R}^+$ we have $$ \max_{n\geq x} \left(1-x/n\right)^n\leq e^{-x}\tag{2} $$ and $e^{-x}\cdot e^{x/2}\in L^1(\mathbb{R}^+)$, hence by monotone or dominated convergence ...


1

Hint for the first one: Since $|sin(\theta)| < \theta$ for $\theta>0$: $$|\frac{nsin(x/n)}{x(1+x^2)}| < \frac{n(x/n)}{x(1+x^2)} = \frac{1}{1+x^2}$$ For the second one: Let $$f_n(x) = \frac{\sin(x/n)}{(1+x/n)^n}.$$ We have that $\lim_{n\to\infty} f_n(x) = 0$. But $$|f_n(x)| \leq \frac{1}{(1+x/n)^n} \leq \frac{1}{(1+x/2)^2}$$ for $n\ge 2.$ So,...


0

For the first part, set $f_n(x) := \dfrac{n\sin (x/n)}{x(1+x^2)}$. One has $f_n(x) \sim_{n\infty} \frac{1}{1+x^2}$. Now try to prove that $f_n$ is bounded (uniformly in $n$).


2

Hint: $$\left\{x\in R |f(x)=g(x)\right\}=\left\{x\in R |f(x)-g(x)=0\right\}=(f-g)^{-1}\{0\}.$$ Questions for you: Is the singleton $\{ 0\}$ measurable? Is the sum/difference of measurable function measurable? Once you have established the above, the claim follows by very definition of measurability.


2

Use the dominated convergence theorem with $\max(1,|f(x)|^2)$ as the dominating function.


2

Yes, that's acceptable. I would say: let $\Sigma$ be a σ-algebra. Then $\Sigma$ satisfies the first two semiring properties because, respectively, $\Sigma$ contains the empty set and $\Sigma$ is closed under finite intersections by virtue of being a σ-algebra. For the third property, fix $S, T \in \Sigma$ and let $U$ be the difference $U=S - T \...


1

Take a sequence $\{x_n\}$ such that $x_n \to x$ and $x_j < x $ for each $j$. Then write $\{x\}=\bigcap_{n} (x_n,x]$. By continuity from above, $m(\{x\})=\lim_{n\to \infty} m((x_n,x])=\lim_{n\to \infty} \alpha(x)-\alpha(x_n)=\alpha(x)-\alpha(x-)$


2

The statement is true if $A+B$ is measurable, this is the critical part in the assumption. If $A,B$ are compact, then so is $A+B$ and therefore measureable. Here is a proof for the theorem with the assumption of $A+B$ to be measurable using the statement for compact $A,B$: For arbitrary $\tilde A,\tilde B$ compact with $\tilde A\subset A$ and $\tilde B\...


1

By the pointwise ergodic theorem, we have (by ergodicity) that $$\frac 1n\sum_{j=1}^ng\circ \theta^k(x)\geqslant \frac{\mathbb E\left[g\right]}2$$ for each $n\geqslant n_0(x)$. Therefore, we have $$\left\{n\geqslant n_0(x)\mid b_n\leqslant t\right\} \subset \left\{n\geqslant n_0(x)\mid \frac{g\left(\theta^nx\right)}n \geqslant \frac{\mathbb E\left[g\right]}{...


1

3.25 Examples: a.) If $F:\mathbb{R}\rightarrow \mathbb{R}$ is bounded and increasing, then $F\in BV$ (in fact, $T_F(x) = F(x) - F(-\infty)$). b.) If $F,G\in BV$ and $a,b\in\mathbb{C}$, then $aF + bG\in BV$. c.) If $F$ is differentiable on $\mathbb{R}$ and $F'$ is bounded, then $F\in BV([a,b])$ for $-\infty < a < b < \infty$(by MVT). ...


2

First question is yes. Given $|f|\leq M$ is bounded, we can approximate it uniformly using simple functions by cutting $[-M, M]$ into finitely many intervals $I_n$ with length $\epsilon$ and defining $\phi_n (x) =\sum_n \inf I_n \chi_{f^{-1}(I_n)}(x) $ . We would have $\|f - \phi_n\|_\infty \leq \epsilon$. $$\lim_m\int f d\mu_m = \lim_m \lim_n \int \phi_n ...



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