# Tag Info

7

There is a Borel set $E$ in $\mathbb R^2$ such that $F := \{x-y\colon (x,y) \in E\}$ is not a Borel set. Let $A := \{f \in \mathbf{C}\colon (f(1), f(0)) \in E\}$. Then $A \in \mathcal{B}_{\left[0,\infty\right)}$. How about $T(A)$? In fact $$T(A) = \{g \in \mathbf{C}\colon g(0)=0, g(1) \in F\}$$ and is not Borel. added Mar 10 Why is $T(A)$ not Borel? ...

4

On the same $\Omega$, try $X$ uniform on $\{0,1\}$ and $Y=1-X$, then $\{X\ne Y\}=\Omega$. Edit: Recall that in the probabilistic jargon, a random variable is just a measurable function, here $X:\Omega\to\{0,1\}$ and $Y:\Omega\to\{0,1\}$, that is, for every $\omega$ in $\Omega$, $X(\omega)=0$ or $X(\omega)=1$ and $Y(\omega)=0$ or $Y(\omega)=1$. A notation is ...

3

There isn't, because it doesn't. You are taking three uniformly distributed independent random variables $X,Y,Z$ and forming a weighted average of $Y$ and $Z$ using $X$ as a weight. The result of averaging will fall near $1/2$ more often than near $0$ or $1$. I took $10^6$ samples for illustration: And here is a mathematical derivation of the ...

3

Any monotone function can have only countably many discontinuities: If $x$ is a point of discontinuity, there must exist a rational number $r$ such that $$f(x-) < r < f(x+)$$ where $f(x-0$ and $f(x+)$ are the left- and right-hand limits at $x$, respectively. To see the integral result, we have \begin{align*} \int_0^{\infty} g(y) dy &= ...

3

Elements of the cantor set $\mathcal{C}$ can be written in ternary as $0.d_1d_2d_3d_4\ldots$, where $d_1, d_2, d_3, \ldots \in \{0,2\}$. Elements of $\mathcal{C} \, /\, 2$ can therefore be written as $0.d_1d_2d_3d_4\ldots$, where $d_1, d_2, d_3, \ldots \in \{0,1\}$. Now show that for any $x \in [0,1]$, adding an element of $\mathcal{C}$ to an element of ...

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STEP 1 (Establish a useful equality) :By the fact that $F$ is measurable we have $$m^*(A) = m^*(A\cap F) + m^*(A - F)$$ and since $m^*(A \cap F) \leq m^*(E \cap F) = 0$ we have $$m^*(A) = m^*(A - F)$$ and by the same arguments $$m^*(B) = m^*(B - E)$$ STEP 2 (Actual equality) From the definitions of the measurability of $E$ we have $$m^*(A \cup B) = m^*((A ... 2 Since F is measurable m^*(A)=m^*(A \cap F)+m^*(A \cap F^c) (1) A \cap F \subset E \cap F so m^*(A \cap F)=0 (2) from (1),(2) m^*(A)=m^*(A \cap F^c) (3) again because F is measurable m^*(A \cup B)=m^*((A \cup B)\cap F)+m^*((A \cup B) \cap F^c)=m^*((A \cap F)\cup (B \cap F))+m^*((A \cap F^c) \cup (B \cap F^c)=m^*((A \cap F) \cup B)+m^*(A \cap F^c) ... 2 Recall that the conditional expectation {\rm E}[X\mid Y] has the property that$$ \int_A {\rm E}[X\mid Y]\,\mathrm dP=\int_A X\,\mathrm dP,\quad A\in\sigma(Y). $$Since \{Y=y\}\in \sigma(Y) for all y, we have$$ \int_{\{Y=y\}}{\rm E}[X\mid Y]\,\mathrm dP=\int_{\{Y=y\}}X\,\mathrm dP=\int_\Omega X\mathbf{1}_{\{Y=y\}}\,\mathrm dP. $$At last, recall that ... 2 Suppose that the E_n are increasing (otherwise it is false). If x\in E, then \chi_{E_n}(x) = 0 as long as x\notin E_n, and then \chi_{E_n}(x) = 1. This happens for every x\in E=\cup E_n. If x\notin E, then x\notin E_n and the sequence stays at 0. In every case,$$\chi_{E_n}(x)f(x)\to \chi_{E}(x)f(x) $$2 There are several approaches that can be used here. One is to use the inequality$$ 1+x\le e^x\tag{1} $$for all x. We can substitute x\mapsto\dfrac{2x}{k}$$ \begin{align} 1+\frac{2x}{k}&\le e^{\large\frac{2x}{k}}\tag{2}\\ \left(1+\frac{2x}{k}\right)^{\large\frac k2}&\le e^{x}\tag{3}\\ \left(1+\frac{2x^2}{k}\right)^{\large\frac k2}&\le ...

2

Consider the function $u$ defined on $x\gt0$ by $$u:x\mapsto k\log x-x^2+x.$$ Then $$u'(x)=(k/x)-2x+1,\qquad u''(x)=-(k/x^2)-2.$$ If $k\gt0$, $u''\lt0$ hence $u'$ decreases from $u'(0+)=+\infty$ to $u'(+\infty)=-\infty$, that is, $u$ is increasing then decreasing, in particular $u$ is uniformly upper bounded, that is, $$u(x)\leqslant c$$ for every ...

2

Exercise: Let $H:\mathbb R\to\mathbb R$ denote a non-decreasing function. Show that $H$ is continuous from the right at $a$ if and only if $H(x_n)\to H(a)$ for at least one sequence $(x_n)$ such that $x_n\geqslant a$ for every $n$ and $x_n\to a$ when $n\to\infty$, if and only if $H(a+1/n)\to H(a)$ when $n\to\infty$. Recall that $H$ being continuous from ...

2

Suppose that the are in $\Bbb R$. Let us consider $$f(x) = \int_A 1_{(-\infty,x)}d\lambda$$ $f$ is increasing and $$\lim_{h\to 0^+} f(x+h)-f(x) = \lim_{h\to 0^+}\int_A 1_{[x,x+h)}d\lambda = 0$$ because of the Monotone convergence theorem, and also $$\lim_{h\to 0^-} f(x+h)-f(x) = 0$$ so $f$ is continuous; and as $f(x\to\infty)\to p$, the intermediate ...

2

An important property of Lebesgue measure is that it is nonatomic. A measure $\mu$ on a measurable space $(\Omega,\Sigma)$ is nonatomic if for every $B\in\Sigma$ such that $\mu(B)>0$, there is $A\in\Sigma$ such that $A\subseteq B$ and $0<\mu(A)<\mu(B)$. A nontrivial $\sigma$-finite Borel measure $\mu$ on $\mathbb{R}$ (or any second countable ...

2

Hint: We can say :$$\omega \in C \Leftrightarrow (\forall m \in \Bbb N )(\exists p \in \Bbb N)( \forall k \geq p) \quad \left|\sum_{i=k+1}^{+\infty} X_k(\omega)\right| < \frac 1{m+1}$$ That gives:$$C=\bigcap_{m \in \Bbb N}\bigcup_{p \in \Bbb N}\bigcap_{k \geq p} Y_{m,p,k}$$ Where : $$Y_{m,p,k}=\left\{\omega \in \Omega / \left| \sum_{i=k+1}^{+\infty} ... 2 Using \|F-F_k\|_\infty \le 2^{-k+1} \|F_0-F_1\|_\infty = 2^{-k}/3 you get$$ |F(y)-F(x)| \le 2/3\cdot 2^{-k} + |F_k(y)-F_k(x)| \le (2/3 + 3^k |x-y|) 2^{-k}. $$Now use 3^{-k-1} \le |x-y| < 3^{-k} for some k to obtain$$ |F(y)-F(x)| < \frac{5}{3} 2^{-k} \le \frac{10}{3}|x-y|^{\log_3(2)} Problem 9.12 in my lecture notes contains a hint for an ... 2 We need to check that the random variable X(t) is measurable with respect to \sigma-algebra \mathcal F_t for each t\ge0. If t\in[0,1], the only value that X(t) can take is 0. So we need to find the smallest \sigma-algebra that contains X^{-1}(t)(\{0\})=\Omega. Such \sigma-algebra is \{\emptyset,\Omega\}. If t\in(1,2], the range of ... 2 A set of positive measure does not necessarily have an open subset. An example is the set of irrationals on [0,1], i.e. [0,1] \setminus \mathbb{Q}. So you cannot conclude necessarily that there exists open A on which f > 0. However, it is the case that if a set is measurable, it differs from an open set on a set of measure zero. So you can find ... 2 The key is to show that for some \lambda>0, the set \{x\in (a,b): f(x)>\lambda\} is non-empty. Once you can show such a \lambda exists, then we'd have \begin{align*}\int_{a}^{b}f(x)\ dx\ge \int_{\{x\in (a,b): f(x)>\lambda\}}f(x)\ dx>\lambda\cdot |\{x\in (a,b): f(x)>\lambda\}| >0 \end{align*} (Hint: \{x\in (a,b): ... 1 Hi you have to use properties of Dynkin System, you want to prove that F^c\cap E\in D but F^c\cap E=(\Omega \setminus F )\cap E= (\Omega \cap E ) \setminus (F\cap E ) In the last term both \Omega \cap E and F\cap E belong to D and \Omega \cap E includes F\cap E. Now Dynkin systems are stable by set difference as long as the first term of ... 1 Using f_n=f\chi_{E_n} is a good idea. After slightly rewriting, you have that \int \lim_{n\rightarrow \infty} f_n d\lambda=\lim_{n\rightarrow \infty} \int_{E_n} f d\lambda Note that you have the right hand side of the equality you want to show. What remains is to show that the integrand on the left hand side converges pointwise to the function you ... 1 Introductory Functional Analysis and applications by Erwin Kreszig ?? By the way Brezis is very good in the sense the theory is kind of developed by asking you to do exercises. For measure theory Halmos ? But why don't you try this ? Prove all the propositions and theorems and corollaries by yourself without first looking at the explained text. That ... 1 Rewriting this in terms of X\setminus A_j and using \mu(X)=1, the inequality is equivalent to\mu\left(\bigcup_{j=1}^n(X\setminus A_j)\right)\leqslant \sum_{j=1}^n\mu(X\setminus A_j).$$This can be handled integrating the inequality$$\chi\left(\bigcup_{j=1}^nB_j\right)\leqslant \sum_{j=1}^n\chi(B_j)$$valid for any collection (B_j)_{j=1}^n of ... 1 Hint. f is integrable on X if and only if$$ \int_X \lvert f\rvert\,d\mu=\sum_{n\in\mathbb N}\lvert n^2+n-6\rvert\cdot \lvert n-2\rvert^{-4}<\infty. $$Indeed, f is integrable, as the sum above converges. Note. As Daniel Fischer observed, we should be careful when n=2. At n=2 the function f blows-up, but the measure of \{2\} is zero:$$ ...

1

If $L$ and $U$ are measurable, for is their difference $L-U$ and hence $$E=(L-U)^{-1}(\{0\})$$ is a measurable set, as an inverse of a measurable set. In particular, the characteristic functions of $E$ and $E^c$ are measurable functions. So $$f=\chi_E\cdot L,$$ and thus $f$ is measurable as a product of such.

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I assume that you mean that $x,y$ are two (fixed) elements of $[0,1)$ and that $M$ is the collection of all sets that either contain both $x$ and $y$ or contain neither $x$ nor $y$. We will show that $M$ is a $\sigma$-algebra. It is immediate that $\Omega \in M$ as $\Omega$ contains both $x$ and $y$. Furthermore, if $A \in M$ then $A^c \in M$ since if $A$ ...

1

This $\mu$ is a probability distribution on $[0,\infty)$ (from the case $n=0$). Let $X$ be a random variable with that distribution. Then $n^n$ are the moments of $X$, and the moment generating function of $X$ is $$\mathbb E[e^{tX}] = \sum_{n=0}^\infty \frac{n^n}{n!}\;t^n = \frac{1}{1+W(-t)}$$ where $W$ is the Lambert W function. So our measure is the ...

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I finally got this. Since $f$ is Schwartz, $\hat{f}$ is also Schwartz (in particular, $C^\infty$), so we may take the Taylor series of $\hat{f}$ near the origin. Through interchanging $D_\xi$ and $x$, we can easily show that $\hat{f}(0) = 1$, $(\hat{f})'(0)$ (by $\int xf(x)dx = 0$), and $(\hat{f})''(0) = \int x^2 f(x)dx$ which is some finite constant, say ...

1

The singletons look to me like an unnecessary detour. If $F$ were generated by countably many of its elements, then it would also be generated by a family $G$ of countably many countable sets, because you could just replace any co-countable sets among the original generators by their complements. The union $U$ of all the generators in $G$ is a countable ...

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If $\sum_{n=1}^\infty p_n<\infty$, then you can find for each $\epsilon>0$ some index $N$ such that $\sum_{n=N}^\infty p_n<\epsilon$. In particular, you can do this for some $0<\epsilon<1$. By what you have shown, $$\prod_{n=N}^M (1-p_n)>1-\epsilon>0,$$ and since $\prod_{n=N}^M (1-p_n)$ converges as $M\to\infty$ (it is a decreasing ...

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