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5

There is a Borel set $E$ in $\mathbb R^2$ such that its difference set $F := \{x-y\colon x,y \in E\}$ is not a Borel set. Let $A := \{f \in \mathbf{C}\colon (f(0), f(1)) \in E\}$. 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.

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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 : $$Y_{m,p,k}=\left\{\omega \in \Omega / \left| \sum_{i=k+1}^{+\infty} ... 1 Rewriting this in terms of X\setminus A_j and using \mu(X)=1, the inequality is equivalent to$$\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)$$valid for any collection (B_j)_{j=1}^n of ... 1 Rewrite this as z_n=z_{n-1}y_n where z_n=f(x_n) and y_n=1+\alpha/m+\beta\epsilon_n/\sqrt{m}, thus, (y_n) is i.i.d. such that E(y_1)=\gamma with \gamma=1+\alpha/m and \mathrm{var}(y_1)=\beta^2/m. If x_0 is independent of (\epsilon_n), one gets E(z_n)=E(z_{n-1})E(y_n), that is,$$E(f(x_n))=\gamma^nE(f(x_0)).  Likewise, ...

1

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

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 ...

1

Correct answer with simple proof: $\frac{1}{2}$. Notice that the entire situation is symmetric. Everything happens with the same probability to white as it does to red. Thus $P(Red)=P(white)=\frac{1}{2}$ What's wrong with your answer: "We prove that if all that jugs have same amount of red balls and the same amount of white balls (each one), the probability ...

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Consider the following two scenarios: 1) You have two independent Poisson processes $N_1(t)$ and $N_2(t)$ with rates $\lambda p$ and $\lambda (1-p)$. $N(t) = N_1(t) + N_2(t)$ is their sum. This is a Poisson process with rate $\lambda$. 2) You have a Poisson process $N(t)$ with rate $\lambda$. Each time an event occurs, a coin is tossed: with ...

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