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Your elementary divisors are primary, so powers of irreducible polynomials. Clearly the irreducible (over $\Bbb Q$) polynomials $x$, $x^2-2$ and $x^2+4$ are the only ones relevant here, and they can be considered independently. For each the power in $p_A$ gives the product of the elementary divisors for this irreducible, and the power in $m_A$ gives the ...

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As commenters said, the answer is negative: e.g., $f(x) =|x|^{-1} e^{-|x|}$ is a nonintegrable function such that $|f(x)\sin x|$ is globally integrable. More generally, the following are equivalent: $f$ is integrable whenever $fg$ is; $\operatorname{ess\,inf}|g|>0$ Indeed, if 2) fails then the sets $A_n = \{x: 2^{-n}\le |g(x)|<2^{1-n}\}$ have ...

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$\frac{\sin tu-\sin t(u-x)}{t}$ has a removable singularity at $t=0$, that is $\lim_{t\to 0} \frac{\sin tu-\sin t(u-x)}{t}$ exists and is finite. So no principle value integral is required. It is a double integral of a bounded function over two finite measure spaces, and so Fubini applies.

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If $X$ is a Poisson type random variable i.e. for $k\in \mathbb{N}_0$ and $\lambda>0$ $$P\{X=x_0+k\cdot h\}=\frac{\lambda^k e^{-\lambda}}{k!}$$ its characteristic function is given by $$\varphi_X(t)=\sum_{k=0}^{\infty}e^{it(x_0+kh)}\frac{\lambda^k e^{-\lambda}}{k!}=e^{itx_0}\sum_{k=0}^{\infty}e^{it(kh)}\frac{\lambda^k ... 1 For complex numbers, by definition \log z = \log |z| + i \arg z. Consequently, since I_n(t) \to \log f(t) = \log |f(t)| + i \arg f(t), where only the first term is real, \Re I_n(t) \to \log |f(t)|. 1 The default topology on \{0,1\} is the subspace topology induced from \mathbb{N}, \mathbb{Z}, \mathbb{Q}, \mathbb{R}, i.e. the discrete topology. If a different topology is considered, that is (usually, at least) explicitly mentioned. Typically, characteristic functions are considered in contexts where other real- or complex-valued functions occur too, ... 1 Once you've shown that n^{-1} S_n has the same distribution as X_1, then it's pretty much immediate that n^{-\gamma} S_n does not convergence in distribution for \gamma < 1. If \gamma > 1 the characteristic function argument shows that n^{- \gamma} S_n \stackrel{\text{d}}{\to} 0. So n^{-\gamma} S_n converges in distribution for \gamma ... 0 For each u consider the random variable X_u: \Omega_u\to \Bbb{R}  here \Omega_u= \Bbb{R} but we consider it endowed with the probability \mu_u associated with the characteristic function f(u,\cdot). The probability space is (\Bbb{R}, \mathbb{B}, \mu_u) Extend this to (\mathbb{R}^{(-\infty,\infty)}, \mathcal{B}(\mathbb{R})^{\otimes ... 0 Partial answer: According to a theorem (Gil-Pelaez), if 0 is a point of continuity of F_X, then$$ F_X(0) = \frac{1}{2} - \frac{1}{\pi} \int_0^{\infty} \frac{\mathrm{Im}[\phi_X(t)]}{t} dt, so, since a random variable X is a.s. non-negative, then F_X(0) = 0, \int_0^{\infty} \frac{\mathrm{Im}[\phi_X(t)]}{t} dt = \pi / 2 is the condition for a.s. ... 0 You're saying the pair (X,Y) has the same distribution as the pair (\bar X,\bar Y) and \bar X,\bar Y are independpendent and you want to prove X,Y are independent. \begin{align} & \Pr(X\in A\ \&\ Y\in B) \\[10pt] = {} & \Pr((X,Y)\in A\times B) \\[10pt] = {} & \Pr((\bar X,\bar Y)\in A\times B) & & \text{(since the joint ... 0 Denote by \mathbb{P}_X the distribution of a random variable X and by "\stackrel{d}{=}" equality in distribution. Since \tilde{X} and \tilde{Y} are independent, the distribution of \tilde{Z} equals\mathbb{P}_{\bar{Z}} = \mathbb{P}_{\tilde{X}} \otimes \mathbb{P}_{\tilde{Y}}. Moreover, $\tilde{X} \stackrel{d}{=}X$ and \$\tilde{Y} ...

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