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You have $$\chi_{\bigcup B_n}(x) = \bigvee \chi_{B_n}(x)$$ where $\vee$ is the operation which gives the maximum of its arguments.

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Hint: Since the sets $B_i$ are disjoint by assumption, we have for each point $x_0 \in \mathbb{R}^d$ that $$x_0 \in \cup_{n} B_n \Leftrightarrow \exists\, n_0: x_0 \in B_{n_0}$$ where the RHS implies (due to disjointness) that $x_0 \notin B_n$ for all $n\neq n_0$. Substituting into the characteristic function yields the required equality.

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The statement is not true in general, since $\chi$ takes the values $0,1$ but $\sum\chi$ takes values on (potentially) all of $\mathbb N$. The equality only holds if the $B_n$ are disjoint. Otherwise, suppose $x\in B_1\cap B_2$. Then $\chi_{\bigcup B_n}(x)=1$, but $\sum \chi_{B_n}(x)\geq \chi_{B_1}(x)+\chi_{B_2}(x)=2$. We have a non-$0,1$ value for the sum ...

0

The last formula is nothing more than a tautology. Recall that $d\lambda$ is the Lebesgue measure on the line. Also, the integrand function is continuous, hence Riemann integrable. Therefore the Lebesgue integral reduces to a Riemann integral.

2

(1) The problem lies in the definition of $e^{itx}$ when $x=\pm \infty$, as $\lim_{x\to +\infty} e^{itx}$ is not well defined. We are not bothered by this when $X\in\mathbb R$ almost surely, because we can define $\widetilde X$ a real valued random variable equal to $X$ almost everywhere. In particular, it won't change the integral. (2) More generally, if ...

2

Define $w=y-Cx$, then $$E[\mathrm e^{\mathrm iu^Tw}|\mathcal F^x] =E[\mathrm e^{\mathrm iu^Ty}|\mathcal F^x]\,\mathrm e^{-\mathrm iu^TCx} = \mathrm e^{-u^T \tilde{Q}u/2}$$ is independent of $x$. Thus $w$ is independent of $x$ and centered normal with covariance $\tilde{Q}$.

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I think that, in your second equality, you are stating that: $$\mathbb{E}\left[\mathrm e^{itXY}|Y \right]$$ is some function of $X$. As you condition on the $\sigma$-algebra $\sigma(Y)$, this conditional expectation must be a measurable function of $Y$. For example, in the convenient case where $X$ and $Y$ are independent, we have:  ...

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