Revisions to Prove that $\forall x,y \in \mathbb{R}^d , P_x\{y\in B\mathopen]0,1]\}=0$

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edited Sep 12, 2023 at 14:18

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I'm folowing the proof of corollary 1.8 page 5 of Mörters - Sample path properties of Brownian motion.

I want to show that $$\forall x,y \in \mathbb{R}^d , P_x\{y\in B\mathopen]0,1]\}=0$$ where $B$ is a $d$-dimensional Brownian motion.
We will use the Lévy's theorem that says: $$\mathcal{L}_2(B[0,1])=0 \hspace{0.5cm} \text{almost surely.}$$

in the first step we will show that $$\forall y \in \mathbb{R}^d, P_x\{y\in B[0,1]\}=0 \hspace{0.5cm} \text{for $\mathcal{L}_2$-almost every } x \in \mathbb{R}^d$$ for any fixed $y\in\mathbb{R}^d$\begin{eqnarray*} \int_{\mathbb{R}^2}P_{y}(x\in B[0,1])dx&=&\int_{\mathbb{R}^2}\int_\Omega 1_{\{x\in B[0,1]\}}(\omega)dP_y(\omega)dx\\&=&\int_{\mathbb{R}^2}\int_\Omega 1_{B[0,1](\omega)}(x)dP_y(\omega)dx\\&=&\int_\Omega \int_{\mathbb{R}^2}1_{B[0,1](\omega)}(x)dxdP_y(\omega)\hspace{0.5cm} \text{by Tonelli}\\&=& \int_\Omega \mathcal{L}_2(B[0,1](\omega))dP_y(\omega)\\&=& E_{y}(\mathcal{L}_{2}(B[0,1])=0\\ \end{eqnarray*} so $$P_x\{y\in B[0,1]\}=0 \hspace{0.5cm} \text{for almost every } x \in \mathbb{R}^d$$ (now how can we get rid of the "for almost every $x$"?).
in the second step he shows by symmetry of Brownian motion that: (but I don't know where did he use it) \begin{eqnarray*} P_{x}(y\in B[0,1])&=&P_{0}(y-x\in B[0,1])\\&=&P_{0}(y-x \in -B[0,1])\\&=&P_{y}(x\in B[0,1])=0\\ \end{eqnarray*}
in the third and last step he will get rid of the "for almost every $x$": we will show that $\forall \epsilon >0, $ we have almost surely $P_{B(\epsilon)}\{y\in B[0,1]\}=0$ and we will find the result by sending $\epsilon$ to zero \begin{eqnarray*} P_{x}\{y\in B\mathopen]0,1]\}&=&P_{x}\{y=B_t / t\in \mathopen]0,1]\} \\&=&P_{x}(\bigcup_{\epsilon>0}\{ y=B_t / t\in \mathopen]\epsilon,1]\})\\&=& \lim_{\epsilon\rightarrow 0}P_{x}\{ y=B_t / t\in ]\epsilon,1]\} \\&=& \lim_{\epsilon\rightarrow 0}P_{x}\{ y=B_{t+\epsilon} / t\in \mathopen]0,1-\epsilon]\} \\&=& \lim_{\epsilon\rightarrow 0}P_{x}\{ y=B_{t+\epsilon}-B_\epsilon +B_\epsilon / t\in \mathopen]0,1-\epsilon]\}\\&=& \lim_{\epsilon\rightarrow 0}P_{x}\{ y=B_t+B_\epsilon / t\in \mathopen]0,1-\epsilon]\}\\&=& \underset{\epsilon \rightarrow 0}{\lim}P_{B\epsilon}\{ y=B_t+x / t\in \mathopen]0,1-\epsilon]\}\\&=& \underset{\epsilon \rightarrow 0}{\lim}\int_\Omega 1_{\{y=B_t+x / t\in \mathopen]0,1-\epsilon]\}}(\omega)dP_{B(\epsilon)}(\omega)\\&=& \lim_{\epsilon\rightarrow 0} E_xP_{B(\epsilon)}\{y\in B[0,1]\}. \end{eqnarray*}

I didn't get how he arrived to the last equation: $\lim_{\epsilon \rightarrow 0}\int_\Omega 1_{\{y=B_t+x / t\in \mathopen]0,1-\epsilon]\}}(\omega)dP_{B(\epsilon)}(\omega)= \lim_{\epsilon\rightarrow 0} E_xP_{B(\epsilon)}\{y\in B[0,1]\}$$\underset{\epsilon \rightarrow 0}{\lim}\int_\Omega 1_{\{y=B_t+x / t\in \mathopen]0,1-\epsilon]\}}(\omega)dP_{B(\epsilon)}(\omega)= \underset{\epsilon\rightarrow 0}{\lim} E_xP_{B(\epsilon)}\{y\in B[0,1]\}$.

I'm folowing the proof of corollary 1.8 page 5 of Mörters - Sample path properties of Brownian motion.

I want to show that $$\forall x,y \in \mathbb{R}^d , P_x\{y\in B\mathopen]0,1]\}=0$$ where $B$ is a $d$-dimensional Brownian motion.
We will use the Lévy's theorem that says: $$\mathcal{L}_2(B[0,1])=0 \hspace{0.5cm} \text{almost surely.}$$

in the first step we will show that $$\forall y \in \mathbb{R}^d, P_x\{y\in B[0,1]\}=0 \hspace{0.5cm} \text{for $\mathcal{L}_2$-almost every } x \in \mathbb{R}^d$$ for any fixed $y\in\mathbb{R}^d$\begin{eqnarray*} \int_{\mathbb{R}^2}P_{y}(x\in B[0,1])dx&=&\int_{\mathbb{R}^2}\int_\Omega 1_{\{x\in B[0,1]\}}(\omega)dP_y(\omega)dx\\&=&\int_{\mathbb{R}^2}\int_\Omega 1_{B[0,1](\omega)}(x)dP_y(\omega)dx\\&=&\int_\Omega \int_{\mathbb{R}^2}1_{B[0,1](\omega)}(x)dxdP_y(\omega)\hspace{0.5cm} \text{by Tonelli}\\&=& \int_\Omega \mathcal{L}_2(B[0,1](\omega))dP_y(\omega)\\&=& E_{y}(\mathcal{L}_{2}(B[0,1])=0\\ \end{eqnarray*} so $$P_x\{y\in B[0,1]\}=0 \hspace{0.5cm} \text{for almost every } x \in \mathbb{R}^d$$ (now how can we get rid of the "for almost every $x$"?).
in the second step he shows by symmetry of Brownian motion that: (but I don't know where did he use it) \begin{eqnarray*} P_{x}(y\in B[0,1])&=&P_{0}(y-x\in B[0,1])\\&=&P_{0}(y-x \in -B[0,1])\\&=&P_{y}(x\in B[0,1])=0\\ \end{eqnarray*}
in the third and last step he will get rid of the "for almost every $x$": we will show that $\forall \epsilon >0, $ we have almost surely $P_{B(\epsilon)}\{y\in B[0,1]\}=0$ and we will find the result by sending $\epsilon$ to zero \begin{eqnarray*} P_{x}\{y\in B\mathopen]0,1]\}&=&P_{x}\{y=B_t / t\in \mathopen]0,1]\} \\&=&P_{x}(\bigcup_{\epsilon>0}\{ y=B_t / t\in \mathopen]\epsilon,1]\})\\&=& \lim_{\epsilon\rightarrow 0}P_{x}\{ y=B_t / t\in ]\epsilon,1]\} \\&=& \lim_{\epsilon\rightarrow 0}P_{x}\{ y=B_{t+\epsilon} / t\in \mathopen]0,1-\epsilon]\} \\&=& \lim_{\epsilon\rightarrow 0}P_{x}\{ y=B_{t+\epsilon}-B_\epsilon +B_\epsilon / t\in \mathopen]0,1-\epsilon]\}\\&=& \lim_{\epsilon\rightarrow 0}P_{x}\{ y=B_t+B_\epsilon / t\in \mathopen]0,1-\epsilon]\}\\&=& \underset{\epsilon \rightarrow 0}{\lim}P_{B\epsilon}\{ y=B_t+x / t\in \mathopen]0,1-\epsilon]\}\\&=& \underset{\epsilon \rightarrow 0}{\lim}\int_\Omega 1_{\{y=B_t+x / t\in \mathopen]0,1-\epsilon]\}}(\omega)dP_{B(\epsilon)}(\omega)\\&=& \lim_{\epsilon\rightarrow 0} E_xP_{B(\epsilon)}\{y\in B[0,1]\}. \end{eqnarray*}

I didn't get how he arrived to the last equation: $\lim_{\epsilon \rightarrow 0}\int_\Omega 1_{\{y=B_t+x / t\in \mathopen]0,1-\epsilon]\}}(\omega)dP_{B(\epsilon)}(\omega)= \lim_{\epsilon\rightarrow 0} E_xP_{B(\epsilon)}\{y\in B[0,1]\}$.

I'm folowing the proof of corollary 1.8 page 5 of Mörters - Sample path properties of Brownian motion.

I want to show that $$\forall x,y \in \mathbb{R}^d , P_x\{y\in B\mathopen]0,1]\}=0$$ where $B$ is a $d$-dimensional Brownian motion.
We will use the Lévy's theorem that says: $$\mathcal{L}_2(B[0,1])=0 \hspace{0.5cm} \text{almost surely.}$$

in the first step we will show that $$\forall y \in \mathbb{R}^d, P_x\{y\in B[0,1]\}=0 \hspace{0.5cm} \text{for $\mathcal{L}_2$-almost every } x \in \mathbb{R}^d$$ for any fixed $y\in\mathbb{R}^d$\begin{eqnarray*} \int_{\mathbb{R}^2}P_{y}(x\in B[0,1])dx&=&\int_{\mathbb{R}^2}\int_\Omega 1_{\{x\in B[0,1]\}}(\omega)dP_y(\omega)dx\\&=&\int_{\mathbb{R}^2}\int_\Omega 1_{B[0,1](\omega)}(x)dP_y(\omega)dx\\&=&\int_\Omega \int_{\mathbb{R}^2}1_{B[0,1](\omega)}(x)dxdP_y(\omega)\hspace{0.5cm} \text{by Tonelli}\\&=& \int_\Omega \mathcal{L}_2(B[0,1](\omega))dP_y(\omega)\\&=& E_{y}(\mathcal{L}_{2}(B[0,1])=0\\ \end{eqnarray*} so $$P_x\{y\in B[0,1]\}=0 \hspace{0.5cm} \text{for almost every } x \in \mathbb{R}^d$$ (now how can we get rid of the "for almost every $x$"?).
in the second step he shows by symmetry of Brownian motion that: (but I don't know where did he use it) \begin{eqnarray*} P_{x}(y\in B[0,1])&=&P_{0}(y-x\in B[0,1])\\&=&P_{0}(y-x \in -B[0,1])\\&=&P_{y}(x\in B[0,1])=0\\ \end{eqnarray*}
in the third and last step he will get rid of the "for almost every $x$": we will show that $\forall \epsilon >0, $ we have almost surely $P_{B(\epsilon)}\{y\in B[0,1]\}=0$ and we will find the result by sending $\epsilon$ to zero \begin{eqnarray*} P_{x}\{y\in B\mathopen]0,1]\}&=&P_{x}\{y=B_t / t\in \mathopen]0,1]\} \\&=&P_{x}(\bigcup_{\epsilon>0}\{ y=B_t / t\in \mathopen]\epsilon,1]\})\\&=& \lim_{\epsilon\rightarrow 0}P_{x}\{ y=B_t / t\in ]\epsilon,1]\} \\&=& \lim_{\epsilon\rightarrow 0}P_{x}\{ y=B_{t+\epsilon} / t\in \mathopen]0,1-\epsilon]\} \\&=& \lim_{\epsilon\rightarrow 0}P_{x}\{ y=B_{t+\epsilon}-B_\epsilon +B_\epsilon / t\in \mathopen]0,1-\epsilon]\}\\&=& \lim_{\epsilon\rightarrow 0}P_{x}\{ y=B_t+B_\epsilon / t\in \mathopen]0,1-\epsilon]\}\\&=& \underset{\epsilon \rightarrow 0}{\lim}P_{B\epsilon}\{ y=B_t+x / t\in \mathopen]0,1-\epsilon]\}\\&=& \underset{\epsilon \rightarrow 0}{\lim}\int_\Omega 1_{\{y=B_t+x / t\in \mathopen]0,1-\epsilon]\}}(\omega)dP_{B(\epsilon)}(\omega)\\&=& \lim_{\epsilon\rightarrow 0} E_xP_{B(\epsilon)}\{y\in B[0,1]\}. \end{eqnarray*}

I didn't get how he arrived to the last equation: $\underset{\epsilon \rightarrow 0}{\lim}\int_\Omega 1_{\{y=B_t+x / t\in \mathopen]0,1-\epsilon]\}}(\omega)dP_{B(\epsilon)}(\omega)= \underset{\epsilon\rightarrow 0}{\lim} E_xP_{B(\epsilon)}\{y\in B[0,1]\}$.

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edited Sep 12, 2023 at 14:12

LSpice

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I'm folowing the proof of corollary 1.8 page 5 of these lecture notesMörters - Sample path properties of Brownian motion.

I want to show that $$\forall x,y \in \mathbb{R}^d , P_x\{y\in B\mathopen]0,1]\}=0$$ where $B$ is a $d$-dimensional Brownian motion.
We will use the Lévy's theorem that says: $$\mathcal{L}_2(B[0,1])=0 \hspace{0.5cm} \text{almost surely.}$$

in the first step we will show that $$\forall y \in \mathbb{R}^d, P_x\{y\in B[0,1]\}=0 \hspace{0.5cm} \text{for $\mathcal{L}_2$-almost every} x \in \mathbb{R}^d$$$$\forall y \in \mathbb{R}^d, P_x\{y\in B[0,1]\}=0 \hspace{0.5cm} \text{for $\mathcal{L}_2$-almost every } x \in \mathbb{R}^d$$ for any fixed $y\in\mathbb{R}^d$\begin{eqnarray*} \int_{\mathbb{R}^2}P_{y}(x\in B[0,1])dx&=&\int_{\mathbb{R}^2}\int_\Omega 1_{\{x\in B[0,1]\}}(\omega)dP_y(\omega)dx\\&=&\int_{\mathbb{R}^2}\int_\Omega 1_{B[0,1](\omega)}(x)dP_y(\omega)dx\\&=&\int_\Omega \int_{\mathbb{R}^2}1_{B[0,1](\omega)}(x)dxdP_y(\omega)\hspace{0.5cm} \text{par Tonelli}\\&=& \int_\Omega \mathcal{L}_2(B[0,1](\omega))dP_y(\omega)\\&=& E_{y}(\mathcal{L}_{2}(B[0,1])=0\\ \end{eqnarray*} \begin{eqnarray*} \int_{\mathbb{R}^2}P_{y}(x\in B[0,1])dx&=&\int_{\mathbb{R}^2}\int_\Omega 1_{\{x\in B[0,1]\}}(\omega)dP_y(\omega)dx\\&=&\int_{\mathbb{R}^2}\int_\Omega 1_{B[0,1](\omega)}(x)dP_y(\omega)dx\\&=&\int_\Omega \int_{\mathbb{R}^2}1_{B[0,1](\omega)}(x)dxdP_y(\omega)\hspace{0.5cm} \text{by Tonelli}\\&=& \int_\Omega \mathcal{L}_2(B[0,1](\omega))dP_y(\omega)\\&=& E_{y}(\mathcal{L}_{2}(B[0,1])=0\\ \end{eqnarray*} so $$P_x\{y\in B[0,1]\}=0 \hspace{0.5cm} \text{for almost every} x \in \mathbb{R}^d$$$$P_x\{y\in B[0,1]\}=0 \hspace{0.5cm} \text{for almost every } x \in \mathbb{R}^d$$ (now how can we get rid of the "for almost every x"$x$"?).
in the second step he showshows by symmetry of Brownian motion that: (but I don't know where did he use it) \begin{eqnarray*} P_{x}(y\in B[0,1])&=&P_{0}(y-x\in B[0,1])\\&=&P_{0}(y-x \in -B[0,1])\\&=&P_{y}(x\in B[0,1])=0\\ \end{eqnarray*}
in the third and last step he will get rid of the "for almost every x"$x$": we will show that $\forall \epsilon >0, $ we have almost surely $P_{B(\epsilon)}\{y\in B[0,1]\}=0$ and we will find the result by sending $\epsilon$ to zero \begin{eqnarray*} P_{x}\{y\in B\mathopen]0,1]\}&=&P_{x}\{y=B_t / t\in \mathopen]0,1]\} \\&=&P_{x}(\underset{\epsilon>0}{\cup}\{ y=B_t / t\in ]\epsilon,1]\})\\&=& \underset{\epsilon \rightarrow 0}{\lim}P_{x}\{ y=B_t / t\in ]\epsilon,1]\} \\&=& \underset{\epsilon \rightarrow 0}{\lim}P_{x}\{ y=B_{t+\epsilon} / t\in \mathopen]0,1-\epsilon]\} \\&=& \underset{\epsilon \rightarrow 0}{\lim}P_{x}\{ y=B_{t+\epsilon}-B_\epsilon +B_\epsilon / t\in \mathopen]0,1-\epsilon]\}\\&=& \underset{\epsilon \rightarrow 0}{\lim}P_{x}\{ y=B_t+B_\epsilon / t\in \mathopen]0,1-\epsilon]\}\\&=& \underset{\epsilon \rightarrow 0}{\lim}P_{B\epsilon}\{ y=B_t+x / t\in \mathopen]0,1-\epsilon]\}\\&=& \underset{\epsilon \rightarrow 0}{\lim}\int_\Omega 1_{\{y=B_t+x / t\in \mathopen]0,1-\epsilon]\}}(\omega)dP_{B(\epsilon)}(\omega)\\&=& \underset{\epsilon\rightarrow 0}{\lim} E_xP_{B(\epsilon)}\{y\in B[0,1]\} \end{eqnarray*}\begin{eqnarray*} P_{x}\{y\in B\mathopen]0,1]\}&=&P_{x}\{y=B_t / t\in \mathopen]0,1]\} \\&=&P_{x}(\bigcup_{\epsilon>0}\{ y=B_t / t\in \mathopen]\epsilon,1]\})\\&=& \lim_{\epsilon\rightarrow 0}P_{x}\{ y=B_t / t\in ]\epsilon,1]\} \\&=& \lim_{\epsilon\rightarrow 0}P_{x}\{ y=B_{t+\epsilon} / t\in \mathopen]0,1-\epsilon]\} \\&=& \lim_{\epsilon\rightarrow 0}P_{x}\{ y=B_{t+\epsilon}-B_\epsilon +B_\epsilon / t\in \mathopen]0,1-\epsilon]\}\\&=& \lim_{\epsilon\rightarrow 0}P_{x}\{ y=B_t+B_\epsilon / t\in \mathopen]0,1-\epsilon]\}\\&=& \underset{\epsilon \rightarrow 0}{\lim}P_{B\epsilon}\{ y=B_t+x / t\in \mathopen]0,1-\epsilon]\}\\&=& \underset{\epsilon \rightarrow 0}{\lim}\int_\Omega 1_{\{y=B_t+x / t\in \mathopen]0,1-\epsilon]\}}(\omega)dP_{B(\epsilon)}(\omega)\\&=& \lim_{\epsilon\rightarrow 0} E_xP_{B(\epsilon)}\{y\in B[0,1]\}. \end{eqnarray*}

iI didn't get how he arrivearrived to the last equation: $\underset{\epsilon \rightarrow 0}{\lim}\int_\Omega 1_{\{y=B_t+x / t\in \mathopen]0,1-\epsilon]\}}(\omega)dP_{B(\epsilon)}(\omega)= \underset{\epsilon\rightarrow 0}{\lim} E_xP_{B(\epsilon)}\{y\in B[0,1]\}$$\lim_{\epsilon \rightarrow 0}\int_\Omega 1_{\{y=B_t+x / t\in \mathopen]0,1-\epsilon]\}}(\omega)dP_{B(\epsilon)}(\omega)= \lim_{\epsilon\rightarrow 0} E_xP_{B(\epsilon)}\{y\in B[0,1]\}$.

I'm folowing the proof of corollary 1.8 page 5 of these lecture notes.

I want to show that $$\forall x,y \in \mathbb{R}^d , P_x\{y\in B\mathopen]0,1]\}=0$$ where $B$ is a $d$-dimensional Brownian motion.
We will use the Lévy's theorem that says: $$\mathcal{L}_2(B[0,1])=0 \hspace{0.5cm} \text{almost surely.}$$

in the first step we will show that $$\forall y \in \mathbb{R}^d, P_x\{y\in B[0,1]\}=0 \hspace{0.5cm} \text{for $\mathcal{L}_2$-almost every} x \in \mathbb{R}^d$$ for any fixed $y\in\mathbb{R}^d$\begin{eqnarray*} \int_{\mathbb{R}^2}P_{y}(x\in B[0,1])dx&=&\int_{\mathbb{R}^2}\int_\Omega 1_{\{x\in B[0,1]\}}(\omega)dP_y(\omega)dx\\&=&\int_{\mathbb{R}^2}\int_\Omega 1_{B[0,1](\omega)}(x)dP_y(\omega)dx\\&=&\int_\Omega \int_{\mathbb{R}^2}1_{B[0,1](\omega)}(x)dxdP_y(\omega)\hspace{0.5cm} \text{par Tonelli}\\&=& \int_\Omega \mathcal{L}_2(B[0,1](\omega))dP_y(\omega)\\&=& E_{y}(\mathcal{L}_{2}(B[0,1])=0\\ \end{eqnarray*} so $$P_x\{y\in B[0,1]\}=0 \hspace{0.5cm} \text{for almost every} x \in \mathbb{R}^d$$ (now how can we get rid of the "for almost every x"?)
in the second step he show by symmetry of Brownian motion that: (but I don't know where did he use it) \begin{eqnarray*} P_{x}(y\in B[0,1])&=&P_{0}(y-x\in B[0,1])\\&=&P_{0}(y-x \in -B[0,1])\\&=&P_{y}(x\in B[0,1])=0\\ \end{eqnarray*}
in the third and last step he will get rid of the "for almost every x" we will show that $\forall \epsilon >0, $ we have almost surely $P_{B(\epsilon)}\{y\in B[0,1]\}=0$ and we will find the result by sending $\epsilon$ to zero \begin{eqnarray*} P_{x}\{y\in B\mathopen]0,1]\}&=&P_{x}\{y=B_t / t\in \mathopen]0,1]\} \\&=&P_{x}(\underset{\epsilon>0}{\cup}\{ y=B_t / t\in ]\epsilon,1]\})\\&=& \underset{\epsilon \rightarrow 0}{\lim}P_{x}\{ y=B_t / t\in ]\epsilon,1]\} \\&=& \underset{\epsilon \rightarrow 0}{\lim}P_{x}\{ y=B_{t+\epsilon} / t\in \mathopen]0,1-\epsilon]\} \\&=& \underset{\epsilon \rightarrow 0}{\lim}P_{x}\{ y=B_{t+\epsilon}-B_\epsilon +B_\epsilon / t\in \mathopen]0,1-\epsilon]\}\\&=& \underset{\epsilon \rightarrow 0}{\lim}P_{x}\{ y=B_t+B_\epsilon / t\in \mathopen]0,1-\epsilon]\}\\&=& \underset{\epsilon \rightarrow 0}{\lim}P_{B\epsilon}\{ y=B_t+x / t\in \mathopen]0,1-\epsilon]\}\\&=& \underset{\epsilon \rightarrow 0}{\lim}\int_\Omega 1_{\{y=B_t+x / t\in \mathopen]0,1-\epsilon]\}}(\omega)dP_{B(\epsilon)}(\omega)\\&=& \underset{\epsilon\rightarrow 0}{\lim} E_xP_{B(\epsilon)}\{y\in B[0,1]\} \end{eqnarray*}

i didn't get how he arrive to the last equation: $\underset{\epsilon \rightarrow 0}{\lim}\int_\Omega 1_{\{y=B_t+x / t\in \mathopen]0,1-\epsilon]\}}(\omega)dP_{B(\epsilon)}(\omega)= \underset{\epsilon\rightarrow 0}{\lim} E_xP_{B(\epsilon)}\{y\in B[0,1]\}$

I'm folowing the proof of corollary 1.8 page 5 of Mörters - Sample path properties of Brownian motion.

I want to show that $$\forall x,y \in \mathbb{R}^d , P_x\{y\in B\mathopen]0,1]\}=0$$ where $B$ is a $d$-dimensional Brownian motion.
We will use the Lévy's theorem that says: $$\mathcal{L}_2(B[0,1])=0 \hspace{0.5cm} \text{almost surely.}$$

in the first step we will show that $$\forall y \in \mathbb{R}^d, P_x\{y\in B[0,1]\}=0 \hspace{0.5cm} \text{for $\mathcal{L}_2$-almost every } x \in \mathbb{R}^d$$ for any fixed $y\in\mathbb{R}^d$ \begin{eqnarray*} \int_{\mathbb{R}^2}P_{y}(x\in B[0,1])dx&=&\int_{\mathbb{R}^2}\int_\Omega 1_{\{x\in B[0,1]\}}(\omega)dP_y(\omega)dx\\&=&\int_{\mathbb{R}^2}\int_\Omega 1_{B[0,1](\omega)}(x)dP_y(\omega)dx\\&=&\int_\Omega \int_{\mathbb{R}^2}1_{B[0,1](\omega)}(x)dxdP_y(\omega)\hspace{0.5cm} \text{by Tonelli}\\&=& \int_\Omega \mathcal{L}_2(B[0,1](\omega))dP_y(\omega)\\&=& E_{y}(\mathcal{L}_{2}(B[0,1])=0\\ \end{eqnarray*} so $$P_x\{y\in B[0,1]\}=0 \hspace{0.5cm} \text{for almost every } x \in \mathbb{R}^d$$ (now how can we get rid of the "for almost every $x$"?).
in the second step he shows by symmetry of Brownian motion that: (but I don't know where did he use it) \begin{eqnarray*} P_{x}(y\in B[0,1])&=&P_{0}(y-x\in B[0,1])\\&=&P_{0}(y-x \in -B[0,1])\\&=&P_{y}(x\in B[0,1])=0\\ \end{eqnarray*}
in the third and last step he will get rid of the "for almost every $x$": we will show that $\forall \epsilon >0, $ we have almost surely $P_{B(\epsilon)}\{y\in B[0,1]\}=0$ and we will find the result by sending $\epsilon$ to zero \begin{eqnarray*} P_{x}\{y\in B\mathopen]0,1]\}&=&P_{x}\{y=B_t / t\in \mathopen]0,1]\} \\&=&P_{x}(\bigcup_{\epsilon>0}\{ y=B_t / t\in \mathopen]\epsilon,1]\})\\&=& \lim_{\epsilon\rightarrow 0}P_{x}\{ y=B_t / t\in ]\epsilon,1]\} \\&=& \lim_{\epsilon\rightarrow 0}P_{x}\{ y=B_{t+\epsilon} / t\in \mathopen]0,1-\epsilon]\} \\&=& \lim_{\epsilon\rightarrow 0}P_{x}\{ y=B_{t+\epsilon}-B_\epsilon +B_\epsilon / t\in \mathopen]0,1-\epsilon]\}\\&=& \lim_{\epsilon\rightarrow 0}P_{x}\{ y=B_t+B_\epsilon / t\in \mathopen]0,1-\epsilon]\}\\&=& \underset{\epsilon \rightarrow 0}{\lim}P_{B\epsilon}\{ y=B_t+x / t\in \mathopen]0,1-\epsilon]\}\\&=& \underset{\epsilon \rightarrow 0}{\lim}\int_\Omega 1_{\{y=B_t+x / t\in \mathopen]0,1-\epsilon]\}}(\omega)dP_{B(\epsilon)}(\omega)\\&=& \lim_{\epsilon\rightarrow 0} E_xP_{B(\epsilon)}\{y\in B[0,1]\}. \end{eqnarray*}

I didn't get how he arrived to the last equation: $\lim_{\epsilon \rightarrow 0}\int_\Omega 1_{\{y=B_t+x / t\in \mathopen]0,1-\epsilon]\}}(\omega)dP_{B(\epsilon)}(\omega)= \lim_{\epsilon\rightarrow 0} E_xP_{B(\epsilon)}\{y\in B[0,1]\}$.

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edited Sep 12, 2023 at 14:02

sara

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I'm folowing the proof of corollary 1.8 page 5 of these lecture notes.

I want to show that $$\forall x,y \in \mathbb{R}^d , P_x\{y\in B\mathopen]0,1]\}=0$$ where $B$ is a $d$-dimensional Brownian motion.
We will use the Lévy's theorem that says: $$\mathcal{L}_2(B[0,1])=0 \hspace{0.5cm} \text{almost surely.}$$

in the first step we will show that $$\forall y \in \mathbb{R}^d, P_x\{y\in B[0,1]\}=0 \hspace{0.5cm} \text{for $\mathcal{L}_2$-almost every} x \in \mathbb{R}^d$$ for any fixed $y\in\mathbb{R}^d$\begin{eqnarray*} \int_{\mathbb{R}^2}P_{y}(x\in B[0,1])dx&=&\int_{\mathbb{R}^2}\int_\Omega 1_{\{x\in B[0,1]\}}(\omega)dP_y(\omega)dx\\&=&\int_{\mathbb{R}^2}\int_\Omega 1_{B[0,1](\omega)}(x)dP_y(\omega)dx\\&=&\int_\Omega \int_{\mathbb{R}^2}1_{B[0,1](\omega)}(x)dxdP_y(\omega)\hspace{0.5cm} \text{par Tonelli}\\&=& \int_\Omega \mathcal{L}_2(B[0,1](\omega))dP_y(\omega)\\&=& E_{y}(\mathcal{L}_{2}(B[0,1])=0\\ \end{eqnarray*} so $$P_x\{y\in B[0,1]\}=0 \hspace{0.5cm} \text{for almost every} x \in \mathbb{R}^d$$ (now how can we get rid of the "for almost every x"?)
in the second step he show by symmetry of Brownian motion that: (but I don't know where did he use it) \begin{eqnarray*} P_{x}(y\in B[0,1])&=&P_{0}(y-x\in B[0,1])\\&=&P_{0}(y-x \in -B[0,1])\\&=&P_{y}(x\in B[0,1])=0\\ \end{eqnarray*}
in the third and last step he will get rid of the "for almost every x" we will show that $\forall \epsilon >0, $ we have almost surely $P_{B(\epsilon)}\{y\in B[0,1]\}=0$ and we will find the result by sending $\epsilon$ to zero \begin{eqnarray*} P_{x}\{y\in B\mathopen]0,1]\}&=&P_{x}\{y=B_t / t\in \mathopen]0,1]\} \\&=&P_{x}(\underset{\epsilon>0}{\cup}\{ y=B_t / t\in ]\epsilon,1]\})\\&=& \underset{\epsilon \rightarrow 0}{\lim}P_{x}\{ y=B_t / t\in ]\epsilon,1]\} \\&=& \underset{\epsilon \rightarrow 0}{\lim}P_{x}\{ y=B_{t+\epsilon} / t\in \mathopen]0,1-\epsilon]\} \\&=& \underset{\epsilon \rightarrow 0}{\lim}P_{x}\{ y=B_{t+\epsilon}-B_\epsilon +B_\epsilon / t\in \mathopen]0,1-\epsilon]\}\\&=& \underset{\epsilon \rightarrow 0}{\lim}P_{x}\{ y=B_t+B_\epsilon / t\in \mathopen]0,1-\epsilon]\}\\&=& \underset{\epsilon \rightarrow 0}{\lim}P_{B\epsilon}\{ y=B_t+x / t\in \mathopen]0,1-\epsilon]\}\\&=& \underset{\epsilon \rightarrow 0}{\lim}\int_\Omega 1_{\{y=B_t+x / t\in \mathopen]0,1-\epsilon]\}}(\omega)dP_{B(\epsilon)}(\omega)\\ \end{eqnarray*}\begin{eqnarray*} P_{x}\{y\in B\mathopen]0,1]\}&=&P_{x}\{y=B_t / t\in \mathopen]0,1]\} \\&=&P_{x}(\underset{\epsilon>0}{\cup}\{ y=B_t / t\in ]\epsilon,1]\})\\&=& \underset{\epsilon \rightarrow 0}{\lim}P_{x}\{ y=B_t / t\in ]\epsilon,1]\} \\&=& \underset{\epsilon \rightarrow 0}{\lim}P_{x}\{ y=B_{t+\epsilon} / t\in \mathopen]0,1-\epsilon]\} \\&=& \underset{\epsilon \rightarrow 0}{\lim}P_{x}\{ y=B_{t+\epsilon}-B_\epsilon +B_\epsilon / t\in \mathopen]0,1-\epsilon]\}\\&=& \underset{\epsilon \rightarrow 0}{\lim}P_{x}\{ y=B_t+B_\epsilon / t\in \mathopen]0,1-\epsilon]\}\\&=& \underset{\epsilon \rightarrow 0}{\lim}P_{B\epsilon}\{ y=B_t+x / t\in \mathopen]0,1-\epsilon]\}\\&=& \underset{\epsilon \rightarrow 0}{\lim}\int_\Omega 1_{\{y=B_t+x / t\in \mathopen]0,1-\epsilon]\}}(\omega)dP_{B(\epsilon)}(\omega)\\&=& \underset{\epsilon\rightarrow 0}{\lim} E_xP_{B(\epsilon)}\{y\in B[0,1]\} \end{eqnarray*}

I'm stuck here,i didn't get how did he find the result, how can I use the transfert formulaarrive to find the last equation: $E_x$?$\underset{\epsilon \rightarrow 0}{\lim}\int_\Omega 1_{\{y=B_t+x / t\in \mathopen]0,1-\epsilon]\}}(\omega)dP_{B(\epsilon)}(\omega)= \underset{\epsilon\rightarrow 0}{\lim} E_xP_{B(\epsilon)}\{y\in B[0,1]\}$

I'm folowing the proof of corollary 1.8 page 5 of these lecture notes.

I want to show that $$\forall x,y \in \mathbb{R}^d , P_x\{y\in B\mathopen]0,1]\}=0$$ where $B$ is a $d$-dimensional Brownian motion.
We will use the Lévy's theorem that says: $$\mathcal{L}_2(B[0,1])=0 \hspace{0.5cm} \text{almost surely.}$$

in the first step we will show that $$\forall y \in \mathbb{R}^d, P_x\{y\in B[0,1]\}=0 \hspace{0.5cm} \text{for $\mathcal{L}_2$-almost every} x \in \mathbb{R}^d$$ for any fixed $y\in\mathbb{R}^d$\begin{eqnarray*} \int_{\mathbb{R}^2}P_{y}(x\in B[0,1])dx&=&\int_{\mathbb{R}^2}\int_\Omega 1_{\{x\in B[0,1]\}}(\omega)dP_y(\omega)dx\\&=&\int_{\mathbb{R}^2}\int_\Omega 1_{B[0,1](\omega)}(x)dP_y(\omega)dx\\&=&\int_\Omega \int_{\mathbb{R}^2}1_{B[0,1](\omega)}(x)dxdP_y(\omega)\hspace{0.5cm} \text{par Tonelli}\\&=& \int_\Omega \mathcal{L}_2(B[0,1](\omega))dP_y(\omega)\\&=& E_{y}(\mathcal{L}_{2}(B[0,1])=0\\ \end{eqnarray*} so $$P_x\{y\in B[0,1]\}=0 \hspace{0.5cm} \text{for almost every} x \in \mathbb{R}^d$$ (now how can we get rid of the "for almost every x"?)
in the second step he show by symmetry of Brownian motion that: (but I don't know where did he use it) \begin{eqnarray*} P_{x}(y\in B[0,1])&=&P_{0}(y-x\in B[0,1])\\&=&P_{0}(y-x \in -B[0,1])\\&=&P_{y}(x\in B[0,1])=0\\ \end{eqnarray*}
in the third and last step he will get rid of the "for almost every x" we will show that $\forall \epsilon >0, $ we have almost surely $P_{B(\epsilon)}\{y\in B[0,1]\}=0$ and we will find the result by sending $\epsilon$ to zero \begin{eqnarray*} P_{x}\{y\in B\mathopen]0,1]\}&=&P_{x}\{y=B_t / t\in \mathopen]0,1]\} \\&=&P_{x}(\underset{\epsilon>0}{\cup}\{ y=B_t / t\in ]\epsilon,1]\})\\&=& \underset{\epsilon \rightarrow 0}{\lim}P_{x}\{ y=B_t / t\in ]\epsilon,1]\} \\&=& \underset{\epsilon \rightarrow 0}{\lim}P_{x}\{ y=B_{t+\epsilon} / t\in \mathopen]0,1-\epsilon]\} \\&=& \underset{\epsilon \rightarrow 0}{\lim}P_{x}\{ y=B_{t+\epsilon}-B_\epsilon +B_\epsilon / t\in \mathopen]0,1-\epsilon]\}\\&=& \underset{\epsilon \rightarrow 0}{\lim}P_{x}\{ y=B_t+B_\epsilon / t\in \mathopen]0,1-\epsilon]\}\\&=& \underset{\epsilon \rightarrow 0}{\lim}P_{B\epsilon}\{ y=B_t+x / t\in \mathopen]0,1-\epsilon]\}\\&=& \underset{\epsilon \rightarrow 0}{\lim}\int_\Omega 1_{\{y=B_t+x / t\in \mathopen]0,1-\epsilon]\}}(\omega)dP_{B(\epsilon)}(\omega)\\ \end{eqnarray*}

I'm stuck here, how did he find the result, how can I use the transfert formula to find the $E_x$?

I'm folowing the proof of corollary 1.8 page 5 of these lecture notes.

I want to show that $$\forall x,y \in \mathbb{R}^d , P_x\{y\in B\mathopen]0,1]\}=0$$ where $B$ is a $d$-dimensional Brownian motion.
We will use the Lévy's theorem that says: $$\mathcal{L}_2(B[0,1])=0 \hspace{0.5cm} \text{almost surely.}$$

in the first step we will show that $$\forall y \in \mathbb{R}^d, P_x\{y\in B[0,1]\}=0 \hspace{0.5cm} \text{for $\mathcal{L}_2$-almost every} x \in \mathbb{R}^d$$ for any fixed $y\in\mathbb{R}^d$\begin{eqnarray*} \int_{\mathbb{R}^2}P_{y}(x\in B[0,1])dx&=&\int_{\mathbb{R}^2}\int_\Omega 1_{\{x\in B[0,1]\}}(\omega)dP_y(\omega)dx\\&=&\int_{\mathbb{R}^2}\int_\Omega 1_{B[0,1](\omega)}(x)dP_y(\omega)dx\\&=&\int_\Omega \int_{\mathbb{R}^2}1_{B[0,1](\omega)}(x)dxdP_y(\omega)\hspace{0.5cm} \text{par Tonelli}\\&=& \int_\Omega \mathcal{L}_2(B[0,1](\omega))dP_y(\omega)\\&=& E_{y}(\mathcal{L}_{2}(B[0,1])=0\\ \end{eqnarray*} so $$P_x\{y\in B[0,1]\}=0 \hspace{0.5cm} \text{for almost every} x \in \mathbb{R}^d$$ (now how can we get rid of the "for almost every x"?)
in the second step he show by symmetry of Brownian motion that: (but I don't know where did he use it) \begin{eqnarray*} P_{x}(y\in B[0,1])&=&P_{0}(y-x\in B[0,1])\\&=&P_{0}(y-x \in -B[0,1])\\&=&P_{y}(x\in B[0,1])=0\\ \end{eqnarray*}
in the third and last step he will get rid of the "for almost every x" we will show that $\forall \epsilon >0, $ we have almost surely $P_{B(\epsilon)}\{y\in B[0,1]\}=0$ and we will find the result by sending $\epsilon$ to zero \begin{eqnarray*} P_{x}\{y\in B\mathopen]0,1]\}&=&P_{x}\{y=B_t / t\in \mathopen]0,1]\} \\&=&P_{x}(\underset{\epsilon>0}{\cup}\{ y=B_t / t\in ]\epsilon,1]\})\\&=& \underset{\epsilon \rightarrow 0}{\lim}P_{x}\{ y=B_t / t\in ]\epsilon,1]\} \\&=& \underset{\epsilon \rightarrow 0}{\lim}P_{x}\{ y=B_{t+\epsilon} / t\in \mathopen]0,1-\epsilon]\} \\&=& \underset{\epsilon \rightarrow 0}{\lim}P_{x}\{ y=B_{t+\epsilon}-B_\epsilon +B_\epsilon / t\in \mathopen]0,1-\epsilon]\}\\&=& \underset{\epsilon \rightarrow 0}{\lim}P_{x}\{ y=B_t+B_\epsilon / t\in \mathopen]0,1-\epsilon]\}\\&=& \underset{\epsilon \rightarrow 0}{\lim}P_{B\epsilon}\{ y=B_t+x / t\in \mathopen]0,1-\epsilon]\}\\&=& \underset{\epsilon \rightarrow 0}{\lim}\int_\Omega 1_{\{y=B_t+x / t\in \mathopen]0,1-\epsilon]\}}(\omega)dP_{B(\epsilon)}(\omega)\\&=& \underset{\epsilon\rightarrow 0}{\lim} E_xP_{B(\epsilon)}\{y\in B[0,1]\} \end{eqnarray*}

i didn't get how he arrive to the last equation: $\underset{\epsilon \rightarrow 0}{\lim}\int_\Omega 1_{\{y=B_t+x / t\in \mathopen]0,1-\epsilon]\}}(\omega)dP_{B(\epsilon)}(\omega)= \underset{\epsilon\rightarrow 0}{\lim} E_xP_{B(\epsilon)}\{y\in B[0,1]\}$