Pointwise modification can destroy path continuity

Let $\mathcal{F}=\{A\cap \mathrm{\Omega }:A\in \mathcal{L}(\mathbb{R})\}$ and define $P(E)={\lambda }_1(E)$ for $E\in \mathcal{F}$. Because $\mathrm{\Omega }$ is Lebesgue measurable, every $E\in \mathcal{F}$ is Lebesgue measurable. The trace family $\mathcal{F}$ contains $\mathrm{\Omega }$, is closed under complements relative to $\mathrm{\Omega }$ and under countable unions, so it is a sigma-algebra. Countable additivity of $P$ is inherited from ${\lambda }_1$, and $P(\mathrm{\Omega })={\lambda }_1([0,1])=1$ by [F1]. Thus $(\mathrm{\Omega },\mathcal{F},P)$ is a probability space by [F3]; this is normalized Lebesgue measure on the interval.

Every path $t\mapsto X_t(\omega )$ is the constant zero function and is continuous by [F4]. Fix any $\omega \in \mathrm{\Omega }$. Its $Y$ path equals one at $t=\omega$ and zero at every other time. Test continuity at $\omega$ with $\epsilon =1\mathrm{/}2$. Given any $\delta >0$, if $\omega <1$ set $h=\min \{\delta \mathrm{/}2,(1-\omega )\mathrm{/}2\}$ and $s=\omega +h$; then $0<h<\delta$, $s\in [0,1]$, and $s\mathrm{\ne }\omega$. If $\omega =1$, set $h=\min \{\delta \mathrm{/}2,1\mathrm{/}2\}$ and $s=1-h$; the same conclusions hold. In either case $\mathrm{\mid }s-\omega \mathrm{\mid }<\delta$ but $\mathrm{\mid }{Y}_s(\omega )-Y_{\omega }(\omega )\mathrm{\mid }=\mathrm{\mid }0-1\mathrm{\mid }=1≮1\mathrm{/}2$. Thus [F4] makes the path discontinuous at its spike. The first case includes the one-sided endpoint $\omega =0$ and all interior points; the second is the one-sided endpoint $\omega =1$.

The identity $U:\mathrm{\Omega }\to \mathbb{R}$ is measurable: for Borel $B\subseteq \mathbb{R}$, $U^{-1}(B)=B\cap \mathrm{\Omega }\in \mathcal{F}$ by [F2]. For fixed $t\in [0,1]$, the event $\{U=t\}=\{t\}$ is measurable by [F1], so $Y_t={\mathbf{1}}_{\{t\}}$ is measurable by [F2]; the constant $X_t=0$ is the indicator of the empty event and is measurable as well. Hence both displayed families are genuine real stochastic processes on the same probability space.

The simultaneous-equality event is $D=\{\omega :X_t(\omega )=Y_t(\omega )\text{ for every }t\in [0,1]\}\mathrm{.}$ For each $\omega \in \mathrm{\Omega }$, take the already given time $t=\omega$. Then $X_{\omega }(\omega )=0$ but $Y_{\omega }(\omega )=1$, so no outcome belongs to $D$ and $D=\mathrm{\varnothing }$. It is measurable and has probability zero, not one; hence [F3] shows that $X$ and $Y$ are not indistinguishable.

Fix $t\in [0,1]$. The equality event is $\{X_t=Y_t\}=\mathrm{\Omega }\setminus \{t\}$, which is measurable, and disjoint additivity with [F1] gives $P(X_t=Y_t)=P(\mathrm{\Omega })-P(\{t\})=1-0=1$. Since this holds for every fixed $t$, [F3] says that $Y$ is a modification of $X$. In particular every finite-dimensional law of either process is the point mass at the all-zero vector: only the finite null set of outcomes equal to one of the selected times can produce a nonzero coordinate for $Y$. If a displayed tuple repeats a time, its repeated coordinates agree and the same all-zero almost-sure conclusion holds.

Steps 1.2--3.1 prove every asserted contrast on the uncountable index set [F5]. The values zero and one, total probability one, the empty simultaneous-equality event, both endpoints, and every interior spike are explicit. There is no biconditional. Countable choice is assumed exactly for the Lebesgue-measure suppliers in [F1]--[F2]; setting $t=\omega$ in step 2.2 and the explicit nearby point in step 1.2 make no choice from an indexed family.