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Let $C[0,1]$ the space of continuous functions $x:[0,1]\to \mathbb{R}$ with $x(0)=0$. Equip $C[0,1]$ with the sigma algebra generated by the cylinder sets $$B_E=\{x\in C:(x(t_1),x(t_2),\dots,x(t_n))\in E\in B(\mathbb{R}^n)\}$$ which coincides with the Borel sigma algebra. Let $W$ denote the Wiener measure on $B(C[0,1])$ and consider the measure space $(C[0,1],B(C[0,1]),W)$. Then the functionals $$f_t:C[0,1]\to f_t(x)=x(t)$$ are normally distributed random variables $N(0,t)$. For $t>s$, $f_t-f_s$ is $N(0,t-s)$. And finally for $t_3>t_2>t_1$, $f_{t_3}-f_{t_2}$ is stochastically independent of $f_{t_2}-f_{t_1}$.

We see that $(f_t)_{t\in[0,1]}$ satisfies the properties of Brownian motion.

Is $f_t(\omega)=B(t,\omega)$ a Brownian motion for $\omega\in C[0,1]$? How do I show almost sure continuity?

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Isn't it there in the very definition? For every fixed $x \in C[0,1]$ the sample path corresponding to it simply the function $x$ on $[0,1]$ which is continuous.

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  • $\begingroup$ Is this the correct definition of the Wiener process? $\endgroup$
    – user515599
    Jul 18, 2019 at 9:22
  • $\begingroup$ Yes. The definition is $B(t,\omega)=\omega (t)$ for $\omega \in C[0,1]$ and $t \in [0,1]$. $\endgroup$ Jul 18, 2019 at 9:23

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