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I am reading the HDR thesis of Joseph Lehec, here:

https://hal.archives-ouvertes.fr/tel-01428644

At page 20 he introduces the Wiener space $(\mathbb W, \mathcal B, \gamma)$, where $\gamma$ is the "Wiener measure", defined as "the law of a standard Brownian motion". Indeed, he introduced a standard n-dim. Brownian motion a few lines before, denoting it by $(B_t)$. For completeness, let me state that $\mathbb W$ is defined as the space of continuous functions $u\colon [0, 1]\to \mathbb R^n$ such that $u(0)=0$, with topology of uniform convergence, and $\mathcal B$ is the associated Borel $\sigma$-algebra.

Unfortunately I am not familiar with stochastic processes. My understanding is that this measure $\gamma$ is defined as follows; $$ \int_{\mathbb W} F(w)\, \gamma(dw)=\int_\Omega F(B(\cdot, \omega))dP(\omega),$$ for all $F\colon \mathbb W\to [0, \infty]$ that is $\mathcal B$-measurable. Here $(\Omega, \mathcal F, dP)$ is the probability space that supports the Brownian motion $(B_t)$.

Is my understanding correct?

NOTE. I changed the question after Kavi Rama Murthy made me notice it was meaningless as stated. Thank you. Kavi Rama Murthy's answer refers to the formula $$\int_{\mathbb W} F(w)\, \gamma(dw)={\sf E}\left[\int_0^1F(B_t)\, dt\right].$$

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  • $\begingroup$ Equivalently, for any Borel set $A \subset \mathbb{W}$, we have $\gamma(A) = P(B \in A) = P(\{\omega : B(\cdot, \omega) \in A\})$. This is just what the word "law" always means. $\endgroup$ Commented Aug 4, 2021 at 21:43
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    $\begingroup$ Some nice examples of sets $A \subset \mathbb{W}$ to keep in mind: the set of paths which equal 0 at time 1/2; the set of paths that are inside the unit ball at time 1/2; the set of paths which remain inside the unit ball for all times $0 \le t \le 1$; the set of paths which are nowhere differentiable; etc. $\endgroup$ Commented Aug 4, 2021 at 21:46

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$F(B_t)$ does not make sense. For a fixed $\omega$, $B_t(\omega)$ is just a number and not an element of $\mathbb W$. So you have to evalute $F$ at the entire function $t \to B_t$ and then take the expectation. There is no integral involved here.

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  • $\begingroup$ Right, thanks a lot. I edited the question. I think it is OK now. $\endgroup$ Commented Aug 4, 2021 at 12:38
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    $\begingroup$ It is correct now. $\endgroup$ Commented Aug 4, 2021 at 12:57

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