# Do I correctly understand this definition of Wiener measure?

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 $$\int_{\mathbb W} F(w)\, \gamma(dw)={\sf E}\left[\int_0^1F(B_t)\, dt\right].$$

• 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. Commented Aug 4, 2021 at 21:43
• 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. Commented Aug 4, 2021 at 21:46

$$F(B_t)$$ 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.