For any $t_1,\dots ,t_n\in {\mathbb{R}}_{+}$, the distribution of $(X_{t_1},\dots ,X_{t_n})$ is given by a finite-dimensional distribution of $P_B$, and therefore is Gaussian.

The map

is a continuous linear map, and $(X_s,X_t)$ is Gaussian by Lemma 6.2, therefore $X_t-X_s$ is Gaussian by Lemma 3.7. By definition of $P_B$, we have

and

Therefore $X_t-X_s$ is Gaussian with mean $0$ and variance $|t-s|$.

$X_t-X_s$ is a Gaussian random variable with mean $0$ and variance $|t-s|$ (Lemma 6.3). Thus, by Lemma 3.3, we have

The pre-Brownian process is a Gaussian process by Lemma 6.2. The Brownian motion is a modification of the pre-Brownian process by Definition 6.5. Thus, the Brownian motion is a Gaussian process as well by Lemma 3.32.

Consider the cover of ${\mathbb{R}}_{+}$ given by $T_n:=[0,n+1)$. By Lemma 5.82, this cover has bounded covering numbers with exponent $1$. Moreover, by Lemma 6.4, the pre-Brownian process satisfies the Kolmogorov condition with exponents $(2n+4,n+2)$ for all $n\in \mathbb{N}$. In particular, for all $n\in \mathbb{N}$, $2n+4>1$. Applying Theorem 5.84 we deduce that the modification used in Definition 6.5 has locally Hölder continuous paths for all orders $\gamma \in (0,\underset{n}{sup}(n+1)/(2n+4))$. Clearly for all $n\in \mathbb{N}$ we have $(n+1)/(2n+4)⩽2$, and this quantity tends to $2$ as $n$ goes to infinity. Therefore the Brownian motion has Hölder continuous paths of order $\gamma$ for all $\gamma \in (0,1/2)$.

The paths are $1/4$-Hölder continuous by Lemma 6.7 because $0<1/4<1/2$, therefore they are continuous.

The Brownian motion $B$ is a modification of the pre-Brownian process, and therefore has same finite dimensional distributions. By Definition 4.10, $B_t$ has law $\mathcal{N}(0,t)$.

The Brownian motion $B$ is a modification of the pre-Brownian process, and therefore has same finite dimensional distributions. By Lemma 6.3, $B_t-B_s$ has law $\mathcal{N}(0,|t-s|)$.

Let $t_1\le t_2\le \dots \le t_n$ be $n$ times in ${\mathbb{R}}_{+}$. Then $(B_{t_2}-B_{t_1},B_{t_3}-B_{t_2},\dots ,B_{t_n}-B_{t_{n-1}})$ is a linear transformation of $(B_{t_1},\dots ,B_{t_n})$, hence it is Gaussian. We can compute its mean (which is $0$) and its covariance matrix. The covariance matrix is diagonal, which means the the random variables are uncorrelated. Because they are jointly Gaussian, this implies that they are independent.

We prove that this holds for $C(X,Y)$ as long as $X$ and $Y$ are second-countable topological spaces endowed with their Borel sigma-algebra, $X$ is locally compact (i.e. each point of $X$ has a basis of compact neighbourhoods), and $Y$ is regular (i.e. for any $C\subseteq Y$ closed and $y\notin C$, there exist disjoint open subsets $U,V\subseteq Y$ such that $C\subseteq U$ and $y\in V$). The proof is taken from this stackexchange question.

We denote by ${\mathcal{T}}_1$ and ${\mathcal{T}}_2$ the two sigma-algebras.

First of all, the evaluation maps are continuous for the compact-open topology, and therefore measurable for ${\mathcal{T}}_1$. We deduce that ${\mathcal{T}}_2\subseteq {\mathcal{T}}_1$.

Let us turn to the converse direction. For any $A\subseteq X$ and $B\subseteq Y$, denote

The compact-open topology is generated by the sets of the form $M(K,U)$, for $K$ compact and $U$ open. Because $X$ and $Y$ are second-countable and $X$ is locally compact, $C(X,Y)$ is second-countable. Therefore it is enough to show that for any $K$ compact and $U$ open, $M(K,U)\in {\mathcal{T}}_2$. We now fix $K$ and $U$.

Consider $(V_n{)}_{n\in \mathbb{N}}$ a countable family of subsets of $Y$ which generates the topology (it exists by second-countability assumption). Clearly we have that

Conversely, consider $y\in U$. Because $Y$ is regular, the set of $\overline{V_n}$ such that $y\in V_n$ is a basis of neighbourhoods of $y$. Indeed, consider $A$ an open neighbourhood of $y$. Then $A^c$ is a closed set which does not contain $y$. Therefore there exist disjoint open sets $B$ and $C$ such that $y\in B$ and $A^c\subseteq C$. There exists $n\in \mathbb{N}$ such that $x\in V_n\subseteq B$. Therefore $V_n\subseteq C^c$, so $\overline{V_n}\subseteq C^c\subseteq A$, and we are done. This proves that we have in fact

Consider now $f\in M(K,U)$. We then have

Because $K$ is compact and $f$ is continuous, $f(K)$ is compact. But each $V_n$ is open, so there exists a finite set $\hat{f}$ of natural integers such that for any $n\in \hat{f}$, $\overline{V_n}\subseteq U$, and

Conversely, if there exists a finite set $\hat{f}$ such that for any $n\in \hat{f}$, $\overline{V_n}\subseteq U$, and

then $f(K)\subseteq U$. We can therefore write that

This is a countable union, so we just have to prove that each term is measurable.

We now fix a finite set $I$ as in the union above. Because $X$ is second countable, there exists $Q$ a countable dense subset of $K$. For any $f$ continuous, because $Q$ is dense and $\bigcup _{i\in I}\overline{V_i}$ is closed, we have

In other words, $M(K,\bigcup _{i\in I}\overline{V_i})=M(Q,\bigcup _{i\in I}\overline{V_i})$. We can write

and it is enough to show that each term in the intersection is measurable. Because $\bigcup _{i\in I}\overline{V_i}$ is closed, it is measurable. It therefore remains to prove that $f\mapsto f(q)$ is measurable for all $q\in Q$, but this is obvious from the definition of ${\mathcal{T}}_2$. This concludes the proof. □