Brownian bridge from Brownian motion

Fix $n\ge 1$, times $t_1,\dots ,t_n\in [0,1]$, and coefficients $a_1,\dots ,a_n$. Then ${\sum }_{j=1}^n{a}_j{\beta }_{t_j}={\sum }_{j=1}^n{a}_j{B}_{t_j}-\left({\sum }_{j=1}^n{a}_j{t}_j\right)B_1\mathrm{.}$ This is a finite linear combination of Brownian values, with time $1$ appended if necessary, so [F1] makes it normal; repeated occurrences of time $1$, repeated $t_j$, and zero coefficients are allowed. Its mean is zero by finite linearity because all Brownian values are centered. Hence $\beta$ is a centered Gaussian process.

For $s,t\in [0,1]$, covariance bilinearity gives $\mathrm{Cov}({\beta }_s,{\beta }_t)=\min (s,t)-t\min (s,1)-s\min (1,t)+st\mathrm{Var}(B_1)\mathrm{.}$ Since $s,t\le 1$ and $\mathrm{Var}(B_1)=1$, this is $\min (s,t)-ts-st+st=\min (s,t)-st$.

Let $A$ be the probability-one event on which $t\mapsto B_t(\omega )$ is continuous on $[0,\mathrm{\infty })$, and let $A_0=\{B_0=0\}$. Their intersection has probability one by [F5]. For $\omega \in A\cap A_0$, the map $t\mapsto t{B}_1(\omega )$ is continuous and [F4] makes $t\mapsto {\beta }_t(\omega )$ continuous on $[0,1]$. On this event ${\beta }_0=B_0=0$, while for every $\omega$ one has ${\beta }_1=B_1-B_1=0$.

Steps 1.1--1.3 establish Gaussianity, centering, the covariance, path continuity, and both endpoints. The cases $s=0$, $t=0$, $s=t$, and $s=t=1$ follow directly from the same covariance formula, including its zero endpoint variances. The empty finite-dimensional list, if admitted, has the unique empty-tuple law. AC is used only through [F1]; the deterministic linear transformation and continuity argument make no new choice.