Brownian bridge expression for a Brownian motion

If $f$ is any continuous function, then L'Hopital's rule gives \[ \lim_{t\to1}\; (1-t)\int_0^t\frac{f(s)}{(1-s)^2}\,ds = f(1). \] In terms of generalized functions, this says that if \[ \mu_t(s) = \frac{1-t}{(1-s)^2}1_{[0,t]}(s), \] then $\mu_t\to\delta_1$. The Brownian bridge can also be written in terms of generalized functions: \[ Y_t= a(1 - t) + bt + \langle\partial B,\nu_t\rangle, \] where \[ \nu_t(s) = \frac{1-t}{1-s}1_{[0,t]}(s), \] and $\partial B$ is the (random) distributional derivative of the Brownian sample path. Note that $\langle\partial B,\nu_t\rangle = -\langle\partial\nu_t,B\rangle$, and \[ \partial\nu_t = (1-t)\delta_0 + \mu_t - \delta_t. \] Hence, \begin{align*} \lim_{t\to1}\;\langle\partial B,\nu_t\rangle &= \lim_{t\to1}\;-\langle(1-t)\delta_0 + \mu_t - \delta_t,B\rangle\\ &= \lim_{t\to1}\;\langle\delta_t - \mu_t,B\rangle\\ &= \langle\delta_1-\delta_1,B\rangle = 0. \end{align*}

Well, without generalized functions, the proof is based on Doob's inequality for martingales: If $\{X_t\}_{t \in [0,T]}$ is continuous quadratic integrable martingale, then for all $C>0$ we have: $$P\left(\sup\limits_{t \in [0,T]}|X_t|\geq C\right)\leq \dfrac{EX^2(T)}{C^2}$$ So if we denote $X_t=\int_{0}^t\dfrac{dB_s}{1-s}$, then $$P\left(\sup\limits_{1-2^{-n}\leq t\leq 1-2^{-n-1}}(1-t)|X_t|>C\right)\leq 2 C^{-2}2^{-n},$$ take $C=2^{-n/4}$ and apply the Borel-Cantelli lemma. The rest is almost evident ($X_t \to 0$ a.s.)