Brownian bridge expression for a Brownian motion Let $B_t$ be a standard Brownian motion in $\mathbb R$, then the Brownian bridge on $[0,1]$ is defined as
$$
Y_t = a(1-t)+bt+(1-t)\int\limits_0^t\frac{\mathrm dB_s}{1-s}
$$ 
for $0\leq t<1$. Here $Y_0 = a$ and $\lim\limits_{t\to 1} Y_t = b$ a.s. The latter implies
$$
\lim\limits_{t\to 1}\;(1-t)\int\limits_0^t\frac{\mathrm dB_s}{1-s} = 0\text{ a.s.}
$$
and using integration by parts:
$$
\lim\limits_{t\to 1}\;(1-t)\int\limits_0^t\frac{B_s}{(1-s)^2}\mathrm ds = B_1 \text{ a.s.}
$$
I wonder if the latter formula has been shown to have a particular interesting meaning. Maybe there  a known relationship with a Cauchy's integral formula.
 A: 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*}
A: 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.)
