Why is the Ito Isometry used to prove the limit of the Brownian Bridge?

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The 1-dimensional equation for a Brownian Bridge, $dY_t=\frac{b-Y_t}{1-t}dt + dB_t$; $0\le t < 1$, $Y_0=a$ has a solution $Y_t=a(1-t)+bt+(1-t)\int_0^t\frac{dBs}{1-s}$; $0 \le t < 1$.

Solutions on this site like Brownian bridge sde and https://math.stackexchange.com/a/410220/311948 use Ito's Isometry $\mathbb{E}\{X_t^2\}$ as the starting point for their proof that $\lim\limits_{t\to 1}Y_t=b$ a.s.

While I understand how Ito's Isometry can be used in general, and I understand how the remainder of the limit argument works afterwards, what is the motivation to start with $\mathbb{E}\{X_t^2\}$ ? It seems to come from nowhere without any explanation. Thank you

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First to be clear, in Brownian bridge they are computing the second moment in order to prove $L^{2}$-convergence.

The a.s. convergence cannot be done just with second moment because Why does $L^2$ convergence not imply almost sure convergence.

However, as mentioned in the comment of Brownian bridge sde, one simply observes that

$$(Y_{t})_{t\in [0,1)}\stackrel{law}{=}(B_{t}-tB_{1}))_{t\in [0,1)}$$

as Gaussian processes by comparing their covariances and use that Gaussian processes are uniquely determined by their covariance (Gaussian Process uniquely determined by Covariance and mean?). And from here the a.s. convergence is immediate.