Brownian Bridge conditional probability

The problem is to show that the density $P[W_{t_1} \in dx_1,...,W_{t_n}\in dx_n | W_T = b]$ is the density of a Brownian bridge from $a$ to $b$. $W$ is Brownian motion.

The density of a Brownian bridge from $a$ to $b$ is defined to be $\prod_{i=1}^n p(t_i-t_{i-1},x_i,x_{i-1}) \frac{p(T-t_n,x_n,b)}{p(T,b,a)}$ where $0=t_0<t_1<...<t_n=T$,$a=x_0,b=x_n$.

My problem is that I have no idea how to work with the object $P[W_{t_1} \in dx_1,...,W_{t_n}\in dx_n | W_T = b]$.

I believe that its definition is the following: We have by the Doob Dynkin Lemma an $f$ such that $f(W_T) = P[W_{t_1}\leq x_1,...,W_{t_n}\leq x_n |\sigma(W_T)]$.

Now $f(W_T)$ should have a density (although this also is mysterious to me but I accept it) we can denote by $df(W_T) = P[W_{t_1} \in dx_1,...,W_{t_n}\in dx_n |\sigma(W_T)]$

Finally the object we started with is defined to be $df(b)$ where $df$ is defined in the line above.

I dont think I can apply any markov properties to this beast so I have no known way of manipulating it outside of nonsense intuitive arguments. Can someone please shed some light on this or at least give a good reference where one can learn how to use these objects.Thank you

If $Y$ has a density $g$ and $(X,Y)$ has a density $f$ then $$P(X\in\mathrm dx\mid Y=y)=\frac{f(x,y)}{g(y)}.$$ Apply this to $$X=(W_{t_k})_{1\leqslant k\leqslant n},\qquad Y=W_T.$$
\begin{align*} \int_{B \times A} f(x,y)dxdy &= P[Y \in B, X \in A] \\ &= \int_{Y^{-1}(B)} 1_{A}(X) dP \\ &= \int_{Y^{-1}(B)} E[1_{A}(X)|Y]dP \\ &= \int_{Y^{-1}(B)} \phi(Y) dP \\ &= \int_{B} \phi(y) \mu(dy) \\ &= \int_{B} E[1_{A}(X)|Y=y]\mu(dy)\\ &= \int_{B} E[1_{A}(X)|Y=y]g(y)dy \\ &=\int_{B \times A} E[X \in dx|Y=y]g(y)dxdy \end{align*}