piecewise weak convergence in $C[0,1]$ Let $P$ and the sequence $P_n$ be probability measures on $C[0,1]$ with the uniform metric. Fix $0<u<1$ and let $\Pi_1$ and $\Pi_2$ be the projections from $C[0,1]$ onto $C[0,u]$ and $C[u,1]$, respectively. Now suppose that $P_n\circ\Pi_1^{-1}$ and $P_n\circ\Pi_2^{-1}$ both converge weakly to $P\circ\Pi_1^{-1}$ and $P\circ\Pi_2^{-1}$. What additional requirement needs to be imposed to ensure that $P_n$ converges weakly to $P$? This and the obvious generalization to finitely many pieces seems like it should be a standard result but I couldn't find it in Billingsley.
 A: It isn't true as stated.  The problem is that separate weak convergence  doesn't keep track of correlations between what's going on the two halves of the interval.
Indeed, if we just consider the case of a constant sequence where $P_n = Q$ for every $n$, the question becomes: 

If $P \circ \Pi_1^{-1} = Q \circ \Pi_1^{-1}$ and $P \circ \Pi_2^{-1} = Q \circ \Pi_2^{-1}$, does $P=Q$?

And in general the answer is no.  Here's a simple counterexample.
Take $u = \frac{1}{2}$ and consider the following four paths $\omega_i \in C([0,1])$:
$$\begin{align*}
\omega_1(t) &= t - \frac{1}{2} \\
\omega_2(t) &= \frac{1}{2}-t \\
\omega_3(t) &= \left|t - \frac{1}{2}\right| \\
\omega_4(t) &=  -\left|t - \frac{1}{2}\right|.
\end{align*}$$
(I recommend sketching these paths.)  Note that
$$\begin{align*}\omega_1|_{[0, \frac{1}{2}]} &= \omega_4|_{[0, \frac{1}{2}]} \\
\omega_2|_{[0, \frac{1}{2}]} &= \omega_3|_{[0, \frac{1}{2}]} \\
\omega_1|_{[ \frac{1}{2},1]} &= \omega_3|_{[ \frac{1}{2},1]} \\
\omega_2|_{[ \frac{1}{2},1]} &= \omega_4|_{[ \frac{1}{2},1]}.
\end{align*}$$
Now let $P$ be the measure that places probability $1/2$ each on $\omega_1, \omega_2$, and $Q$ be the measure that places probability $1/2$ each on $\omega_3, \omega_4$.  Then you can easily see that $P \circ \Pi_1^{-1} = Q \circ \Pi_1^{-1}$ and $P \circ \Pi_2^{-1} = Q \circ \Pi_2^{-1}$ but $P \ne Q$.
A: A first observation is that the sequence $(P_n)_{n\geqslant 1}$ is tight in $C[0,1]$. Indeed, for a fixed $\varepsilon$, there exists a compact set $A_1$ of $C[0,u]$ such that for each $n$, $P_n\circ \pi_1^{-1}(A_1)\gt 1-\varepsilon$. Similarly, we can find a compact set $A_2$ such that for each $n$, $P_n\circ \pi_2^{-1}(A_2)\gt 1-\varepsilon$. The set 
$$A:=\{x\in C[0,1], \pi_1(f)\in A_1\mbox{ and } \pi_2(f)\in A_2  \} $$
is a compact subset of $[0,1]$ since it is both bounded and equi-continuous. Since $P_n(A)\geqslant 1-2\varepsilon$,  the sequence $(P_n)_{n\geqslant 1}$ is tight in $C[0,1]$.
This sequence will converge to $P$ if the finite-dimensional distributions converge to that of $P$. But it is not clear that if $0\leqslant v\lt u\lt w\leqslant 1$, then $\mathbb P_n\{x\in C[0,1] , (x(v),x(w))\in B\}\to 
\mathbb P\{x\in C[0,1] , (x(v),x(w))\in B\}$ for each $B$ in the Borel $\sigma$-algebra of $\mathbb R^2$ such that $\mathbb P\{x\in C[0,1] , (x(v),x(w))\in\partial B\}=0$.
