TITLE: Donsker Theorem Billingsley
QUESTION [5 upvotes]: Theorems $16.1$ and $16.3$ in Billingsley Convergence of measures.
$16.1$ reads : Random variables $u_1,\ldots$ on $(\Omega,\mathcal{B},\mathbb P)$
and are i.i.d. with $0$ mean and finite variance $\sigma^{2}$, define $X_n(t,\omega) = \frac{1}{\sigma\sqrt{n}}S_{[nt]}(\omega)$ (here $t\in [0,1]$). Then $X_n \to W$ (converges in law to Wiener process).
$16.3$ Theorem $16.1$ remains valid if $\mathbb P$ is replaced by an arbitrary probability measure $\mathbb P_0$ such that $\mathbb P_0 \ll\mathbb  P$.
I have a problem understanding why is Theorem $16.3$ needed at all?$16.1$ makes no assumption about the measure $\mathbb P$. If someone has a copy of the great book by Billingsley, can you please look into it? This has been troubling me for quite sometime and I have to admit I am on my own with no advanced probability course being offered in my school. don't delete it quickly because it dont meet your standards....let someone into probability theory give a short reply then delete it

REPLY [5 votes]: I think the statement of Theorem 16.3 Billingsley meant is 

Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space, $(\xi_i,i\geqslant 1)$ be i.i.d. zero mean random variables (for $\mathbb P$), and $X_n(t,\omega):=\frac 1{\sigma\sqrt n}\sum_{i=1}^{\lfloor nt\rfloor}\xi_i(\omega)$. Assume that $\mathbb P_0\ll\mathbb P$. Then 
  $$X_n\to W\mbox{ in distribution with respect to }\mathbb P_0,$$
  where $W$ is a standard Brownian motion. 

The result is non trivial because the random variables $(\xi_i,i\geqslant 1)$ are not independent with respect to $\mathbb P_0$ in general. So it's not only Theorem 16.1 used with $\mathbb P_0$ instead of $\mathbb P$.
We thus have that if $g$ is a non-negative integrable function with respect to $\mathbb P$, and $F\colon D[0,1]\to\mathbb R$ is continuous and bounded, then 
$$\int_\Omega F(X_n(\cdot,\omega))g(\omega)\mathrm d\mathbb P\to \int_\Omega F(W(\cdot,\omega))g(\omega)\mathrm d\mathbb P(\omega).$$ 
Theorem 16.1 gave it for $g=1$ while Theorem 16.3 extends to each $g\in L^1_+(\mathbb P)$.