Proof of Strong Markov Property: why are finite dimensional distributions determined by bounded continuous functions

Question

I am looking at the proof of the strong Markov property of Brownian motion which starts by considering $T<\infty$ a.s. and fix $A\in \mathscr{F}_T$. And take $0\le t_1 < \cdots < t_p$ and let $F$ be a bounded continuous function from $\mathbb{R}^p$ into $\mathbb{R}_+$. Then the proof verifies that $$E[1_AF(B_{t_1}^{(T)},\dots, B_{t_p}^{(T)})]=P(A)E[F(B_{t_1},\dots, B_{t_p})].$$ Here we set $B_t^{(T)}=1_{\{T<\infty\}}(B_{T+t}-B_T).$

The proof seems to be arguing that the finite distributions are determined by the set of bounded continuous functions. I know this is true for bounded measurable functions but how does this extend to continuous functions?

Answer 1

Here are two possible ways to see this: For $ℝ^p$-valued random variables,

1. You can approximate indicator functions of cubes by bounded continuous functions, and use that two probability measures agreeing on all cubes are equal by the uniqueness theorem for measures.

   In detail: Let $μ $ and $ν$ probability measures on $ℝ^p$. For every $a,b ∈ ℝ^p$ with $a \leq b$ (componentwise), we consider the closed cube $[a,b] := [a_1,b_1] \times \cdots \times [a_p,b_p]$. The set of all such cubes is an intersection stable generator of $\mathcal B(ℝ^p)$ and there exists a sequence $Q_n$ of cubes with $Q_n ↗ ℝ^p$. Hence by the uniqueness theorem for measures, if we can show that $μ(Q) = ν(Q) $ for every cube $Q$, it follows that $μ = ν$.

   To see that $ ∫ f dμ = ∫f dν $ for all $f ∈ C_b(ℝ^p)$ implies $μ(Q) = ν(Q)$ for every cube $Q$, consider first the case $p =1 $. For every $ε > 0$, we can approximate $ 1_{[a,b]}$ by a continuous, piecewise linear function $f_ε$ supported on $[a-ε,b+ε]$ as shown in the following plot:

   Now as $ε → 0$, it holds that $f_ε → 1_{[a,b]} $ pointwise. From dominated convergence, $$ μ([a,b]) = \lim_{n → ∞} ∫ f_{1/n} dμ = \lim_{n → ∞} ∫f_{1/n} dν = ν([a,b]), $$ proving the case $p = 1$. In higher dimensions, the argument is exactly the same, noting that if for $i = 1,\ldots p$, $f_ε^i $ approximates $1_{[a_i,b_i]}$ as before, then $ \prod_{i=1}^p f_ε^i $ approximates $1_{[a,b]}$.

2. Less elementary: you can use that if $μ,ν$ are probability measures on $ℝ^p,\ $ then $ ∫ f dμ = ∫f dν $ for all $f ∈ C_b(ℝ^p)$ implies equality of their characteristic functions, which implies $μ = ν$.