Show $\lim\limits_{\delta \to 0} E(\vert F((X_{t_{i}}+t_{i}B_{1})_{1 \leq i \leq n})-F(X)\vert A_{\delta}]=0$

Question

Show $\lim\limits_{\delta \to 0} E(\vert F((X_{t_{i}}+t_{i}B_{1})_{1 \leq i \leq n})-F(X)\vert \vert A_{\delta}]=0$

Let B be a brownian motion and X is the brownian bridge on $[0,1]$, i.e. $X_{t}=B_{t}-tB_{1}, t\in [0,1]$. Secondly let $A_{\delta}:=\{\vert B_{1}\vert \leq \delta \}$

Now, let $F: \mathbb R^{n} \to \mathbb R$ be a bounded, uniformly continuous function

Show that $\lim\limits_{\delta \to 0} E(\vert F((X_{t_{i}}+t_{i}B_{1})_{1 \leq i \leq n})-F(X)\vert \vert A_{\delta}]=0$

My idea:

Let $\epsilon >0$. We know that for small enough $\delta$ that $\vert \vert (X_{t_{i}}+t_{i}B_{1})_{1\leq i \leq n}-X\vert\vert \leq \delta$

implies $\vert F((X_{t_{i}}+t_{i}B_{1})_{1 \leq i \leq n})-F(X)\vert\leq \epsilon$

and on $A_{\delta}$ we know that $\vert \vert (X_{t_{i}}+t_{i}B_{1})_{1\leq i \leq n}-X\vert\vert \leq n\delta $. So if we choose a $\overline{\delta}:=\frac{\delta}{n}$. Thus we know:

for small enough $\overline{\delta}$ that for $\vert \vert (X_{t_{i}}+t_{i}B_{1})_{1\leq i \leq n}-X\vert\vert \leq \overline{\delta}$ that

implies $\vert F((X_{t_{i}}+t_{i}B_{1})_{1 \leq i \leq n})-F(X)\vert\leq \epsilon$ on $A_{\delta}$ and hence $E(\vert F((X_{t_{i}}+t_{i}B_{1})_{1 \leq i \leq n})-F(X)\vert \vert A_{\delta})=\frac{E(\vert F((X_{t_{i}}+t_{i}B_{1})_{1 \leq i \leq n})-F(X)\vert 1_{A_{\delta}})}{P(A_{\delta})}\leq \frac{\epsilon}{P(A_{\delta})}$

I cannot lose the $P(A_{\delta})$ term

Note that $F(X)=F((X_{t_{i}})_{1\leq i \leq n})$

Answer 1

You are almost done, since $\vert F((X_{t_{i}}+t_{i}B_{1})_{1 \leq i \leq n})-F(X)\vert\leq \epsilon$ on $A_\delta$ translates as $$ \vert F((X_{t_{i}}+t_{i}B_{1})_{1 \leq i \leq n})-F(X)\vert \mathbf 1_{A_\delta}\leq \epsilon\mathbf 1_{A_\delta}. $$