By the <Tonelli theorem> and the planar <Brownian transition density>,
$$
\mathbb E[A_t]
=\int_0^t\mathbb E[f(X_s)]\,ds.
$$
For $s\geq1$,
$$
\mathbb E[f(X_s)]
=\int_{\mathbb R^2}f(y)\frac1{2\pi s}
e^{-|y-X_0|^2/(2s)}\,dy
\leq\frac1{2\pi s},
$$
while for $0\leq s\leq1$ it is at most $\|f\|_\infty$. Hence
$$
\mathbb E[A_t]\leq\|f\|_\infty+\frac{\log t}{2\pi}
$$
for $t\geq1$, and consequently
$$
\boxed{\mathbb E[A_t/t]\longrightarrow0.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 201 / 5 / b / Solution

{c}
{title2=$p_t(x,y)$}

The transition density of standard Brownian motion in $\mathbb R^d$ is the <heat kernel>
$$
p_t(x,y)=\frac{1}{(2\pi t)^{d/2}}
\exp\left(-\frac{|y-x|^2}{2t}\right),
\qquad t>0.
$$
Thus $\mathbb P_x(X_t\in A)=\int_Ap_t(x,y)\,dy$.

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