= Solution
Set $W^-_t=B_{h-t}-B_h$ and $W^+_t=B_{h+t}-B_h$, $0\leq t\leq h$, with $h=1/2$. Both start at zero and are continuous. Increments of $W^-$ over consecutive disjoint time intervals are negatives of Brownian increments over corresponding disjoint past intervals, read in reverse order. They are independent centered <normal random variables> with covariance equal to the interval length times the identity. The same properties hold for $W^+$ using future increments. Every past increment is independent of all the future increments after deterministic time $h$, so the two whole processes are independent, by finite-dimensional <independence> and generation by rational-time evaluations. They are the <independent Brownian arms at a deterministic time>.
Let $R^-$ and $R^+$ be their ranges. Pathwise,
$$
R^-=A_1-B_h,\qquad R^+=A_2-B_h,\qquad
\lambda(R^-\cap R^+)=\lambda(A_1\cap A_2).
$$
Although $W^-$ need not be independent of $B_h$, this last equality uses only translation invariance of <Lebesgue measure>. Put $p(y)=\mathbb P(T_y\leq h)$ for a standard <Brownian motion> started at zero. <Independence> of the centered processes and <Tonelli theorem> give
$$
0=\mathbb E\lambda(R^-\cap R^+)=\int_{\mathbb R^2}\mathbb P(y\in R^-)\mathbb P(y\in R^+)\,dy
=\boxed{\int_{\mathbb R^2}p(y)^2\,dy}.
$$
Since $p\geq0$, it follows that $p(y)=0$ for Lebesgue-almost every $y$. Therefore
$$
\mathbb E\lambda(A_1)=\int_{\mathbb R^2}p(y)\,dy=0.
$$
The scaling relation in part (a) gives $\mathbb E\lambda(R)=2\mathbb E\lambda(A_1)=0$. A nonnegative random variable of <expectation> zero vanishes almost surely, proving the <area of a planar Brownian path> result
$$
\boxed{\lambda(\{B_t:0\leq t\leq1\})=0\quad\text{almost surely}.}
$$
The integral argument establishes the needed almost-everywhere spatial assertion; it does not require a prior theorem that individual points are polar.
Back to article page
