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A planar Brownian motion range has zero two-dimensional Lebesgue measure almost surely. Assuming finite expected area, Brownian scaling gives half that expected area for each half-time range. Inclusion-exclusion then makes their expected overlap zero. Translate their ranges by the midpoint position; the independent Brownian arms at a deterministic time make the expected overlap the integral of the squared point-hitting probabilities for a half-time range. Those probabilities vanish almost everywhere, so the half-time expected area, and hence the full-time expected area, vanish. This argument does not assert zero hitting probability for every individual point solely from an almost-everywhere integral conclusion.

Set $W_t^{-}=B_{h-t}-B_h$ and $W_t^{+}=B_{h+t}-B_h$, $0\le t\le h$, 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,Although $W^{-}$ need not be independent of $B_h$, this last equality uses only translation invariance of Lebesgue measure. Put $p(y)=\mathbb{P}(T_y\le h)$ for a standard Brownian motion started at zero. Independence of the centered processes and Tonelli theorem giveSince $p\ge 0$, it follows that $p(y)=0$ for Lebesgue-almost every $y$. ThereforeThe 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 resultThe integral argument establishes the needed almost-everywhere spatial assertion; it does not require a prior theorem that individual points are polar.