A planar Brownian motion has area zero I'm looking for proofs of Paul Lévy's theorem that a planar Brownian motion has Lebesgue measure $0$. I know of only two proofs: one is in Lévy's original paper (Théorème 12, p. 532) and the other is in Mörters & Peres's "Brownian Motion" (Theorem 2.24, p. 46).
Unfortunately, Mörters & Peres's proof leaves out many technical details, without which I find the proof hopelessly difficult to understand.
I have not attempted to read Lévy's original proof, partly because it's in French (though I can read French, with effort) and partly because it's in the middle of a long article that was not written for students, but rather for professional mathematicians, and hence I suspect that it would be a pain for me to read it.
I have not been able to find any other sources containing a proof of this theorem. If someone knows of any such sources, in English, French or German, ideally textbooks (but any other source will be helpful too), please let me know.
Thank you.
 A: Unfortunately, I'm not aware of an alternative proof (or a more detailed version of the proof by Mörters/Peres). So, instead of providing you with a source, I'll follow the lines of Paul Lévy; hopefully filling the gaps in his proof.

Let $(B(t))_{t \geq 0}=((B^1(t),B^2(t)))_{t \geq 0}$ be a two-dimensional Brownian motion and denote by $\lambda$ the (two-dimensional) Lebesgue measure.
First, we show that $L := \lambda(B([0,1]))$ satisfies $\mathbb{E}L<\infty$. To this end, note that
$$\begin{align*} \{L>4r^2\} &\subseteq \left\{ \max_{t \in [0,1]}(|B^1(t)|,|B^2(t)|) > r \right\} \\ &= \left\{ \max_{t \in [0,1]} B^1(t) > r \right\} \cup \left\{ \max_{t \in [0,1]} B^2(t) > r \right\} \\ &\quad \cup \left\{ \min_{t \in [0,1]} B^1(t) < -r \right\}\cup \left\{ \min_{t \in [0,1]} B^2(t) < -r \right\}.\end{align*}$$
From the reflection principle we know that $\max_{t \in [0,1]} B^j(t) \sim -\min_{t \in [0,1]} B^j(t) \sim |B^j(1)|$ for $j=1,2$. Hence,
$$\mathbb{P}(L>4r^2) \leq 4 \sqrt{\frac{2}{\pi}} \int_r^{\infty} \exp \left( - \frac{y^2}{2} \right) \, dy \leq \frac{4 \sqrt{2}}{\pi} \frac{1}{r} \exp \left(- \frac{r^2}{2} \right).$$
Combining this estimate with the fact that $\mathbb{E}L = \int_0^{\infty} \mathbb{P}(L >r) \, dr$, we get $\mathbb{E}L<\infty$.
Next, we note that for the restarted Brownian motion $W_t := B_{t+1}-B_1$, we have $\lambda(W([0,1])) = \lambda(B([1,2]))$ and, since $W \sim B$, we conclude that $$\mathbb{E}(\lambda(B([0,1]))) = \mathbb{E}(\lambda(B([1,2]))).$$ Similarly, as $\tilde{W}_t := \frac{1}{\sqrt{2}} B_{2t}$ is a Brownian motion, we have
$$\lambda(\tilde{W}([0,1])) = \lambda \left( \frac{1}{\sqrt{2}} B([0,2]) \right) = \frac{1}{2} \lambda(B([0,2])$$
i.e. $\lambda(B([0,2])) \sim 2 \lambda(B([0,1]))=2L$. Therefore,
$$\begin{align*} 2\mathbb{E}L = \mathbb{E}(\lambda(B([0,2]) &= \mathbb{E}(\lambda(B([0,1])) + \mathbb{E}(\lambda(B([1,2])) - \mathbb{E}(\lambda(B([0,1]) \cap B([1,2]))) \\ &= 2\mathbb{E}L + \mathbb{E}(\lambda(B([0,1]) \cap B([1,2]))). \end{align*}$$
Hence, $$ \mathbb{E}\big(\lambda(B([0,1]) \cap B([1,2]))\big)=0. \tag{1}$$
By Fubini's theorem, we have
$$\mathbb{E}L = \int \int 1_{B([0,1])}(x,y) d\lambda(x,y) \, d\mathbb{P} = \int \underbrace{\mathbb{P}((x,y) \in B([0,1]))}_{=:p(x,y)} \, d\lambda(x,y). \tag{2}$$
If we set $W_t := B_{1-t}-B_1$ and $\tilde{W}_t := B_{t+1}-B_1$, then both processes are Brownian motions and
$$W([0,1]) = B([0,1])-B_1 \qquad \qquad \tilde{W}([0,1]) = B([1,2])-B_1.$$
Using that $(W_t)_{t \geq 0}$ and $(\tilde{W}_t)_{t \geq 0}$ are independent, we get
$$\begin{align*}\mathbb{E}(\lambda(B([0,1]) \cap B([1,2]))) &= \mathbb{E}(\lambda(W([0,1]) \cap \tilde{W}([0,1]))) \\ &= \int \mathbb{P}((x,y) \in W([0,1]) \cap \tilde{W}([0,1])) \, d\lambda(x,y) \\ &= \int \mathbb{P}((x,y) \in W([0,1])) \mathbb{P}((x,y) \in \tilde{W}([0,1])) \, d\lambda(x,y) \\ &= \int p^2(x,y) \, d\lambda(x,y). \end{align*}$$
Now $(1)$ implies $p(x,y)=0$ $\lambda$-almost everywhere and therefore, by $(2)$, $\mathbb{E}L=0$. Hence, as $L \geq 0$, we finally conclude $L=0$ almost surely.
Remark It is not obvious that $\omega \mapsto L(\omega)$ and $((x,y),\omega) \mapsto 1_{B([0,1],\omega)}((x,y))$ are random variables; see this question and Evan Aad's answer for the measurability of $L$ and this question for the measurability of the second mapping. (Note that measurability of $((x,y),\omega) \mapsto 1_{B([0,1],\omega)}((x,y)$ entails the measurability of $L$, by Fubini's theorem.)
