Levy’s characterisation theorem for Brownian motion states that for a local martingale $X$ with $X\_0 = 0$, $X$ is a Brownian motion if and only if it has quadratic variation $\langle X, X \rangle\_t = t$.


The usual proofs of this fact use the characteristic function of the normal distribution. I am seeking an alternate proof in order to improve my intuition about the theorem.


**Question:** Is there a proof of this that does not go through the characteristic function?



Alternate proof of Levy’s characterisation of Brownian motion
Fix $A>0$; we will be proving that $X\_t\stackrel{\mathcal D}= B\_t$ on $[0;A].$ Let $T^{(m)}$ be an a.s. increasing sequence of stopping times such that $T^{(m)}\to\infty$ almost surely, and for each $m$, $X\_{n\wedge T^{(m)}}$ is a martingale. Then, $Y\_n=X\_{\epsilon n\wedge T^{(m)}}$, $n=0,1,\dots,$ is a discrete-time martingale satisfying $$\mathbb{E}((Y\_{n+1}-Y\_n)^2|\mathcal{F}\_{\epsilon n})=\mathbb{E}(\epsilon\wedge(T^{(m)}-\epsilon n)\_+|\mathcal{F}\_{\epsilon n})=:\epsilon\_n.$$


Put $t\_n:=\sum\_{i=0}^{n-1}\epsilon\_n$. Note that $0\leq\epsilon\_n\leq \epsilon$ almost surely, hence $\epsilon n-t\_n$ is non-negative and increasing. We infer that for $N=[A\epsilon^{-1}]$, $$\mathbb{P}(\max\_{n\leq N}|\epsilon n-t\_n|\geq \delta)=\mathbb{P}(\epsilon N-t\_N\geq \delta)\leq \delta^{-1}\mathbb{E}(\epsilon N-t\_N)=\delta^{-1}\mathbb{E}((N\epsilon-T^{(m)})\_+)\leq\delta^{-1}\mathbb{E}((A+1)-T^{(m)})\_+).$$ By monotone convergence, the right-hand side tends to zero as $n\to\infty,$ note that it also does not depend on $\epsilon.$ Therefore, given $\delta>0$, we can choose $m$ such that for all $\epsilon$,
$$
\mathbb{P}(\max\_{n\leq N}|\epsilon n-t\_n|\geq \delta)\leq \delta.
$$


By Skorokhod embedding theorem *[A. V. Skorohod, Studies in the theory of random processes, Addison-Wesley, Reading, Mass.,1965]*, there exists a Brownian motion $\{B\_t\}\_{t\geq 0}$, which we can take independent of $\{X\_t\}\_{t\geq 0}$, and a sequence of stopping times $0=\tau\_0<\tau\_1<\tau\_2<\dots$ such that $\{B\_{\tau\_n}\}\_{n\geq 0}$ has the same distribution as $\{Y\_n\}\_{n\geq 0}.$ Since both $B\_t$ and $X\_t$ are continuous, it is enough to show that $\max\_{n\leq A\epsilon^{-1}}|\tau\_n - n\epsilon|\to 0$ in probability as $\epsilon\to 0$ and then $m\to\infty$. In view of the above discussion, it is enough to show that for any fixed $m$, $\max\_{n\leq A\epsilon^{-1}}|\tau\_n - t\_n|\to 0$ in probability as $\epsilon\to 0$.


Recall that the construction in Skorokhod embedding theorem runs iteratively: conditionally on $\mathcal{F'\_n}=\sigma(\mathcal{F}\_{\epsilon n},\tau\_n, \{B\_{t}\}\_{0\leq t\leq\tau\_n})$, we construct $\tau\_{n+1}-\tau\_{n}$ as a stopping time for the Brownian motion $B\_{\tau\_n+t}-B\_{\tau\_n}$ such that $B\_{\tau\_{n+1}}-B\_{\tau\_n}$ has the same distribution as the conditional distribution of $Y\_{n+1}-Y\_{n}$. This implies:


$$\mathbb{E}(\tau\_{n+1}-\tau\_n|\mathcal{F}'\_n)=\mathbb{E}(B\_{\tau\_{n+1}}^2-B\_{\tau\_n}^2|\mathcal{F}'\_n)=\mathbb{E}(Y\_{n+1}^2-Y\_{n}^2|\mathcal{F}'\_n)=\epsilon\_n.$$ That is, $\tau\_n-t\_n$ is in fact a martingale. In particular, by Doob's inequality, we have $$\mathbb{P}(\max\_{n\leq N}|\tau\_n-t\_n|>\delta)\leq\delta^{-2}\mathbb{E}(\tau\_N-t\_N)^2,$$ which we will apply to $N=[A\epsilon^{-1}]$. In fact, Skorokhod embedding theorem guarantees that
$\mathbb{E}(\tau\_{n+1}-\tau\_n)^2\leq C\mathbb{E}(Y\_{n+1}-Y\_{n})^4$, thus
$$
\mathbb{E}(\tau\_{N}-t\_N)^2\leq \sum\_{n=0}^{N-1}\mathbb{E}(\tau\_{n+1}-\tau\_n)^2\leq \sum\_{n=0}^{N-1}\mathbb{E}(Y\_{n+1}-Y\_{n})^4.
$$


It remains to show that the right-hand side tends to zero as $\epsilon\to 0$. This is well-known for bounded, continuous local martingale, see Karatzas-Shreve, Lemma 5.10. We can reduce to this case by localization: since $X\_t$ is a.s. bounded on any finite interval, we have that $\hat{T}^{(m)}:=\min\{t:|X\_t|\geq m\}\to \infty$ almost surely, we can replace $T^{(m)}$ by $\hat{T}^{(m)}\wedge T^{(m)}$.


