I want to construct a counterexample to the converse of the following statement:
If $X$ is a square integrable martingale with $\sup_{n \in \mathbb N} \langle X \rangle_n < \infty$ almost surely, then $X$ converges almost surely.
Here, $\langle X \rangle$ is the square variation process of $X$ given by $$ \langle X \rangle_n = \sum_{k=1}^n \mathbb E \left[\left(X_k - X_{k-1}\right)^2\,\big|\, \mathcal F_{k-1}\right] $$ so that $X^2 - \langle X \rangle$ is a martingale (with $\mathbb F = \left( \mathcal F_n\right)_{n \in \mathbb N_0}$ is the filtration for $X$). So specifically, I want to construct a square integrable martingale $X$ that converges almost surely, but with $\mathbb P\left[\sup_{n \in \mathbb N} \langle X \rangle_n = \infty\right] \neq 0$. (By “square integrable” I mean that each $X_n$ is in $\mathcal L^2(\mathbb P)$, not that $\sup_{n \geq 0} \mathbb E\left[X_n^2\right] < \infty$.)
What I've tried: My general strategy for constructing martingales is to start with a symmetric random walk $X$ on $\mathbb Z$, so that $Y_k := X_k - X_{k-1}$ are independent and $\mathrm{Rad}_{1/2}$-distributed, and either (a) define a stopping time $\tau$ and consider the stopped process $X^\tau$, with $(X^\tau)_n = X_{\tau \wedge n}$; or (b) construct a locally bounded $\mathbb F$-predictable process $H$ and consider the discrete stochastic integral $H \cdot X$, with $(H \cdot X)_n = \sum_{k=1}^n H_k(X_k - X_{k-1})$.
In strategy (a), we simply need $X_\tau < \infty$ a.s. in order for $X^\tau$ to be an a.s. convergent process. In this case, we can easily compute $$ \langle X^\tau\rangle_n = \langle X \rangle_{\tau \wedge n} = \sum_{k=1}^{\tau \wedge n} \mathbb E\left[Y_k^2\,\big|\,\mathcal F_{k-1}\right] = \tau \wedge n $$ so in order for $\mathbb P\left[\sup_{n \in \mathbb N} \langle X \rangle_n = \infty\right] \neq 0$, we simply need $\mathbb P\left[\tau = \infty\right] > 0$. But if $X_\tau < \infty$ a.s., that means $X$ must converge a.s., which of course a symmetric random walk on $\mathbb Z$ doesn't. I've also tried letting $\sigma = \inf\{n \geq 0 : X_n \leq -1\}$, and letting $\tau = \inf\{n \geq 0 : X_{\sigma \wedge n} \geq 1\}$, which has $\mathbb P\left[\tau = \infty\right] > 0$, but $\mathbb E\left[\left(X_{\sigma \wedge k} - X_{\sigma \wedge (k-1)}\right)^2\,\big|\,\mathcal F_{k-1}\right]$ is harder to compute, so I'm not sure what $\langle X^\sigma\rangle_{\tau \wedge n}$ would be.
In strategy (b), we need that $H$ is almost surely summable in order for $H \cdot X$ to be a convergent martingale (this includes the case where $H$ is a.s. eventually $0$). The square variation process is $$ \langle H \cdot X\rangle_n = \sum_{k=1}^n \mathbb E\left[(H \cdot X)_k - (H \cdot X)_{k-1} \,\big|\,\mathcal F_{k-1}\right] = \sum_{k=1}^n \mathbb E \left[ \left(H_k Y_k\right)^2 \,\big|\,\mathcal F_{k-1}\right] = \sum_{k=1}^n H_k^2 $$ since $H$ is predictable and $Y^2_k \equiv 1$. But if $H$ is a.s. summable, then $\sum_{k=1}^n H_k^2$ a.s. approaches a finite number as $n \to \infty$, so $\sup_{n \in \mathbb N} \langle H \cdot X\rangle_n <\infty$ a.s.
Is there a better way to construct a martingale for this problem? Is there a classical example of an a.s. convergent square integrable martingale $X$ with $\langle X\rangle_n$ unbounded in $n$?