I am a bit confused about the following result I read: Let $(\Omega,\mathfrak{A},\mathfrak{F},\mathbb{P})$ be a filtered probability space. Let $\left\{X_t\right\}_{t\in[0,T]}$ be a continuous and adapted process on this space taking values in $\mathbb{R}$. Then $X$ is locally bounded in the sense that there is a localizing sequence $(\tau_n)_{n\in\mathbb{N}}$ of stopping times (which satisfies $\tau_k\leq\tau_{k+1}$ for $k\in\mathbb{N}$ and for each $\omega\in\Omega$ there is some $n_\omega\in\mathbb{N}$ such that $\tau_n(\omega)=T$ for all $n\ge n_\omega$) such that the stopped process $X(\cdot\wedge\tau_k)$ is bounded (in fact: $|X(t\wedge\tau_k)|\leq k$).
Now, my confusion comes: Let's take the filtration $\mathfrak{F}$ to be constant $\mathfrak{A}$. Suppose there is some $Z\in L^{1}(\mathbb{P})\setminus L^2(\mathbb{P})$. If I then define the continuous, adapted process $X$ by $X_t\equiv Z$ I arrive at the local boundedness of this process: For the described sequence of stopping times I get $|Z|=|X(t\wedge\tau_k)|\le k$. But doesn't this imply that $Z$ is square-integrable which I excluded?
Can somebody tell me why and where I am wrong? Thank you very much in advance!