[proofplan] We show that $H_t = X_{t-}$ is left-continuous, adapted, and previsible. Left-continuity follows from the definition of cadlag paths: the left-limit of left-limits equals the left-limit. Adaptedness follows from the fact that $X_{t-}$ is the limit of $X_s$ as $s \uparrow t$, and each $X_s$ is $\mathcal{F}_s$-measurable with $\mathcal{F}_s \subseteq \mathcal{F}_t$. For previsibility, we approximate $H$ by a sequence of simple previsible processes $H^n$ that converge pointwise to $H$. Each $H^n$ is constructed by sampling $H$ at dyadic points and holding the value constant on half-open intervals, making it a finite sum of indicator functions of sets in the previsible $\sigma$-algebra $\mathcal{P}$. The pointwise limit of previsible processes is previsible since $\mathcal{P}$ is a $\sigma$-algebra. [/proofplan] [step:Verify that $H_t = X_{t-}$ is left-continuous] For $t > 0$, we must show $\lim_{s \uparrow t} H_s = H_t$, i.e., $\lim_{s \uparrow t} X_{s-} = X_{t-}$. Fix $\varepsilon > 0$. Since $X$ is cadlag, the left-limit $X_{t-} = \lim_{u \uparrow t} X_u$ exists. Choose $\delta > 0$ such that $|X_u - X_{t-}| < \varepsilon$ for all $u \in (t - \delta, t)$. For any $s \in (t - \delta/2, t)$, the left-limit $X_{s-} = \lim_{v \uparrow s} X_v$ satisfies $|X_{s-} - X_{t-}| \leq \varepsilon$, since for $v$ sufficiently close to $s$ (with $v \in (t - \delta, s)$), we have $|X_v - X_{t-}| < \varepsilon$, and therefore $|X_{s-} - X_{t-}| = |\lim_{v \uparrow s} X_v - X_{t-}| = \lim_{v \uparrow s} |X_v - X_{t-}| \leq \varepsilon$. Since $\varepsilon > 0$ was arbitrary, $\lim_{s \uparrow t} X_{s-} = X_{t-}$. At $t = 0$, we have $H_0 = X_{0-} := X_0$ by convention, and no left-continuity condition is required. [/step] [step:Verify that $H$ is adapted to $(\mathcal{F}_t)_{t \geq 0}$] For $t > 0$, choose a sequence $s_n \uparrow t$ with $s_n < t$ for all $n$. Then $H_t = X_{t-} = \lim_{n \to \infty} X_{s_n}$. Each $X_{s_n}$ is $\mathcal{F}_{s_n}$-measurable since $X$ is adapted, and $\mathcal{F}_{s_n} \subseteq \mathcal{F}_t$ since the filtration is increasing and $s_n < t$. Therefore each $X_{s_n}$ is $\mathcal{F}_t$-measurable, and the pointwise limit $H_t = \lim_n X_{s_n}$ is $\mathcal{F}_t$-measurable. For $t = 0$, $H_0 = X_0$ is $\mathcal{F}_0$-measurable since $X$ is adapted. [/step] [step:Approximate $H$ by simple previsible processes] For each $n \geq 1$, define the process \begin{align*} H_t^n := X_0 \cdot \mathbb{1}_{\{0\}}(t) + \sum_{k=1}^{n \cdot 2^n} X_{(k-1)2^{-n}}\, \mathbb{1}_{((k-1)2^{-n},\, k \cdot 2^{-n}]}(t). \end{align*} We verify that each $H^n$ is previsible. The previsible $\sigma$-algebra $\mathcal{P}$ on $\Omega \times [0,\infty)$ is generated by the collection of sets \begin{align*} \Pi = \bigl\{ E \times (u, v] : E \in \mathcal{F}_u,\ 0 \leq u < v \bigr\} \cup \bigl\{ E \times \{0\} : E \in \mathcal{F}_0 \bigr\}. \end{align*} The term $X_0 \cdot \mathbb{1}_{\{0\}}(t)$ is $\mathcal{P}$-measurable since $X_0$ is $\mathcal{F}_0$-measurable and $\{0\}$ is a predictable set ($\Omega \times \{0\} \in \mathcal{P}$). For each $k \geq 1$, $X_{(k-1)2^{-n}}$ is $\mathcal{F}_{(k-1)2^{-n}}$-measurable (since $X$ is adapted), and the set $\Omega \times ((k-1)2^{-n}, k \cdot 2^{-n}]$ belongs to $\Pi$ with $u = (k-1)2^{-n}$ and $v = k \cdot 2^{-n}$. Therefore the product $X_{(k-1)2^{-n}}(\omega) \cdot \mathbb{1}_{((k-1)2^{-n}, k \cdot 2^{-n}]}(t)$ is $\mathcal{P}$-measurable. Since $H^n$ is a finite sum of $\mathcal{P}$-measurable functions, $H^n$ is previsible. [guided] Why does this construction produce previsible processes? The previsible $\sigma$-algebra $\mathcal{P}$ is generated by "predictable rectangles" of the form $E \times (u,v]$ where $E \in \mathcal{F}_u$, together with $E \times \{0\}$ where $E \in \mathcal{F}_0$. Intuitively, a previsible process at time $t$ can only depend on information strictly before time $t$ (except at $t = 0$). The process $H^n$ evaluates $X$ at the left endpoint $(k-1)2^{-n}$ of each dyadic interval and holds this value constant on the half-open interval $((k-1)2^{-n}, k \cdot 2^{-n}]$. At any time $t \in ((k-1)2^{-n}, k \cdot 2^{-n}]$, the value $H_t^n = X_{(k-1)2^{-n}}$ depends only on information available at time $(k-1)2^{-n} < t$ -- exactly the "looking backward" property that characterizes previsibility. Formally: $X_{(k-1)2^{-n}} \in \mathcal{F}_{(k-1)2^{-n}}$ and the indicator $\mathbb{1}_{((k-1)2^{-n}, k \cdot 2^{-n}]}(t)$ defines a time interval starting strictly after $(k-1)2^{-n}$, so their product $(\omega, t) \mapsto X_{(k-1)2^{-n}}(\omega) \cdot \mathbb{1}_{((k-1)2^{-n}, k \cdot 2^{-n}]}(t)$ is measurable with respect to $\mathcal{P}$, because the set $\{(\omega, t) : \omega \in E,\, t \in (u,v]\}$ with $E \in \mathcal{F}_u$ is a generator of $\mathcal{P}$. [/guided] [/step] [step:Show $H^n_t \to H_t$ pointwise and conclude previsibility] Fix $(\omega, t) \in \Omega \times (0, \infty)$. For each $n$ large enough that $t \leq n$, there exists a unique $k_n$ such that $t \in ((k_n - 1)2^{-n}, k_n \cdot 2^{-n}]$, and \begin{align*} H_t^n(\omega) = X_{(k_n - 1)2^{-n}}(\omega). \end{align*} Since $(k_n - 1)2^{-n} < t \leq k_n \cdot 2^{-n}$ and both $(k_n - 1)2^{-n}$ and $k_n \cdot 2^{-n}$ lie within distance $2^{-n}$ of $t$, we have $(k_n - 1)2^{-n} \uparrow t$ as $n \to \infty$. By the cadlag property of $X$, the left-limit $X_{t-}$ exists and \begin{align*} H_t^n(\omega) = X_{(k_n - 1)2^{-n}}(\omega) \to X_{t-}(\omega) = H_t(\omega) \quad \text{as } n \to \infty. \end{align*} At $t = 0$, $H_0^n = X_0 = H_0$ for all $n$. Since each $H^n$ is $\mathcal{P}$-measurable and $H^n \to H$ pointwise on $\Omega \times [0,\infty)$, the limit $H$ is $\mathcal{P}$-measurable. Therefore $H$ is previsible. [guided] The key point is that $(k_n - 1)2^{-n} \to t$ from the left, not from the right. Since $t \in ((k_n-1)2^{-n}, k_n \cdot 2^{-n}]$, the left endpoint $(k_n - 1)2^{-n}$ is strictly less than $t$, so the convergence $(k_n - 1)2^{-n} \uparrow t$ approaches from below. This is essential because we need $X_{(k_n-1)2^{-n}} \to X_{t-}$ (the left-limit), and this convergence holds precisely when the evaluation points approach $t$ from the left. If the evaluation points approached from the right, we would get $X_{t+} = X_t$ (by right-continuity) instead of $X_{t-}$, which would give the wrong process. **Why is the pointwise limit of previsible processes previsible?** The previsible $\sigma$-algebra $\mathcal{P}$ is a $\sigma$-algebra on the product space $\Omega \times [0,\infty)$. A process is previsible if and only if it is $\mathcal{P}$-measurable as a function on this product space. The pointwise limit of $\mathcal{P}$-measurable functions is $\mathcal{P}$-measurable (this is a general property of $\sigma$-algebras: the class of measurable functions is closed under pointwise limits). Therefore $H = \lim_n H^n$ is $\mathcal{P}$-measurable, i.e., previsible. [/guided] [/step] [step:Deduce that every continuous adapted process is previsible] If $X$ is continuous and adapted, then $X_{t-} = X_t$ for all $t \geq 0$ (since continuity implies left-continuity). The process $H_t = X_{t-} = X_t$ is therefore previsible by the argument above. [/step]