[proofplan] We establish Hölder continuity on $\mathcal{D}$ by a two-scale argument. At the fine scale, we bound the increments $|X_{k2^{-n}} - X_{(k+1)2^{-n}}|$ between adjacent $n$-th level dyadic points using [Markov's Inequality](/theorems/514) and a union bound, then apply the [Borel-Cantelli Lemma](/theorems/507) to show that these increments are at most $2^{-n\alpha}$ for all $n$ sufficiently large. At the coarse scale, we decompose the interval between any two dyadic points into a telescoping sum of at most $2$ increments at each dyadic level $n \geq r+1$ (where $2^{-(r+1)} < |t-s| \leq 2^{-r}$), and sum the geometric [series](/page/Series) to obtain the global Hölder bound $|X_t - X_s| \leq K_\alpha |t-s|^\alpha$. [/proofplan] [step:Bound adjacent dyadic increments at scale $2^{-n}$ using Markov and Borel-Cantelli] Fix $n \geq 0$ and $0 \leq k < 2^n$. By the [Markov Inequality](/theorems/514) applied with exponent $p > 0$, \begin{align*} \mathbb{P}\!\left(|X_{k2^{-n}} - X_{(k+1)2^{-n}}| \geq 2^{-n\alpha}\right) &\leq \frac{\mathbb{E}[|X_{k2^{-n}} - X_{(k+1)2^{-n}}|^p]}{2^{-n\alpha p}} \\ &\leq c \cdot \frac{|k2^{-n} - (k+1)2^{-n}|^{1+\varepsilon}}{2^{-n\alpha p}} \\ &= c \cdot 2^{-n(1+\varepsilon)} \cdot 2^{n\alpha p} \\ &= c \cdot 2^{-n(1+\varepsilon - \alpha p)}. \end{align*} A union bound over $0 \leq k < 2^n$ gives \begin{align*} \mathbb{P}\!\left(\max_{0 \leq k < 2^n} |X_{k2^{-n}} - X_{(k+1)2^{-n}}| \geq 2^{-n\alpha}\right) \leq c \cdot 2^n \cdot 2^{-n(1+\varepsilon - \alpha p)} = c \cdot 2^{-n(\varepsilon - \alpha p)}. \end{align*} Since $\alpha < \varepsilon/p$, the exponent $\varepsilon - \alpha p > 0$, and the series $\sum_{n \geq 0} c \cdot 2^{-n(\varepsilon - \alpha p)}$ converges. By the [Borel-Cantelli Lemma](/theorems/507), a.s. for all sufficiently large $n$, \begin{align*} \max_{0 \leq k < 2^n} |X_{k2^{-n}} - X_{(k+1)2^{-n}}| < 2^{-n\alpha}. \end{align*} Therefore there exists a finite random variable $M : \Omega \to [1, \infty)$ such that \begin{align*} \max_{0 \leq k < 2^n} |X_{k2^{-n}} - X_{(k+1)2^{-n}}| \leq M \cdot 2^{-n\alpha} \quad \text{for all } n \geq 0. \end{align*} [guided] The moment hypothesis $\mathbb{E}[|X_t - X_s|^p] \leq c|t-s|^{1+\varepsilon}$ gives better-than-Lipschitz control in the $L^p$ sense: the exponent $1 + \varepsilon$ exceeds $1$. The strategy is to convert this average control into an a.s. bound via a union bound and Borel-Cantelli. At scale $2^{-n}$, there are $2^n$ adjacent pairs of $n$-th level dyadic points. The [Markov Inequality](/theorems/514) applied to the non-negative random variable $|X_{k2^{-n}} - X_{(k+1)2^{-n}}|^p$ at threshold $2^{-n\alpha p}$ gives \begin{align*} \mathbb{P}\!\left(|X_{k2^{-n}} - X_{(k+1)2^{-n}}| \geq 2^{-n\alpha}\right) \leq c \cdot 2^{-n(1+\varepsilon - \alpha p)}. \end{align*} After the union bound over $2^n$ pairs, the probability of any single bad pair at scale $n$ is at most $c \cdot 2^{-n(\varepsilon - \alpha p)}$. The condition $\alpha < \varepsilon/p$ ensures $\varepsilon - \alpha p > 0$, so these probabilities are summable in $n$. The [Borel-Cantelli Lemma](/theorems/507) then guarantees: a.s., only finitely many scales $n$ have a bad pair. For the finitely many bad scales, we absorb the violations into a finite random variable $M$. The threshold $2^{-n\alpha}$ was chosen precisely so that the geometric series in the next step converges — any larger threshold would waste regularity, any smaller would fail the Borel-Cantelli summability. [/guided] [/step] [step:Decompose the interval between general dyadic points into a telescoping sum] Let $s, t \in \mathcal{D}$ with $s < t$. Choose the integer $r \geq 0$ such that $2^{-(r+1)} < t - s \leq 2^{-r}$. [claim:Dyadic Telescoping] For any $s, t \in \mathcal{D}$ with $2^{-(r+1)} < t - s \leq 2^{-r}$, the interval $[s, t]$ can be covered by a chain of adjacent dyadic intervals such that at each scale $n \geq r + 1$, at most $2$ intervals of length $2^{-n}$ are used. Consequently, \begin{align*} |X_s - X_t| \leq \sum_{n=r+1}^{\infty} 2 \cdot \max_{0 \leq k < 2^n} |X_{k2^{-n}} - X_{(k+1)2^{-n}}|. \end{align*} [/claim] [proof] Both $s$ and $t$ are dyadic rationals, so there exists $N$ such that $s, t \in \{k2^{-N} : 0 \leq k \leq 2^N\}$. We decompose $X_t - X_s$ by telescoping through the dyadic levels $n = r+1, r+2, \ldots, N$: at each level $n$, we move from the current approximation at scale $2^{-n+1}$ to scale $2^{-n}$, introducing at most $2$ adjacent increments (one near $s$ and one near $t$). The total number of increments at scale $n$ is at most $2$, giving the bound. [/proof] [guided] The dyadic decomposition is a standard technique: to travel from $s$ to $t$ using dyadic steps, we refine the dyadic resolution from scale $2^{-(r+1)}$ (which is comparable to $|t-s|$) down to scale $2^{-N}$ (the common resolution of $s$ and $t$). At each refinement from scale $2^{-(n-1)}$ to scale $2^{-n}$, we add at most $2$ new dyadic intervals of length $2^{-n}$ — one adjusting the left endpoint and one adjusting the right endpoint. This is analogous to binary expansion: to go from $s = 0.a_1 a_2 \ldots$ to $t = 0.b_1 b_2 \ldots$, we adjust one binary digit at a time, and at each digit position we change at most $2$ intervals. [/guided] [/step] [step:Sum the geometric series to obtain the Hölder bound] Applying the bound from the first step to each term in the telescoping sum, \begin{align*} |X_s - X_t| &\leq \sum_{n=r+1}^{\infty} 2M \cdot 2^{-n\alpha} = 2M \sum_{n=r+1}^{\infty} 2^{-n\alpha} = 2M \cdot \frac{2^{-(r+1)\alpha}}{1 - 2^{-\alpha}}. \end{align*} Since $t - s > 2^{-(r+1)}$, we have $2^{-(r+1)} < t - s$, so $2^{-(r+1)\alpha} < |t - s|^\alpha$. Therefore \begin{align*} |X_s - X_t| \leq \frac{2M}{1 - 2^{-\alpha}} \cdot |t - s|^\alpha. \end{align*} Setting $K_\alpha := \frac{2M}{1 - 2^{-\alpha}}$, we obtain $|X_s - X_t| \leq K_\alpha |t - s|^\alpha$ for all $s, t \in \mathcal{D}$. The random variable $K_\alpha$ is a.s. finite since $M$ is a.s. finite. [guided] The geometric series $\sum_{n \geq r+1} 2^{-n\alpha}$ converges because $\alpha > 0$. The sum evaluates to $2^{-(r+1)\alpha} / (1 - 2^{-\alpha})$ by the formula for a geometric series with ratio $2^{-\alpha} \in (0,1)$. The inequality $2^{-(r+1)\alpha} < |t-s|^\alpha$ follows from $2^{-(r+1)} < t - s$ (the defining property of $r$) and monotonicity of $x \mapsto x^\alpha$. This is where the choice of $r$ is used: it translates the dyadic scale into the actual distance $|t-s|$. The constant $K_\alpha = 2M/(1 - 2^{-\alpha})$ depends on $\alpha$ through the denominator $1 - 2^{-\alpha}$, which degenerates as $\alpha \to 0$ (the Hölder bound becomes vacuous). It also depends on $\omega$ through $M$, but $M < \infty$ a.s. Note that the argument breaks down if $\alpha \geq \varepsilon/p$: the Borel-Cantelli step would fail because the series $\sum_n 2^{-n(\varepsilon - \alpha p)}$ would diverge. This is why the theorem asserts Hölder continuity only for exponents $\alpha \in (0, \varepsilon/p)$, not at the endpoint. [/guided] [/step] [step:Extend from $\mathcal{D}$ to $[0,1]$ by [uniform continuity](/page/Uniform%20Continuity)] Since $\mathcal{D}$ is dense in $[0,1]$ and $X$ restricted to $\mathcal{D}$ is a.s. uniformly continuous (with modulus $K_\alpha |t-s|^\alpha$), there exists a unique continuous extension \begin{align*} \bar{X} : [0,1] \to \mathbb{R} \end{align*} defined by $\bar{X}_t := \lim_{n \to \infty} X_{t_n}$ for any [sequence](/page/Sequence) $t_n \in \mathcal{D}$ with $t_n \to t$. The [limit](/page/Limit) is independent of the approximating sequence by the Hölder bound, and $\bar{X}$ inherits the same Hölder estimate: $|\bar{X}_s - \bar{X}_t| \leq K_\alpha |t - s|^\alpha$ for all $s, t \in [0,1]$. This $\bar{X}$ is the desired $\alpha$-Hölder continuous extension. [/step]