Completeness of the Adapted Continuous Process Space

Why Completeness Matters for SDE Arguments

Many existence arguments for stochastic differential equations construct a sequence of adapted continuous processes and show that the sequence is Cauchy. To obtain a process as its limit, one needs more than a pointwise limit at each time: the limit must remain adapted, have continuous paths (at least almost surely), and belong to the same space in which the Cauchy estimate was proved. Completeness is the result that guarantees all three properties together.

Fix $T<\infty$ and a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$ relative to the filtration. The result below concerns processes on $[0,T]$; it does not require a stochastic integral. All process equalities and path properties are understood up to events of probability zero when appropriate.

Definition: Let $\mathcal{C}_{\mathrm{ad}}^2([0,T])$ consist of adapted processes $X=(X_t)_{0\le t\le T}$ with almost surely continuous paths and $\mathbb{E}[\sup_{0\le t\le T}|X_t|^2]<\infty$. Identify two processes if they are indistinguishable, meaning their paths agree for every $t\in[0,T]$ on one event of probability one. The norm is $$ \|X\|_{2,\infty}=\left(\mathbb{E}\left[\sup_{0\le t\le T}|X_t|^2\right]\right)^{1/2}. $$ For an almost surely continuous process, the supremum can equivalently be taken over the countable dense set $\mathbb{Q}\cap[0,T]$, with the endpoint $T$ included if needed; this makes its measurability explicit.

Indistinguishability is the appropriate identification because changing a process on a null event should not change its norm or its role as a candidate solution. The norm is finite by definition, and if it is zero then the time supremum is zero almost surely, so the two processes are indistinguishable. Thus this is a norm on the stated equivalence classes.

Uniform Convergence in Probability

Convergence in probability at each fixed time only controls one time at a time. The supremum norm controls the largest error over the entire interval. The associated probabilistic notion is uniform convergence in probability: $X^n$ converges uniformly in probability to $X$ if, for each $\varepsilon>0$, the probability that the largest timewise error exceeds $\varepsilon$ tends to zero.

Proposition: If $X^n,X\in\mathcal{C}_{\mathrm{ad}}^2([0,T])$ and $\|X^n-X\|_{2,\infty}\to0$, then $X^n$ converges uniformly in probability to $X$. In particular, $$ \mathbb{P}\left(\sup_{0\le t\le T}|X^n_t-X_t|>\varepsilon\right) \le \frac{\|X^n-X\|_{2,\infty}^2}{\varepsilon^2}, \qquad \varepsilon>0. $$

Proof. The random variable $Y_n=\sup_{0\le t\le T}|X^n_t-X_t|$ is nonnegative and has finite second moment. Markov’s inequality applied to $Y_n^2$ gives $\mathbb{P}(Y_n>\varepsilon)=\mathbb{P}(Y_n^2>\varepsilon^2)\le \mathbb{E}[Y_n^2]/\varepsilon^2$. The numerator is exactly $\|X^n-X\|_{2,\infty}^2$, which tends to zero. This proves the claim for every fixed $\varepsilon>0$. $\square$

The same estimate applies to differences $X^n-X^m$. Consequently, a Cauchy sequence in this norm is Cauchy in uniform convergence in probability as well. The converse is not asserted: convergence in probability alone need not control second moments of the supremum.

Worked Example: A Deterministic Cauchy Sequence

On $[0,1]$, let $X^n_t=(1-2^{-n})t$, for integers $n\ge1$. These deterministic processes are adapted and continuous. For $m>n$, their difference is $X^m_t-X^n_t=(2^{-n}-2^{-m})t$. Since the coefficient is positive and $0\le t\le1$,

\|X^m-X^n\|_{2,\infty} =\sup_{0\le t\le1}(2^{-n}-2^{-m})t =2^{-n}-2^{-m} \le 2^{-n}.

Given $\varepsilon>0$, choosing $n$ large enough that $2^{-n}<\varepsilon$ makes this bound less than $\varepsilon$ for every $m>n$. Thus the sequence is Cauchy. Its candidate limit is $X_t=t$, and direct calculation gives $\|X^n-X\|_{2,\infty}=2^{-n}\to0$. Check: at $t=1$, the error is $2^{-n}$, which attains the supremum; at $t=0$, it is zero. The calculated norm therefore uses the actual largest error on the interval.

The Completeness Theorem

Theorem: The space $\mathcal{C}_{\mathrm{ad}}^2([0,T])$, with the norm $\|\cdot\|_{2,\infty}$ and identification by indistinguishability, is complete. That is, every Cauchy sequence in this norm converges in the norm to a process in the same space.

Proof. Let $(X^n)_{n\ge1}$ be Cauchy in $\|\cdot\|_{2,\infty}$, and choose adapted, almost surely continuous representatives. Because there are only countably many processes, there is one event of probability one on which every representative has continuous paths. For each $k\ge1$, choose indices $n_k$ increasing so that $\|X^{n_{k+1}}-X^{n_k}\|_{2,\infty}\le 2^{-2k}$. This is possible by the Cauchy property.

Set $D_k=\sup_{0\le t\le T}|X^{n_{k+1}}_t-X^{n_k}_t|$. Each $D_k$ is measurable: for continuous paths its supremum equals the supremum over a countable dense set, outside a null event. Moreover, $(\mathbb{E}[D_k^2])^{1/2}\le2^{-2k}$. For finite sums, Minkowski’s inequality gives $(\mathbb{E}[(\sum_{k=1}^N D_k)^2])^{1/2}\le\sum_{k=1}^N2^{-2k}\le1/3$. By Fatou’s lemma, $\mathbb{E}[(\sum_{k=1}^{\infty}D_k)^2]\le1/9$. In particular, $\sum_kD_k<\infty$ almost surely.

On the event where this sum is finite and all the paths are continuous, for any $r>s$, $\sup_{t\le T}|X^{n_r}_t-X^{n_s}_t|\le\sum_{k=s}^{r-1}D_k$. The tail of the convergent series tends to zero, so the subsequence is uniformly Cauchy in time on that event. It therefore has a uniform limit there, and a uniform limit of continuous functions is continuous.

Define $X_t$ to be the pointwise limit of $X^{n_k}_t$ when that limit exists, and define $X_t=0$ otherwise. For each fixed $t$, the event that the real sequence $X^{n_k}_t$ converges belongs to $\mathcal{F}_t$, since every $X^{n_k}_t$ is $\mathcal{F}_t$-measurable and convergence can be expressed using countably many Cauchy inequalities. The limit, set equal to zero off that event, is also $\mathcal{F}_t$-measurable. Hence $X$ is adapted. On the probability-one event of uniform convergence just established, this pointwise definition agrees at every time with the uniform limit, so $X$ has continuous paths almost surely.

It remains to verify convergence in the norm and finiteness of the limit’s norm. On the same probability-one event, for every $s$, $\sup_{t\le T}|X^{n_s}_t-X_t|\le\sum_{k=s}^{\infty}D_k$. Minkowski’s inequality for finite tails, followed by Fatou’s lemma, yields \[ \|X^{n_s}-X\|_{2,\infty} \le \sum_{k=s}^{\infty}\|D_k\|_{L^2} \le \sum_{k=s}^{\infty}2^{-2k}\longrightarrow0. \] In particular, $X$ has finite norm: $X=X^{n_1}+(X-X^{n_1})$, and both terms have finite $L^2$ supremum norm. Finally, for any $\varepsilon>0$, the original Cauchy property gives $N$ such that $\|X^n-X^m\|_{2,\infty}<\varepsilon/2$ whenever $n,m\ge N$. Choose $s$ with $n_s\ge N$ and $\|X^{n_s}-X\|_{2,\infty}<\varepsilon/2$. The triangle inequality then gives $\|X^n-X\|_{2,\infty}<\varepsilon$ for every $n\ge N$. Thus the full sequence converges to $X$ in the space. $\square$

The subsequence is used to obtain almost sure uniform convergence, which preserves continuity. Adaptedness is checked separately at each fixed time. This distinction matters: a pathwise limit can be continuous without being adapted unless its measurability with respect to each $\mathcal{F}_t$ is verified.

Worked Example: A Random Adapted Limit

Suppose the filtered space includes an $\mathcal{F}_0$-measurable random variable $\xi$ with $\mathbb{P}(\xi=1)=\mathbb{P}(\xi=-1)=1/2$. On $[0,1]$, set $X^n_t=(1-2^{-n})t\xi$. Each process is adapted because $\xi$ is already known at time zero, and each path is continuous. Since $|\xi|=1$ surely, for $m>n$,

\|X^m-X^n\|_{2,\infty} =\left(\mathbb{E}\left[(2^{-n}-2^{-m})^2|\xi|^2\right]\right)^{1/2} =2^{-n}-2^{-m}.

Thus the sequence is Cauchy, and its limit is $X_t=t\xi$. Substituting the limit into the norm gives $\|X^n-X\|_{2,\infty}=(\mathbb{E}[2^{-2n}|\xi|^2])^{1/2}=2^{-n}\to0$. The limit is adapted since $\xi$ is $\mathcal{F}_0$-measurable, and its paths are continuous. Check: for any $\varepsilon>0$, the time supremum of the error is exactly $2^{-n}$. Therefore the probability that it is strictly greater than $\varepsilon$ is $1$ if $2^{-n}>\varepsilon$, and $0$ if $2^{-n}\le\varepsilon$, in agreement with uniform convergence in probability.

Why Pointwise Convergence Is Not Enough

Worked Example: Narrow Continuous Peaks

For $n\ge1$, let $c_n=4^{-n}$ and define the deterministic continuous process $h_n(t)=\max(1-4|t-c_n|/c_n,0)$ on $[0,1]$. Its support is $[3c_n/4,5c_n/4]$ and its maximum is $1$ at $t=c_n$. The supports are disjoint: the next one lies in $[3c_n/16,5c_n/16]$, below the current support’s lower endpoint $3c_n/4$.

For each fixed $t>0$, the upper endpoints $5c_n/4$ tend to zero, so $h_n(t)=0$ eventually; also $h_n(0)=0$. Hence $h_n(t)\to0$ at every fixed time. But for $m\ne n$, at $t=c_n$ we have $h_n(c_n)=1$ and $h_m(c_n)=0$, so $\|h_n-h_m\|_{2,\infty}=1$. Check: the supremum cannot exceed $1$, since both functions take values in $[0,1]$; it equals $1$ at $c_n$. The sequence is not Cauchy, and $\mathbb{P}(\sup_t|h_n(t)-0|>1/2)=1$ for every $n$.

Common mistake. The incorrect calculation is “$h_n(t)\to0$ for every $t$, therefore $\sup_t|h_n(t)|\to0$.” It exchanges the pointwise limit and the supremum. In this example, $\sup_t\lim_n h_n(t)=0$, whereas $\lim_n\sup_t h_n(t)=1$. The supremum norm detects peaks whose locations move with $n$; pointwise convergence does not.

Completeness provides a reliable setting for convergence arguments: it turns a Cauchy estimate in the $L^2$ supremum norm into a process with the required path and measurability properties. The associated Markov bound also gives uniform convergence in probability. Neither conclusion follows merely from convergence at each fixed time.

Check Your Understanding

Use the definition of the norm and the proof of completeness to answer the following.

Why is the supremum of an almost surely continuous process measurable when computed over a countable dense set of times?
What does the bound $\mathbb{P}(\sup_{t\le T}|X^n_t-X_t|>\varepsilon)\le\|X^n-X\|_{2,\infty}^2/\varepsilon^2$ imply as $n$ tends to infinity?
In the completeness proof, why does summability of the $L^2$ norms of $D_k$ give an almost surely uniformly Cauchy subsequence?
How is adaptedness of the limit established without assuming that a global pathwise convergence event belongs to every $\mathcal{F}_t$?
For the narrow-peak example, compute $\sup_t|h_n(t)|$, and explain why pointwise convergence to zero does not make the sequence Cauchy in the stated norm.

Completeness of the Adapted Continuous Process Space

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