Proof of continuity of stochastic processes defined by Ito integrals

Question

I'm currently trying to understand the proof of Theorem 4.6.2 in Kuo, Hui-Hsiung: Introduction to Stochastic Integration:

Suppose $f \in L^2_{ad} ([a,b] \times \Omega )$, then the stochastic process $$X_t = \int_a^t f(s) dB(s), \hspace{3mm} a \leq t \leq b$$ is continuous, namely, all of its sample paths are continuous functions on the interval $[a,b]$.

I hope the notation is clear. This theorem is first proven for stochastic step functions, i.e. if $ \{ \mathcal{F}_t | a \leq t \leq b \} $ is the natural filtration given by the Brownian motion $B(t)$ and $\{ t_0, t_1, \dots , t_n \}$ is a partition of the interval $[a,b]$, let $f(t,\omega) = \sum_{k = 1}^n \xi_{k-1}(\omega) \cdot 1_{(t_{i-1},t_i]}$ with $\xi_{k-1}$ being $\mathcal{F}_{t_i}$-measurable.

The particular step of the proof that strikes me as odd is the following:

We know that for the given $f$, there exists a sequence of stochastic step functions $f_n$ with $$\lim_{n \to \infty} \int_a^b E(|f(t) - f_n(t)|)^2 dt = 0. $$Define for every $n \in \mathbb{N}, a \leq t \leq b$: $$ X_t^{(n)} := \int_a^t f_n(s) dB(s), \hspace{3mm} X_t:= \int_a^t f(s) dB(s).$$ Then, $X_t - X_t^{(n)}$ is a martingale, as we already know that such an Ito integral is a martingale with respect to the filtration $\{\mathcal{F}_t \}$. We can then apply Doob's submartingale inequality: $$ P \lbrace \sup_{a \leq t \leq b} |X_t - X_t^{(n)}| \geq \frac{1}{n} \rbrace \leq n E|X_b - X_b^{(n)}|$$

What this however totally seems to ignore, is that in order to be able to use this inequality, we need our martingale to be right continuous i.e. almost every sample path needs to be right continuous, which is not mentioned in this proof. Is it somehow obvious that the given martingale is right continuous, by continuity of the step functions $f_n$? Or by some other means? Or is this just not allowed? I have no idea. I hope someone can help me, if there is some uncommon notation that I forgot to explain, please let me know.

Answer 1

Two ways to fix this:

• Any martingale with respect to a nice filtration (i.e. a filtration satisfying the usual conditions) has a càdlàg modification. This means that we can assume without loss of generality that the process has càdlàg sample paths. In the case of Brownian motion, the completion of the canonical filtration satisfies indeed the usual conditions.
• Instead of applying Doob's inequality to $X_t-X_t^{(n)}$ apply it to $X_t^{(m)}-X_t^{(n)}$. Then we get that $$\sup_{a \leq t \leq b} |X_t^{(n)}-X_t^{(m)}| \to 0$$ in $L^2$. By the completeness of $L^2$, there exists a process $Y$ such that $$\sup_{a \leq t \leq b} |X_t^{(n)}-Y_t| \to 0$$ in $L^2$. Choosing a convergent subsequence, we find in particular that $(Y_t)_{t \in [a,b]}$ has continuous sample paths (as a uniform limit of continuous functions). On the other hand, we know from the construction of the Itô integral that $(X_t^{(n)})$ converges to $(X_t)_t$ and therefore, by the uniqueness of limits, $(X_t)=(Y_t)_t$. (Sorry that this part is a bit hand-waving; I don't own the mentioned book, so I don't know how Kuo/Hui-Hsiung define the stochastic integral. For more details see e.g. René Schilling/Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes.)