There is a small difference between the two processes. Note that process (1) is left continuous with right limits (caglad process), while process (2) is right continuous with left limits (cadlag process).

For integration with respect to continuous martingales (such as Brownian motion), which one you take as a simple process will not matter, as mentioned by Will in his answer. Indeed, Oksendal (2010) defines the stochastic integral by constructing the integral first for the processes of type (2), while Cont and Tankov (2004) or Karatzas and Shreve (1998) consider first processes of type (1).

In a more general theory of stochastic integration based on semimartingales (which can be discontinuous), one typically considers integration of caglad processes. Therefore, one starts to build integrals with simple processes of type (1). I found two justifications for it in the literature.

(1) As shown in Protter (2004), if we consider a stochastic integral of a cadlag process with respect to semimartingale then the resulting integral may not be a martingale even if the process with respect to which we define the integral is. On the other hand, an integral of a caglad process with respect to a martingale will be a martingale.

As an example (page 65), he takes $M_t=N_t-\lambda t$, a compensated Poisson process with intensity $\lambda$, with jump times denoted by $(T_i)_{i\geq 1}$ and an adapted process $H_t=1_{[0,T_1)}$. In that case, we have

$$ \int_0^t H_s dM_s = -\lambda(t\land T_1),$$

which is not a martingale. But if we consider $H_t=1_{(0,T_1]}$ then the above integral is equal to $$ \int_0^t H_s dM_s = N_{t\land T_1}-\lambda(t\land T_1),$$ which is a martingale (since a stopped martingale is a martingale).

(2) Cont and Tankov (2004) motivate restriction to caglad functions by showing that if we allow cadlag functions to be trading strategies then in the presence of jumps in the price of a stock there would be arbitrage opportunity and one would be able to earn an infinite rate of return (see Example 8.1, page 250). This justifies restriction to caglad processes when considering a suitable class of functions which we want to integrate.

PS. I am self-learning stochastic integrations so my apologies for any inaccuracies in my answer. And I am curious to hear answers' of other users!

## Best Answer

First of all, note that the limit

$$\int_0^t H(s) \, dB_s = \lim_{n \to \infty} \sum_{i=1}^n H_{t_i} (B_{t_i}-B_{t_{i-1}}) \tag{1}$$

is a limit in $L^2$ (or, alternatively, in probability); the limit does, in general,

notexist in a pointwise sense. This means, in particular, that the existence of this limit doesnotimply that the function $t \mapsto H_t(\omega)$ is Riemann-integrable for each $\omega \in \Omega$. This is really something you should be careful about when working with such limits: in which sense does the limit exist?Let's turn to the question why we cannot introduce the Itô integral in a pointwise sense. There is the following general statement which is a direct consequence of the Banach-Steinhaus theorem (see below for a detailed proof):

This means that if we define a stochastic integral with respect to a stochastic process $(X_t)_{t \geq 0}$ in a pointwise sense

$$\int_0^t f(s) \, dX_s(\omega), \qquad \omega \in \Omega$$

as a Riemann-Stieltjes integral, then this integral is only well-defined for all continuous functions if $t \mapsto X(t,\omega)$ is of bounded variation (on compacts) for all $\omega \in \Omega$. However, it is well-known that the sample paths of a Brownian motion are almost surely of unbounded variation, and therefore the definition of a stochastic integral in a pointwise sense is not a good idea: the class of functions which we can integrate would not even include the continuous functions.

Proof of the above theorem:The idea for this proof is taken fromR. Schilling & L. Partzsch: Brownian Motion - An Introduction to Stochastic Processes, Corollary A.41.Consider the normed spaces $$(X,\|\cdot\|_X) := (C[a,b],\|\cdot\|_{\infty}) \qquad \text{and} \qquad (Y,\|\cdot\|_Y) := (\mathbb{R},|\cdot|).$$ For a partition $\Pi = \{a=t_0<\ldots<t_n=b\}$ of the interval $[a,b]$ define $I^{\Pi}:X \to Y$ by

$$I^{\Pi}(f) := \sum_{t_j \in \Pi} f(t_{j-1}) (\alpha(t_j)-\alpha(t_{j-1})).$$

Since, by assumption, $I^{\Pi}(f) \to I(f)$ as $|\Pi| \to 0$ for all $f \in C[a,b]$, we have

$$\sup_{\Pi} |I^{\Pi}(f)| \leq c_f < \infty$$

for all $f \in C[a,b]$. Applying the Banach Steinhaus theorem we find

$$\sup_{f \in C[a,b], \|f\|_{\infty} \leq 1} \sup_{\Pi} I^{\Pi}(f) < \infty$$

which is equivalent to

$$\sup_{\Pi} \sup_{f \in C[a,b],\|f\|_{\infty} \leq 1} |I^{\Pi}(f)| < \infty. \tag{2}$$

For a partition $\Pi$ let $f_{\Pi}:[a,b] \to \mathbb{R}$ be a piecewise linear continuous function such that

$$f_{\Pi}(t_{j-1}) = \text{sgn} \, (\alpha(t_j)-\alpha(t_{j-1}))$$

for all $j=1,\ldots,n$. By the very choice of $f_{\Pi}$, we have

$$I^{\Pi}(f_{\Pi}) = \sum_{t_j \in \Pi} |\alpha(t_j)-\alpha(t_{j-1}))| \leq \sup_{f \in X, \|f\|_X \leq 1} |I^{\Pi}(f)|. \tag{3}$$

Combining $(2)$ and $(3)$, we conclude

$$\text{VAR}_1(\alpha,[a,b]) = \sup_{\Pi} I^{\Pi}(f_{\pi}) \leq \sup_{\Pi} \sup_{f \in C[a,b],\|f\|_{\infty} \leq 1} |I^{\Pi}(f)| < \infty.$$