# Riemann-Stieltjes integral of simple functions

I quote Øksendal (2003).

Let us consider a probability space $$\left(\Omega,\mathbb{P},\mathcal{A},\right)$$ and a class of functions $$f:\left[0,\infty\right]\times\Omega\mapsto\mathbb{R}$$.
For $$0\le S, $$\left(B(t)\right)_{t\ge0}$$ a Brownian motion and $$f(t,\omega)$$ given, we want to define: $$\int_S^T f(t,\omega)dB(t)(\omega)$$ It is reasonable to start with a definition for a simple class of functions $$f$$ and then extend by some approximation procedure. First assume that $$f$$ has the form: $$\phi(t,\omega)=\sum_{j\ge0}e_j(\omega)\cdot1_{[j\cdot2^{-n}, (j+1)2^{-n})}(t)$$ where $$1$$ denotes the indicator function and $$n$$ is a natural number.
For such functions, it is reasonable to define: $$\int_S^T\phi(t,\omega)dB_t(\omega)=\sum_{j\ge0}e_j(\omega)\left[B_{t_{j+1}}-B_{t_j}\right](\omega)\tag{1}$$ where: $$t_k=t_k^{(n)}=\begin{cases}k\cdot 2^{-n}\hspace{0.3cm}\text{if } S\le k\cdot 2^{-n}\le T\tag{2}\\ S\hspace{0.3cm}\text{if } k\cdot 2^{-n}T \end{cases}$$

My doubts concern the part in italics. Namely:

Questions

1. Why, according to $$(1)$$: \begin{align}\int_S^T\phi(t,\omega)dB_t(\omega)&=\int_S^T\sum_{j\ge0}e_j(\omega)\cdot1_{[j\cdot2^{-n}, (j+1)2^{-n})}(t)dB_t(\omega)\\&=\sum_{j\ge0}e_j(\omega)\left[B_{t_{j+1}}-B_{t_j}\right](\omega)\end{align}?

Is that due to the fact that $$\sum_{j\ge0}e_j(\omega)\cdot1_{[j\cdot2^{-n}, (j+1)2^{-n})}(t)$$ do not depend on the variable of integration $$B_{t}(\omega)$$, hence they go outside sign of integration and one has: \begin{align}\int_S^T\phi(t,\omega)dB_t(\omega)&=\int_S^T\sum_{j\ge0}e_j(\omega)\cdot1_{[j\cdot2^{-n}, (j+1)2^{-n})}(t)dB_t(\omega)\\&=\sum_{j\ge0}e_j(\omega)\left[B_{t_{j+1}}-B_{t_j}\right](\omega)\end{align} with $$t_k$$ as specified in $$(2)$$ and $$\sum_{j\ge0}\left[B_{t_{j+1}}-B_{t_j}\right]=B_{T}-B_{S}$$?
2. Besides, could you please detail the reason why $$(2)$$ is defined that way? In particular, is there in place the choice of left-end point of every time interval? Why does the value $$t_k$$ depend on whether $$k\cdot2^{-n}$$ is positioned? What I would expect instead is something like: $$t_k=t_k^{(n)}=\begin{cases}t_k\hspace{0.4cm}\text{if } k\cdot 2^{-n}\le t_k \le (k+1)\cdot2^{-n}\tag{2.bis}\\ 0\hspace{0.5cm}\text{otherwise} \end{cases}$$
• I am writing an answer, but prior to that, I just want to say that if you read the rest of that same chapter, you could get some insight into question $2$. The intention of stochastic calculus may not always coincide with the intention of usual calculus, so that particular observation of yours is important. – Teresa Lisbon Nov 3 '20 at 5:10
• (1) It is the definition of a stochastic integral for elementary functions w.r.t. a BM. (2) Your "definition" of $t_k$ doesn't make sense (it's circular). – d.k.o. Nov 3 '20 at 11:23
• OK, could you please just explain why is $(2)$ originally defined that way? I cannot understand the undelying "logic" @d.k.o. – Strictly_increasing Nov 3 '20 at 11:24
• @Strictly_increasing While I was halfway to writing an answer to this question, the other answer came up, and to be honest I think it is fantastic, so I will not answer the question. Thanks for giving me the opportunity, though. I would still like to provide you with references : "Brownian motion and stochastic calculus" by Schilling and Partzsch is an excellent starting book : it covers both the Ito and Stratonovich integral in separate chapters. Then the introduction to Chapter $3$ of Karatzas-Shreve taks in detail about why this definition is reasonable. – Teresa Lisbon Nov 8 '20 at 8:34
• Oksendal is the book to read if you want a quick introduction to the concept of Stochastic calculus. But it is not comprehensive, and does not indulge in discussion beyond a certain point. Better books (i.e. harder to read, but more discussion-based and with better exercises) are the two books I referenced above. They will give you a solid foundation if you want to get into research in the subject as I have done. – Teresa Lisbon Nov 8 '20 at 8:36

(1) It is the definition of a stochastic integral for elementary functions w.r.t. a BM. (See the beginning of the next section.) Why is it reasonable? Consider a discrete-time analogue. Let $$\{X_n\}$$ be a martingale adapted to $$\{\mathcal{F}_n\}$$ and let $$\{H_n\}$$ be a bounded, previsible process, i.e., $$H_n\in\mathcal{F}_{n-1}$$. Then we define $$(H\cdot X)_n:=\sum_{i=1}^n H_i \Delta X_i,\quad (H\cdot X)_0=0$$
as our discrete time stochastic integral (in fact, it is called the martingale transform of $$X$$). The standard example is that if you bet 1\$each time (i.e., $$H_n=1$$), your total gain/loss at time $$n$$ is exactly $$(H\cdot X)_n$$. A nice property of this process is that it is a martingale (is it crucial that $$H$$ is predictable; take, for example, $$H_n=\operatorname{sgn}(\Delta X_n)$$). The corresponding processes in your case are $$H_n=e_{n-1}$$ and $$X_n=B_{t_n}$$ (setting $$S=0$$). (2) The "logic" behind the definition of $$t_k$$ is related to the definition of elementary functions. For each $$k$$, such a function is constant on $$[k 2^{-n},(k+1)2^{-n})$$ and $$(B_t)$$ is "sampled" at the corresponding end-points. • My question$(1)$specifically refers to the "passage"$\int_S^T\sum_{j\ge0}e_j(\omega)\cdot1_{[j\cdot2^{-n}, (j+1)2^{-n})}(t)dB_t(\omega)=\sum_{j\ge0}e_j(\omega)\left[B_{t_{j+1}}-B_{t_j}\right](\omega)$. Why does that hold true mathematically speaking?  As to my question$(2)$, what I mean is: why$t_k=S$if$S>k2^{-n}$and$t_k=T$if$T<k2^{-n}$. That sounds "counterintuitive" to me (if I try, for example, to visually imagine the situation) – Strictly_increasing Nov 3 '20 at 22:54 • (1) It holds by definition. (2) Why is it "counterintuitive"? To visually imagine the situation look at the graph of$t\mapsto 2^{-n}\lfloor t 2^n\rfloor$between$S$and$T$. – d.k.o. Nov 3 '20 at 23:43 • yeah, you said "Look at the graph of ....$\color{red}{\text{between }S \text{ and }T}$" and that sounds good to me. But the point is exactly that I cannot understand why one has to focus on the points$k2^{-n}$outside the interval$[S,T]$.  What's the point in having a focus on$S>k2^{-n}$(for which$t_k=S$) and on$T<k2^{-n}$(for which$t_k=T$)? – Strictly_increasing Nov 7 '20 at 11:51 • I see. The sequence$\{k2^{-n}:k\ge 0\}$creates a partition of$\mathbb{R}_{\ge 0}$. For small and large$k$, most of these points lie outside of$[S,T]$and need to be truncated. This partition could be equivalently defined as follows:$\{S, k_{min}2^{-n}, \ldots, k_{max}2^{-n},T\}$, where$k_{min}$($k_{max}$) is the minimal (maximal)$k$s.t.$k2^{-n}>S$($k2^{-n}<T$). – d.k.o. Nov 7 '20 at 13:04 • No. Suppose that$t_k\le S$for$k=0, \ldots, K$. Then$B_{t_{k+1}}-B_{t_{k}}=B_S-B_S=0$for all$k=0,\ldots, K-1$and so$\sum_{j=0}^{K-1}e_j(B_{t_{j+1}}-B_{t_{j}})=0\$. – d.k.o. Nov 7 '20 at 15:30