# Why do we unavoidably (or not) use Riemann integral to define Itô integral?

https://en.wikipedia.org/wiki/Itô_calculus

Define $$\int_0^tH_tdB_t\equiv \lim_{n\rightarrow\infty}\sum_{i=1}^nH_{t_i}(B_{t_i}-B_{t_{i-1}})$$

But I'm wondering why not defining this using Lebesgue Integral?

It looks more consistent, meaning we can somehow 'obsolete' Riemann integral after knowing Lebesgue Integral. Moreover, we can integrate processes that are hard to integrate. Isn't this good?

Quote from wikipedia: Suppose that $B$ is a Wiener process (Brownian motion) and that $H$ is a right-continuous (cadlag), adapted and locally bounded process. If $\{π_n\}$ is a sequence of partitions of $[0, t]$ with mesh going to zero, then the Itô integral of H with respect to B up to time t is a random variable. It can be shown that this limit converges in probability...

Or maybe because the above condition that the limit converge is strong enough in most situations?

• How would you define this in terms of Lebesgue-integrals? The Brownian motion has almost surely unbounded variation, so you can't write it as the difference of two monotone functions. Dec 12 '16 at 13:57
• But this is clearly a Riemann sum, when the Riemann sum converges then the function is Riemann integrable and hence Lebesgue integrable?
– ZHU
Dec 12 '16 at 13:59
• This is not a Riemann-sum, but a Riemann-Stieltjes sum. Your statement does not hold in this case. Dec 12 '16 at 14:03
• This is not the problem. The problem is how you would define the measure with respect to which you want to integrate in the sense of Lebesgue. This will in general only make sense for functions of bounded variation. Dec 12 '16 at 14:17
• @monotonic Exactly this construction does not work because of the unbounded variation of the Brownian motion. Dec 12 '16 at 15:07

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, not exist in a pointwise sense. This means, in particular, that the existence of this limit does not imply 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):

Theorem Let $\alpha:[a,b] \to \mathbb{R}$ be a mapping. If the Riemann-Stieltjes integral $$I(f) := \int_a^b f(t) \, d\alpha(t)$$ exists for all continuous functions $f:[a,b] \to \mathbb{R}$, then $\alpha$ is of bounded variation.

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 from R. 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

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

• Could you elaborate a bit more on the application of the Banach-Steinhaus theorem? Dec 12 '16 at 15:39
• @Dominik See my edited answer.
– saz
Dec 12 '16 at 15:55
• Thanks, this is a very interesting application of the Banach Steinhaus theorem. Dec 12 '16 at 16:01
• If the brownian motions are of unbounded variation almost surely, that means the integral doesn't exist almost surely. But how can it still converge in probability?
– ZHU
Dec 13 '16 at 1:03
• @ZHU Why not? Convergence in probability is weaker than pointwise convergence.
– saz
Dec 13 '16 at 8:08