Why Pointwise Convergence Does Not Justify Moving a Limit Through an Itô Integral

Summary

The document examines a sequence of deterministic step functions that approximates the function t on the unit interval. It asks whether the corresponding Itô integrals against Brownian motion converge to the integral with integrand t, and whether the limit can be moved inside the integral. The questioner notes that the step functions are not monotone at every point, so the monotone convergence theorem does not directly apply as proposed.

This is a mathematical question, not a supplied proof or answer. It contrasts the stochastic integral with a Riemann–Stieltjes integral against an individual Brownian sample path, but does not establish that this pathwise integral is defined or that it agrees with the Itô integral. To justify convergence of Itô integrals, one generally needs an appropriate convergence result for the integrands, such as convergence in the relevant square-integrable norm; pointwise convergence alone is insufficient. The document supplies no such argument, estimates, or conclusion, so readers should treat the interchange as unresolved here.

Key ideas

• The step functions approximate the deterministic integrand t pointwise.
• The sequence is not monotone everywhere, so monotone convergence is not justified by the stated observation.
• Convergence of Itô integrals requires suitable control of the integrands, not only pointwise convergence.
• A pathwise Riemann–Stieltjes integral against Brownian motion is not automatically equivalent to an Itô integral.

Tags

# Proof: Deterministic Ito Integral (Thomas Mikosh Chapter 2)


# Proof: Deterministic Ito Integral (Thomas Mikosh Chapter 2)












I'm referencing Elementary Stochastic Calculus with Finance in View by Thomas Mikosch between chapters of Shreve's Volume II text. In one section Mikoshch text makes the following claim without proof:

Let ${f_n(t)}$ be a piecewise deterministic function of ${0\leq t\leq 1}$ such that: $${f_n(t) = \begin{cases} \frac{i-1}{n}, \frac{i-1}{n} \leq t \leq \frac{i}{n} \end{cases} \text{for $1 \leq i \leq n$ }}$$

The text claims ${\lim_{n\rightarrow \infty} \int_0^t f_n(s)\ dB_s(\omega) = \int_0^t lim_{n\rightarrow\infty}f_n(s)\ dB_s(\omega)=\int_0^ts\ dB_s(\omega)}$, where ${B_s(\omega)}$ is a Brownian sample path function, so the right most expression is a Riemann-Stieltjes integral which certainly converges since ${f(s)=s}$ has bounded first order variation on ${[0,1]}$.

But, how do we know the limit can be moved inside the integral? I am thinking it has to do with Monotone Convergence because ${f_n(t) \leq f_{n+1}(t)}$, but there are there are little intersections where ${f_n(t) > f_{n+1}(t)}$ for example ${f_2 > f_3}$ when ${t\in[1/3, 1/2]}$. Does the Monotone Convergence Theorem still work here?

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