Argument in Lemma 3.2.4 of Karatzas & Shreve - approximating bounded progressively measurable process by a continuous process The following is part of an argument for Lemma 3.2.4 from Karatzas and Shreve's Brownian Motion and Stochastic Calculus.
Let $X$ be a bounded, progressively measurable, $\mathscr{F}_t$-adapted process.
For each fixed $T>0$, we wish to approximate $X$ by by a bounded continuous process.
Consider the continuous, progressively measurable processes
$$F_t(\omega) = \int_0^{t \wedge T} X_s(\omega)ds; \; \tilde{X}_t^{(m)} = m [F_t(\omega)-F_{(t-(1/m))^+}(\omega)]; \; m \ge 1,$$ for $t \ge 0, \omega \in \Omega$. From a result earlier in the proof, there exists for each $m \ge 1,$ a sequence of simple processes $\{\tilde{X}^{(m,n)}\}_{n=1}^\infty$ such that $\lim_{n\to \infty} E\int_0^T |\tilde{X}_t^{(m,n)} - \tilde{X}_t^{(m)}|^2 dt = 0.$ Let us consider the $\mathscr{B}([0,T]) \otimes \mathscr{F}_T$-measurable product set
$$A = \{(t,\omega) \in [0,T] \times \Omega; \; \lim_{m \to \infty} \tilde{X}_t^m (\omega) \neq X_t(\omega)\}.$$ For each $\omega \in \Omega$, the cross section $A_\omega = \{t \in [0,T];(t, \omega) \in A\}$ is $\mathscr{B}([0,T])$-measurable and, according to the fundamental theorem of calculus, has measure zero. The bounded convergence theorem now gives $\lim_{m\to \infty} E\int_0^T |\tilde{X}_t^{(m)} - X_t|^2 dt = 0$.
Question: In the definition of $\tilde{X}_t^{(m)}$ above, why do we consider $F_{(t-(1/m))^+}$ instead of $F_{(t-(1/m))}$? Since $F$ is continuous, aren't they the same?
Next, by the fundamental theorem of calculus, doesn't $A_\omega$ equal the whole $[0,T]$?
Finally, to apply the bounded convergence theorem, we need $\tilde{X}^{(m)}$ to be a bounded sequence. How do we ensure this from the definition?
I have been struggling to understand these points for some time and I would greatly appreciate some help solving these questions.
 A: *

*I believe the use of the plus sign there is short hand for the maximum function with zero, i.e. $(t-\frac{1}{m})^+=\max(t-\frac{1}{m},0)$. Since $t$ is allowed to take the value of zero, $t-\frac{1}{m}<0$ for all $m\geq1$ and so we need to make sure the interval $[t-\frac{1}{m},t]\subset[0,T]$.


*The use of the "Fundamental Theorem of Calculus" here is quite a poor choice of words. In reality what they are using is something called Lebesgue's Differentiation Theorem and Lebesgue Points, which does have some connection to the Fundamental Theorem of Calculus, but you will likely understand the argument much better knowing this. Essentially the Lebesgue Differentiation Theorem implies that by the construction of $\tilde{X}^{(m)}_t$ it converges for each $\omega\in\Omega$ to $X_t$ for almost every $t\in[0,T]$.


*From the definition is tricky and you don't need it. Note, by the Lebesgue Differentiation Theorem and by construction,  $\tilde{X}^{(m)}_t$ converges to $X_t$ for each $\omega\in\Omega$ and for almost every $t\in[0,T]$ and so must be bounded $\lambda\otimes P$ - almost every $(t,\omega)\in[0,T]\times\Omega$ as it is a convergent sequence. $\lambda$ here being the Lebesgue measure.
Then, since $\tilde{X}^{(m)}_t$ converges for each $\omega\in\Omega$ to $X_t$ for almost every $t\in[0,T]$ (i.e. we also have $\lambda\otimes P$ - almost everywhere $(t,\omega)\in[0,T]\times\Omega$ convergence) and $|\tilde{X}^{(m)}_t-X_t|$ is bounded $\lambda\otimes P$ - almost everywhere $(t,\omega)\in[0,T]\times\Omega$, by the Bounded Convergence Theorem, $|\tilde{X}^{(m)}_t-X_t|$ converges to zero under the product measure $\lambda\otimes P$, i.e.
$$\lim_{m\rightarrow\infty}\int_{[0,T]\times\Omega}|\tilde{X}^{(m)}_t-X_t|\>(\lambda\otimes P)(dt,d\omega)=\lim_{m\rightarrow\infty}E\int_0^T|\tilde{X}^{(m)}_t-X_t|\>dt=0.$$
Hope that makes sense?
