This is a lemma in the classical book Brownian Motion and Stochastic calculus by Karatzas and Shreve. It is on page 133.
He makes an assertion that a set has Lebesgue measure 0, and that this follows from the fundamental theorem of calculus. But why is this? Isn't the fundamental theorem of calculus something that deals with Riemann integration? We are working with Lebesgue integrals.
The problem is showing that for fixed $\omega$, the lebegsue measure of t's where $\widetilde{X}^{(m)}(t,\omega)$ converges to $X(t,\omega)$ is 0. But since we have an m in the integrand which makes up $\widetilde{X}^{(m)}(t,\omega)$ I can not see that we can use the DCT directly, because it is not bounded. So it seems like he does something with the fundamental theorem of calculus instead?
It is a long proof, I put up everything here. But I highlighted what is relevant for the question in green, and the thing I do not know in red. Do you see how he can say what he says?
The fundamental theorem of calculus says that, for fixed $\omega$, the function $$F_t(\omega) = \int_0^{t\wedge T}X_s(\omega) ds$$ is absolutely continuous with respect to $t$, and $\frac{dF_t(\omega)}{dt} = X_t(\omega)$ almost everywhere over the interval $[0, \, T]$. That is, almost everywhere, \begin{align*} \lim_{m\rightarrow \infty}\widetilde{X}_t^m(\omega) &= \lim_{m\rightarrow \infty} m\left(F_t(\omega) - F_{(t-1/m)^+}(\omega) \right) = X_t(\omega). \end{align*} Therefore, $A_{\omega}$ has Lebesgue measure zero.