= Solution
Fix a sample path on the probability-one event where $X$ is continuous and the <total-variation process> of $A$ satisfies $V_\infty\leq K$. The signed measure $dA$ has absolute variation measure $dV$. Thus
$$
\sup_{t\geq0}\left|\int_0^t\bigl(f_n'(X_s)-\operatorname{sgn}_-(X_s)\bigr)dA_s\right|\leq\int_0^\infty|f_n'(X_s)-\operatorname{sgn}_-(X_s)|dV_s.
$$
The integrand tends to zero at every time, including times when $X_s=0$, and is bounded by $2$. Since $dV$ is a finite measure on this path, the <dominated convergence theorem> makes the right-hand side tend to zero. Therefore
$$
\boxed{\int_0^t f_n'(X_s)dA_s\longrightarrow\int_0^t\operatorname{sgn}_-(X_s)dA_s\quad\text{almost surely for every }t\geq0\text{ simultaneously}.}
$$
In fact the convergence is uniform over the entire time axis under the given total-variation bound. This pathwise estimate establishes one common full-probability event, rather than separate exceptional events for each time.
Back to article page
