Consider two sequences of continuous functions $(f_n)$ and $(g_n)$ for $n \geq 0$ defined by
$$ f_n (x) := \int_0 ^t \Phi\left(\frac{x\Phi ^{-1}(\alpha(s) + \beta_n(s))+\Phi^{-1} (\alpha(s))}{\sqrt{1-x^2}} \right)\lambda'(s) \exp(-\lambda(s)) ~ds $$ and $$ g_n (x) := \int_0 ^t \Phi\left(\frac{x\Phi ^{-1}(\alpha(s) )+\Phi^{-1} (\alpha(s) + \theta_n(s))}{\sqrt{1-x^2}} \right)\lambda'(s) \exp(-\lambda(s)) ~ds $$ on $[0,1]$, where $\lambda \geq 0$, $\mid \lambda \mid <C$, $\Phi$ and $\Phi^{-1}$ are respectively the normal CDF and the inverse normal CDF.
For simplicity sake lets assume $\beta _n(s) \equiv \beta_n$ and $\theta_n(s) \equiv \theta_n$.
Now consider two functions $f,g$ such that $f_n \to f$ and $g_n \to g$ as $ \beta_n ,\theta_n \to 0$ ($f_n$ and $g_n$ are continuous by the same argument).
Here the questions. Are $f, g$ continuous and do we always have $f=g$?
The answer is trivial. Following the dominated convergence theorem the limits are the same and the limit is given by $$ f(x) = g(x)= \int_0 ^t \Phi\left(\frac{x\Phi ^{-1}(\alpha(s))+\Phi^{-1} (\alpha(s))}{\sqrt{1-x^2}} \right)\lambda'(s) \exp(-\lambda(s)) ~ds $$ which is continuous since is given by a composition of continuous functions.
Now the true question. Does $f_n,g_n \overset{\mathcal u}{\to} f$ ? Or alternatively $(f-g)_n \overset{\mathcal u}{\to} 0$.