## Appendix A Proofs of the Two Main Theorems
### Proof of the General Theorem
#### a.1.1 Verifying the Conditions of Proposition 2
As a consequence \(g\circ\theta\) is uniformly continuous on \(\tilde{S}(K,\delta)\), which means that:  
\[\forall\epsilon>0,\,\exists\eta(\epsilon,\delta)>0,\,\forall x,y\in\tilde{S}(K,\delta),\,\|x-y\|\leq\eta(\epsilon,\delta)\implies|g\circ\hat{\theta}(x)-g \circ\hat{\theta}(y)|\leq\epsilon\,. \tag{103}\]  
Let us show that, with \(\alpha:=min(\delta,\eta(\epsilon,\delta))\), \(\forall n\),  
\[|g\circ\hat{\theta}(\tilde{S}_{n}(\theta_{n}))-g\circ\hat{\theta}(\tilde{S}( \theta_{n}))|\mathds{1}_{\theta_{n}\in K}>\epsilon\implies\big{\|}\tilde{S}_{ n}(\theta_{n})-\tilde{S}(\theta_{n})\big{\|}\mathds{1}_{\theta_{n}\in K}>\alpha\,. \tag{104}\]  
To that end, we show that:  
\[\big{\|}\tilde{S}_{n}(\theta_{n})-\tilde{S}(\theta_{n})\big{\|}\mathds{1}_{ \theta_{n}\in K}\leq\alpha\implies|g\circ\hat{\theta}(\tilde{S}_{n}(\theta_{n} ))-g\circ\hat{\theta}(\tilde{S}(\theta_{n}))|\mathds{1}_{\theta_{n}\in K}\leq \epsilon\,.\]Let us assume that \(\|\hat{S}_{n}(\theta_{n})-\hat{S}(\theta_{n})\|\mathds{1}_{\theta_{n}\in K}\leq\alpha\). If \(\theta_{n}\
otin K\), then \(|g\circ\hat{\theta}(\hat{S}_{n}(\theta_{n}))-g\circ\hat{\theta}(\hat{S}(\theta_{ n}))|\mathds{1}_{\theta_{n}\in K}=0\leq\epsilon\). If, in contrary, \(\theta_{n}\in K\), then \(\hat{S}(\theta_{n})\in\hat{S}(K)\subset\hat{S}(K,\delta)\). Since \(\|\hat{S}_{n}(\theta_{n})-\hat{S}(\theta_{n})\|=\big{\|}\hat{S}_{n}(\theta_{n} )-\hat{S}(\theta_{n})\big{\|}\mathds{1}_{\theta_{n}\in K}\leq\alpha\leq\delta\), then \(\hat{S}_{n}(\theta_{n})\in\hat{S}(K,\delta)\). Since \((\hat{S}(\theta_{n}),\hat{S}_{n}(\theta_{n}))\in\hat{S}(K,\delta)^{2}\) and \(\big{\|}\hat{S}_{n}(\theta_{n})-\hat{S}(\theta_{n})\big{\|}\leq\alpha\leq \eta(\epsilon,\delta)\), then we get from Equation (A3)  
\[|g\circ\hat{\theta}(\hat{S}_{n}(\theta_{n}))-g\circ\hat{\theta}(\hat{S}(\theta _{n}))|\mathds{1}_{\theta_{n}\in K}\leq\epsilon\,.\]  
In both cases, we get that:  
\[\big{\|}\hat{S}_{n}(\theta_{n})-\hat{S}(\theta_{n})\big{\|}\mathds{1}_{\theta _{n}\in K}\leq\alpha\implies|g\circ\hat{\theta}(\hat{S}_{n}(\theta_{n}))-g \circ\hat{\theta}(\hat{S}(\theta_{n}))|\mathds{1}_{\theta_{n}\in K}\leq \epsilon\,,\]  
which proves Equation (A4). Sufficient Condition for Convergence  
We use Equation (A4) to find a sufficient condition for (A2). This part differs from [7] as our approximation is not defined as a random sum. Equation (A4) is equivalent to  
\[\forall\text{ compact }K\subseteq\Theta,\quad\lim_{n\to\infty}|g\circ F_{n}( \theta_{n})-g\circ T(\theta_{n})|\mathds{1}_{\theta_{n}\in K}=0\,. \tag{102}\]0  
From that, we get  
\[\forall\text{ compact }K\subseteq\Theta,\quad\lim_{n\to\infty}|g\circ F_{n}( \theta_{n})-g\circ T(\theta_{n})|\mathds{1}_{\theta_{n}\in K}=0\,. \tag{102}\]1  
As a consequence, if  
\[\forall\text{ compact }K\subseteq\Theta,\quad\lim_{n\to\infty}|g\circ F_{n}( \theta_{n})-g\circ T(\theta_{n})|\mathds{1}_{\theta_{n}\in K}=0\,. \tag{102}\]2  
Then  
\[\forall\text{ compact }K\subseteq\Theta,\quad\lim_{n\to\infty}|g\circ F_{n}( \theta_{n})-g\circ T(\theta_{n})|\mathds{1}_{\theta_{n}\in K}=0\,. \tag{102}\]3  
In other, equivalent, words:  
\[ \begin{split}\text{If}&\big{\|}\hat{S}_{n}(\theta_{n})- \hat{S}(\theta_{n})\big{\|}\mathds{1}_{\theta_{n}\in K}\underset{n\to\infty}{ \longrightarrow}0\\ \text{Then}&|g\circ\hat{\theta}(\hat{S}_{n}(\theta_{n}))-g \circ\hat{\theta}(\hat{S}(\theta_{n}))|\mathds{1}_{\theta_{n}\in K} \underset{n\to\infty}{\longrightarrow}0\,.\end{split} \] (A5)  
Hence, having for all compact sets \(K\subset\Theta,\,\,\big{\|}\hat{S}_{n}(\theta_{n})-\hat{S}(\theta_{n})\big{\|} \mathds{1}_{\theta_{n}\in K}\underset{n\to\infty}{\longrightarrow}0\) is sufficient to have the desired condition (A2). Similarly, we find that \(\forall\theta\in K_{0}\):  
\[ \begin{split}&\big{\|}\hat{S}_{n}(\theta)-\hat{S}(\theta)\big{\|} \underset{n\to\infty}{\longrightarrow}0\\ \implies&|g\circ\hat{\theta}(\hat{S}_{n}(\theta))-g \circ\hat{\theta}(\hat{S}(\theta))|\underset{n\to\infty}{\longrightarrow}0\,, \end{split} \] (A6)  
which provides us a sufficient condition for (A1).