My question refers to the notion of good approximation of quasiconformal mappings: Let $G, G' \subseteq \mathbb{C}$ be domains. A sequence $(f_n)_n$ of quasiconformal mappings of $G$ onto $G'$ is called a good approximation of a quasiconformal mapping $f: G \rightarrow G'$ if the $f_n$ converge locally uniformly to $f$ and the corresponding complex dilatations $\mu_n := \mu_{f_n}$ converge to the complex dilatation $\mu_f$ of $f$ almost everywhere (a.e.) in $G$ (for certain reasons, I'm not taking into account whether the quasiconformal mappings $f_n$, $g_n$, $f$ and $g$ have uniformly bounded maximal dilatation $K \in [1, \infty)$). The complex dilatations $\mu_n$ and $\mu$ are supposed to be measurable functions in $G$ (hence, in particular, not necessarily continuous).
Now suppose the sequences of quasiconformal mappings $(f_n)_n$ and $(g_n)_n$ are good approximations of the quasiconformal mappings $f: G \rightarrow G$ and $g: G \rightarrow G$ , respectively, of the same domain $G$ - in other words, the mappings $f_n, g_n, f$ and $g$ are self-mappings of the domain $G$ (mainly for the sake of simplicity). Hence, one may consider the composition $(f_n \circ g_n)_n$ and $f \circ g$ as well as the complex dilatations of the composed mappings: $(\mu_{f_n \circ g_n})_n$ and $\mu_{f \circ g}$. Note that there exists a general formula for the complex dilatation of composed mappings (see for example Lehto/Virtanen, "Quasiconformal Mappings in the Plane", formula (5.6), p. 183): $$ \mu_{f \circ g}(z) = \frac{\mu_g(z) + \mu_f(g(z)) \cdot e^{-2i \operatorname{arg}(g_z(z))}}{1 + \mu_g(z) \cdot \mu_f(g(z))\cdot e^{-2i \operatorname{arg}(g_{\overline{z}}(z))}} \tag{1} $$ for a.e. $z \in G$. My question focuses on the convergence behaviour of the sequence $(\mu_{f_n \circ g_n})_n$:
Under the above-mentioned assumptions, does the sequence of complex dilatations $(\mu_{f_n \circ g_n})_n$ converge pointwise (a.e.) towards the composition $\mu_{f \circ g}$ in $G$?
So far, I tried to use the mentioned formula (1). More precisely, my intention was to show that the term $\mu_{f_n}(g_n(z))$ converges to $\mu_f(g(z))$ as follows: $$ | \mu_{f_n}(g_n(z)) - \mu_f(g(z)) | \leq | \mu_{f_n}(g_n(z)) - \mu_{f_n}(g(z)) | + | \mu_{f_n}(g(z)) - \mu_f(g(z)) | $$ Clearly, the second term on the right-hand side tends to zero as $n \to \infty$. But I'm having trouble with the first term $| \mu_{f_n}(g_n(z)) - \mu_{f_n}(g(z)) |$. Can we actually show that this term tends to zero as well for $n \to \infty$? I thought of maybe using the modulus of continuity of $\mu_{f_n}$ here somehow, but didn't succeed on this yet (note, however, that the complex dilatations are not supposed to be continuous)...More generally, is there maybe a known result, theorem etc. that solves the described problem?
Thanks in advance for any help!