Solution 1:

In order to prove that $\int_{\Omega}f_ngd\mu\to\int_{\Omega}fgd\mu$, it suffices to deal with the case where $g$ takes the form $\mathbf{1}_{A_i}$, where $A\in \mathfrak B$ and has finite measure. Indeed, let $N\in\mathbb N$ and $c_1,\dots,c_N\in\mathbb R$, $A_1,\dots,A_N\in\mathfrak B$. From $$ \left\lvert \int_{\Omega}f_ngd\mu-\int_{\Omega}fgd\mu\right\rvert \leqslant \left\lvert \int_{\Omega}f_n\left(\sum_{i=1}^Nc_i\mathbf{1}_{A_i}\right)d\mu-\int_{\Omega}f\left(\sum_{i=1}^Nc_i\mathbf{1}_{A_i}\right)d\mu\right\rvert+\left\lvert \int_{\Omega}f_n\left(g-\sum_{i=1}^Nc_i\mathbf{1}_{A_i}\right) d\mu\right\rvert +\left\lvert \int_{\Omega}f\left(g-\sum_{i=1}^Nc_i\mathbf{1}_{A_i}\right) d\mu\right\rvert \\ \leqslant \left\lvert \int_{\Omega}f_n\left(\sum_{i=1}^Nc_i\mathbf{1}_{A_i}\right)d\mu-\int_{\Omega}f\left(\sum_{i=1}^Nc_i\mathbf{1}_{A_i}\right)d\mu\right\rvert+2C\left\lVert g-\sum_{i=1}^Nc_i\mathbf{1}_{A_i}\right\rVert_q, $$ we would get that $$ \limsup_{n\to\infty}\left\lvert \int_{\Omega}f_ngd\mu-\int_{\Omega}fgd\mu\right\rvert \leqslant 2C\left\lVert g-\sum_{i=1}^Nc_i\mathbf{1}_{A_i}\right\rVert_q $$ and the fact the class consisting of functions of the form $\sum_{i=1}^Nc_i\mathbf{1}_{A_i}$ is dense in $\mathbb L^q$ would give the wanted convergence.

You are thus reduced to the case where the measure is finite.