Since $\nu$ does not appear anywhere else in the question, I suppose that $L^1(X)=L^1(\nu)$.

In order that the functional be defined, one should then assume (probably without loss of generality) that $\mu$ is absolutely continuous w.r.t. $\nu$.

In this case, the two conditions in the question are more than sufficient, and far from being necessary.

To prove that they are sufficient, it suffices to show that for each sequence $x_n\to x$ in $L_1(\nu)$ there is a subsequence with $F_f(x_{n_k})\to F_f(x)$. Since $x_n\to x$ in $L_1(\nu)$, there is a subsequence with $x_{n_k}\to x$ $\nu$-a.e., hence by hypothesis $\mu$-a.e. Since $f$ is Carathéodory, it follows that $g_{n_k}(t)=f(t,x_{n_k}(t))\to g(t)=f(t,x(t))$ for $\mu$-a.e. $t$. It remains to apply Lebesgue's dominated convergence theorem with the dominating function being $\sup_y f(\cdot,y)$.

Using Vitali's instead of Lebesgue's dominated convergence theorem, one can replace the strong integral hypotheses by various sorts of growth conditions, depending on $\mu/\nu$. For instance, in case $\mu=\nu$ the growth condition $f(t,y)\le C|y|$ with a constant $C$ is sufficient or, if $\mu(X)$ is finite, even $f(t,y)\le C(1+|y|)$ is sufficient.