Let $A_n:=\{x\colon f_n(x)\le1\}$ and $B_n:=\{x\colon f_n(x)>1\}$, where $f_n:=\frac{d\mu_n}{d\mu}$. Then the total variation of $\mu_n-\mu$ is $$\|\mu_n-\mu\|=\int_{B_n}(f_n-1)d\mu+\int_{A_n}(1-f_n)d\mu=2\int_{A_n}(1-f_n)d\mu\to0$$ by dominated convergence if $f_n\to1$ in measure wrt $\mu$.
Vice versa,
$$\|\mu_n-\mu\|=\int_X|f_n-1|d\mu.$$
So, as noted by Nate Eldredge, $\|\mu_n-\mu\|\to0$ means that $f_n\to1$ in $L^1(\mu)$, which implies, by Markov's inequality, that $f_n\to1$ in measure wrt $\mu$.
Thus, $\mu_n\to\mu$ in total variation iff $\frac{d\mu_n}{d\mu}\to1$ in measure wrt $\mu$.