Let $X$ be a measure space, and suppose $\mu_i$ are probability measures on $X$ that are absolutely continuous with respect to another probability measure $\mu$. Is convergence in total variation norm of $\mu_i$ to $\mu$ equivalent to convergence in measure (wrt $\mu$) of the Radon nikodym derivatives $\frac{d\mu_i}{d\mu}$ to $1$?

Let $X$ be a measure space, and suppose $\mu_i$ are probability measures on $X$ that are absolutely continuous with respect to another probability measure $\mu$. Is strong convergence of $\mu_i$ to $\mu$ equivalent to convergence in measure (wrt $\mu$) of the Radon nikodym derivatives $\frac{d\mu_i}{d\mu}$ to $1$?

Let $X$ be a measure space, and suppose $\mu_i$ are probability measures on $X$ that are absolutely continuous with respect to another probability measure $\mu$. Is strong convergence of $\mu_i$ to $\mu$ equivalent to convergence in measure (wrt $\mu$) of the Radon nikodym derivatives $\frac{d\mu_i}{d\mu}$ to $1$?

Let $X$ be a measure space, and suppose $\mu_i$ are probability measures on $X$ that are absolutely continuous with respect to another probability measure $\mu$. Is convergence in total variation norm of $\mu_i$ to $\mu$ equivalent to convergence in measure (wrt $\mu$) of the Radon nikodym derivatives $\frac{d\mu_i}{d\mu}$ to $1$?