# Does absolute continuity preserve convergence in measure?

Let $\sigma,\ \rho$ be two probability measures on the complex unit circle $\mathbb{T}$ such that $\sigma\ll\rho$. Let $(f_n)$ be some sequence of complex functions in $L_2(\rho)$ such that $f_n\rightarrow1$ in measure with respect to $\rho$. Would it be correct to say that $f_n\rightarrow1$ in measure with respect to $\sigma$ as well? If it is essential, I also know that $|f_n(x)|=1$ for every $x\in\mathbb{T}$ and $n\in\mathbb{N}$.

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1. Claim: for each $\varepsilon>0$, we can find $\delta>0$ such that if $\rho(A)<\delta$ then $\sigma(A)<\varepsilon$.
Indeed, if it was not true, we would be able to find $\varepsilon>0$ such that for all $n$, we can find $A_n$ such that $\rho(A_n)<n^{-2}$ and $\sigma(A_n)\geqslant \varepsilon$. Then $\rho(\limsup_jA_j)=0$, but for each $n$, $\sigma\left(\bigcup_{j\geqslant n}A_j\right)\geqslant \varepsilon$, so $\sigma(\limsup_jA_j)\geqslant \varepsilon$, which yields a contradiction.
2. Fix $r>0$ and $\varepsilon>0$. Consider the associated $\delta$ in the previous claim. We can find $n_0$ such that $\rho\{|f_n-1|>r\}\leqslant \delta$ whenever $n\geqslant n_0$. This gives that for these $n$, $\sigma\{|f_n-1|>r\}\leqslant\varepsilon$.
Notice that we don't need any assumption about the sequence $\{f_n\}$ (except measurability), and that we didn't use particular features of the torus (it actually works in any finite measure space).