if $\nu \ll \mu$ and $f_n \to f$ in measure $\mu$ then it converges to $f$ in measure $\nu$ as well.

Question

Suppose $\nu \ll \mu$ on $(X, \mathcal{A})$ and $\mu(X) = \nu(X) = 1$. Show that if a sequence of measurable functions $f_n$ converges in measure $\mu$ to a function $f$, then it converges to $f$ in measure $\nu$ as well.

$\textbf{My attempt}$: since $\nu \ll \mu$ by Radon-Nikodym we have that $\exists g\in L^1$ such that $\forall A\in \mathcal{A}$; $\nu(A)=\int_A g d\mu <\infty$.

let $\epsilon>0$, define $A_n=\{x\in X : |f_n(x)-f(x)|>\epsilon\}$. Choose $c>0$ , then we have \begin{align} \nu(A_n) & = \int_{A_n} g d\mu = \int_{A_n} g \chi_{\{g\le c\}} d\mu + \int_{A_n} g \chi_{\{g> c\}} d\mu\\ & \le c. \mu(A_n) + \int_{A_n} g \chi_{\{g> c\}} d\mu \\ & \le c. \mu(A_n) + \| g \chi_{\{g> c\}}\|_1 \lim_n\mu(A_n)\\ \lim_n\nu(A_n)& \le c. \lim_n\mu(A_n) + \| g \chi_{\{g> c\}}\|_1 \lim_n\mu(A_n)\\ & \le 0 \end{align}

Answer 1

You have used an inequality of the type $\int_A gd\mu \leq \|g\|_1 \mu(A)$. Such an inequity is not valid.

It is a well known fact that $\nu << \mu$ ($\mu, \nu$ finite measures) implies $\nu (A) \to 0$ as $\mu (A) \to 0$. From this it is immediate that $\nu (A_n) \to 0$ because $\mu (A_n) \to 0$.

Details: Suppose $\nu (A_n) $ does not tend to $0$. There exists $r>0$ and a subsequence ${n_k}$ such that $\mu (A_{n_k}) <\frac 1 {2^{k}}$ and $\nu (A_{n_k}) \geq r$ for all $k$. Let $A=\lim \sup A_{n_k}$. I will leave it to you to show that $\mu (A)=0$ and $\nu (A) \geq r$ which contradicts the fact that $\nu << \mu$.

Answer 2

You already know that $g\in L^{1}$, do you know further that, there exists some $\delta>0$, \begin{align*} \int_{S}gd\mu<\epsilon \end{align*} for any measurable $S$ such that $\mu(S)<\delta$?

If you use this, then simply plug $S=A_{n}$ for large $n$.