Relation between $L^2(\mu)$ convergence and $L^2(\nu)$ convergence. Here's the setting: Let $\mu$ and $\nu$ be two probability measures on $(\mathbb{R},\mathcal{B})$. Assume $\mu$ and $\nu$ are equivalent (that is, $\nu\ll\mu$ and $\mu\ll\nu$). Let $(f_n)$ be a sequence of bounded, measurable functions. Obviously one has both $f_n\in L^2(\mu)$ and $f_n\in L^2(\nu)$. Assume that $\int \vert f_n\vert d\mu \rightarrow 0$. Does this imply $\int \vert f_n\vert d\nu\rightarrow 0$?
Comments: I do not know if the $f_n$ are dominated by an integrable function. I think they are not. Thus a Dominated Convergence approach is likely to fail.
I have been trying to establish uniform $\nu$-integrability of $(f_n)$. This would give the result, by Vitali's Convergence Theorem. Indeed, by this Theorem we have, since $\int \vert f_n\vert d\mu \rightarrow 0$ by hypothesis,


*

*$f_n\rightarrow 0$ in $\mu$-measure.

*for each $\varepsilon>0$ there is a set $E_\varepsilon$ such that $\int_{E_\varepsilon^c}\vert f_n\vert d\mu < \varepsilon$ for all $n$.

*for each $\varepsilon>0$ there is a $\delta>0$ such that $\int_E \vert f_n\vert d\mu <\varepsilon$ for all $n$ whenever $\mu(E)<\delta$.


Obtaining items (1) and (2) with $\nu$ in place of $\mu$ is straightforward. I'm having trouble with (3). Any insights?
 A: No, in general we cannot expect this.
(Counter)Example: For suitable constants $c_1,c_2>0$ 
$$\mu(dx) := c_1 \frac{1}{1+x^4} \, dx \qquad \text{and} \qquad \nu(dx) := c_2 \frac{1}{1+x^2} \, dx$$
are probability measures on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$. Moreover, because the density functions are strictly positive, it is not difficult to see that $\mu$ and $\nu$ are equivalent. Now consider
$$f_n(x) := \begin{cases} x, & 2^n \leq x \leq 2^{n+1}, \\ 0, & \text{otherwise}. \end{cases}$$
Then each $f_n$ is bounded and measurable and
$$\int |f_n| \, d\mu \leq \int_{2^n \leq x \leq 2^{n+1}} \frac{1}{x^3} \, dx \xrightarrow[]{n \to \infty} 0.$$
On the other hand, as
$$\frac{1}{1+x^2} \geq \frac{1}{2} \frac{1}{x^2}$$
for all $|x| \geq 1$, we have
$$\begin{align*} \int |f_n| \, d\nu &= \int_{2^n \leq x \leq 2^{n+1}} \frac{x}{1+x^2} \, dx\\  &\geq \frac{1}{2} \int_{2^n \leq x \leq 2^{n+1}} \frac{1}{x} \, dx \\ &= \frac{1}{2} \left( \log (2^{n+1})- \log (2^n) \right) = \frac{\log 2}{2}; \end{align*}$$
this shows that $\int |f_n| \, d\mu$ does not converge to $0$ as $n \to \infty$.
