Let $X$ be a space with its $\sigma$-algebra $\mathcal{B}$; we are given a finite measure $\mu$ and a sequence of finite measures $\nu\_n$ such that, for every bounded continuous function $f:X\to\mathbb{R}$ we have
$$\int\_X fd\nu\_n\longrightarrow \int\_X fd\nu$$
for some finite measure $\nu$.


By the Lebesgue decomposition theorem, for each $n$ there exist a function $g\_n\in L^1(X, \mu)$ and a measure $\eta\_n\perp \mu$ such that $$\nu\_n=g\_n\mu+\eta\_n\;.$$
Moreover, $\nu=g\mu+\eta$.


Is it true that, for every bounded continuous function $f:X\to\mathbb{R}$, we have
$$\int\_X fg\_nd\mu\longrightarrow \int\_X fgd\mu\;?$$
Or, in other words, is it true that, for any such $f$, $fg\_n\to fg$ in $L^1(X,\mu)$?



Weak continuity of Lebesgue decomposition
The answer is no. For example, you can construct a sequence of $g\_n\in L^1(\mathbb{R}^n)$ converging to the Dirac $\delta$ measure. 


Furthermore, we can also construct a sequence of singular measures converging to an $L^1$ function, e.g. Dirac $\delta$ measures suppored on finitely many points (with suitable weights on each point), which become more and more dense.


