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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 $L^1(X,\mu)$?

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up vote 3 down vote accepted

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.

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