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


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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.