# When does narrow convergence of abs. cont. measures imply convergence in $(L^{\infty})^*$

I am struggling with the following question:

Assume that $f,f_1,f_2,...$ are integrable and non-negative (I am not sure, if this is important) and $f_n d\lambda$ narrowly converges against $fd\lambda$, i.e.

$\lim_{n\rightarrow\infty}\int g \cdot f_n d\lambda=\int g \cdot f d\lambda$

for every bounded continuous function g.

Does this imply:

$\lim_{n\rightarrow\infty}\int g \cdot f_n d\lambda=\int g \cdot f d\lambda$ for every $g \in L^{\infty}$

Here $\lambda$ is the Lebesgue measure. I have proved the assumption, if the $f_n$ are equi-integrable. In this case we can approximate a function $g \in L^{\infty}$ with $g_\delta$, such that $\int |g-g_\delta| \cdot f_n d\lambda<\delta$ for all n.

But what happens in the general case?

Best regards,

Max

According to this MathOverflow thread, the answer is no, even if the $f_n$ are supported in $[0,1]$.
Define $\mu_n:=f_n\lambda$ and $\mu:=f\lambda$: then $\mu_n$ and $\mu$ are Borel measures which are absolutely continuous with respect to Lebesgue measure. Assuming $f\neq 0$, we have that $\mu_n\to \mu$ in distribution (considering $f_n/\lVert f_n\rVert$ in order to deal with probability measures). Then the convergence in $(L^{\infty})^*$ would mean that $\mu_n(A)\to \mu(A)$ for each $A$.
It may happen that we have a sequence of probability measures $(\mu_n)$ with density which converges weakly to $\mu$ (with density), but not setwise.