# Around the weak sequential completeness of $L^1$-spaces

Let $\mu, \nu$ be two mutually singular measures (say, on a segment), $w_n\in L^1(\mu)$ weights ($w_n\ge 0$, $\int w_nd\mu=1$) that $\ast$-weakly tend to $\nu$: $$\int fw_n\,d\mu\to \int f\,d\nu \qquad(\ast)$$ for any continuous function $f$. Then 1) the convergence holds for some discontinuous $f$ (now I am able to prove this if the set of discontinuities of $f$ is a closed set of zero $\nu$-measure), and 2) there exist $f\in L^\infty$ for which the convergence fails (otherwise by the weak sequential completeness of $L^1$ the limit measure must correspond to a function from $L^1(\mu)$; I know an approach allowing to produce examples of such functions $f$ taking values $0, 1$).

Are there any known results about the characterization of those $f\in L^\infty(\mu)$ for which the convergence $(\ast)$ holds/fails?

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How do we make sense of $\int f \ d\nu$ for $f\in L^\infty(\mu)$? As $L^\infty(\mu)$ would usually be the equivalence classes of functions; thus we'd need for example that if $f=0$ a.e. for $\mu$, then $\int f \ d\nu=0$, i.e. $f=0$ a.e. for $\nu$. That's quite problematic for singular measures... – Matthew Daws Mar 26 '12 at 20:42
'The convergence fails' means that the sequence has no limit. – vvk Mar 26 '12 at 21:00
To say that the convergence holds, one should first define the values of $f$ $\nu$-almost everywhere. In my example 1) this is the usual extension by continuity; one can also consider natural generalizations if we manage to establish the fact of convergence of the sequence (to some limit). – vvk Mar 27 '12 at 7:49