I have got a question about the weak-* limit of lebesgue measures.
Assume that there exists a weak-* limit $\mu^*$ of measures $\mu_m$ which $\mu_m$(s) are lebesgue measures.
What could we say about the type of the measure $\mu^*$ ? Are the properties of $\mu_m$(s) preserved under weak-* convergence? Is $\mu^*$ a lebesgue measure too?
No, it is not a Lebesgue measure in general. Let us work on $[0,1]$ and denote by $\lambda$ the Lebesgue measure on the interval. Take for example $\mu_n$ defined as $$ \mu_n(A) = n \mu(A\cap [0,1/n]), \quad \int f d\mu_n = n \int_0^{1/n} f(x) \, dx. $$ The sequence $\mu_n$ is weakly converging to the Dirac measure $\delta_0$. This is not a Lebesgue measure.
Let us prove the convergence. For all continuous $f$, and using the change of variables $y=nx$, $$ \int f\, d\mu_n = n\int_0^{1/n} f(x) dx = \int_0^1 f\left({y\over n}\right) dy \longrightarrow f(0) = \int f\, d\delta_0 $$ by continuity of $f$ at $0$ and the dominated convergence theorem.