Home > Problem Solving, Real Analysis > Compacts convergence implies characteristic convergence

Compacts convergence implies characteristic convergence


Prove that if  the sequence \Omega_n converges in the compact sense to \Omega and if the boundary of \Omega has zero measure, then the sequence \Omega_n converges also in the way of characteristic functions.

Solution: Pick a compact set K which intersects \Omega.  Recalling a theorem of Caratheodory which states that the Lebesgue measure comes from an outer measure, or the regularity property of the Lebesgue measure, we can see that we can cover the part of the boundary of \Omega which is also in K, with a countable collection of open boxes whose union we denote with D with sufficiently small measure, let’s say \varepsilon. Then K\setminus D is closed and therefore compact, and can be written as the union of two compacts: K_1=K \cap \Omega\setminus D and K_2=(K\setminus D)\setminus K_1. Since K_1 \subset \Omega and K_2 \subset \overline{\Omega}^c we have K_1 \subset \Omega_n and K_2 \subset \overline{\Omega_n}^c for n suficiently large.

Then \int_K \chi_{\Omega_n}-\chi_{\Omega}=|K\cap \Omega_n|-|K \cap \Omega|=|K_1|+|\Omega_n \cap D|-|K_1|-|\Omega \cap D|= =|\Omega_n \cap D|-|\Omega \cap D|, which is in [-\varepsilon,\varepsilon]. This means that the considered integral converges to 0.

Now taking \int_K |\chi_{\Omega_n}-\chi_{\Omega}|=|K \cap (\Omega\setminus \Omega_n)|+|K \cap (\Omega_n \setminus \Omega)|. From the above K \cap (\Omega_n \setminus \Omega)\subset D and K \cap (\Omega \setminus \Omega_n) \subset D, which finishes the proof.

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