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.