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## Open sets convergence

Consider the sequence of open sets of $[0,1]$ defined by
a) $\displaystyle \Omega_n:= \bigcup_{k=0}^{2^{n-1}-1} \left(\frac{2k}{2^n},\frac{2k+1}{2^n}\right)$.

b) $\displaystyle \Omega_n:= \bigcup_{k=0}^{2^{n}-1} \left(\frac{k}{2^n},\frac{k+1}{2^n}\right)$.

Study the convergence of the sets $\Omega_n$ in the following types of convergence:

• weak convergence of characteristic functions;
• convergence in the topology defined by the Hausdorff distance;
• convergence in the way of compact sets.

(Each of the previous types of convergence will be presented below.)

weak convergence of characteristic functions

The sequence of measurable sets $( \Omega_n )$ of $\Bbb{R}_N$ is said to converge in the way of characteristic functions towards the measurable set $\Omega$ if the sequence of characteristic functions $\chi_{\Omega_n}$ converges to $\chi_{\Omega}$ in $L^p_\text{loc}(\Bbb{R}^N), \ \forall p \in [1,\infty)$.

The first remark is that $p \in [1,\infty)$ doesn’t matter since $0^p=0,\ 1^p=1$. The second remark is that if the sequence $\chi_{\Omega_n}$ converges in the weak-* topology of $L^\infty (\Bbb{R}^N)$  to $\chi_{\Omega}$ then $(\Omega_n)$ converges in the way of characteristic functions to $\Omega$.

convergence in the topology defined by the Hausdorff distance

Consider a fixed (large enough) compact of $K \subset \Bbb{R}^N$. For two compacts $K_1,K_2$ in $K$, and consider the following definitions:

1. $d(x,K_1):=\inf_{y \in K_1} d(x,y)$;
2. $\rho(K_1,K_2)=\sup_{x \in K_1}d(x,K_2)$;
3. $d^H(K_1,K_2)=\max(\rho(K_1,K_2),\rho(K_2,K_1))$.

A sequence of open sets $(\Omega_n)$ and $\Omega$ included in $B$. We say that $\Omega_n$ converges in the Hausdorff distance to $\Omega$ if $d^H(B\setminus \Omega_n, B\setminus \Omega) \to 0$ for $n \to \infty$. We denote $\Omega_n \xrightarrow[]{H} \Omega$. We can define the same type of definition for compact sets from $K$.

convergence in the way of compact sets

We say that the sequence of open sets $(\Omega_n)$ of $\Bbb{R}^N$ converges in the way of compact sets towards $\Omega$, and we denote $\Omega_n \xrightarrow[]{K} \Omega$ if we have

• $\forall K$ compact $\subset \Omega$, we have $K\subset \Omega_n$ for $n$ large enough;
• $\forall L$ compact $\subset \overline{\Omega}^c$ we have $L \subset \overline{\Omega_n}^c$ for $n$ large enough.

Solution: a) For the characteristic functions convergence we will show that $\chi_{\Omega_n} \to 1/2\$ weakly in $L^p(0,1)$, which gives us an example of sequence of characteristic functions whose limit is not a characteristic function.
We need to show that $\displaystyle \lim_{n\to \infty} \int_{(0,1)} \chi_{\Omega_n} \phi \to \int_{(0,1)} \frac{1}{2} \phi$, for all $\phi \in L^1(0,1)$.

For proving this, we use a trick used in many real analysis proofs. Start with step functions, then extend to positive function, and finally to measurable functions. I will propose a similar, more detailed approach:

• $\phi=\chi_{[a,b]}$ where $a,b$ are fractions with denominators powers of $2$. This is quite easy, given the form of the sets $\Omega_n$.
• extend the last step to all compact intervals, $latex \phi=\chi_{[a,b]}$, (using Beppo Levi’s Theorem)
• extend to step functions
• then to positive functions
• finally to integrable functions

b) $\chi_{\Omega_n}=\chi_{[0,1]}\ a.e.$, therefore, the convergence in the way of characteristic functions is clear enough.

Denote $K_n:=[0,1]\setminus \Omega_n =\bigcup_{k=0}^{2^n} \{\frac{k}{2^n}\}$ which converges in the Hausdorff distance to the interval $[0,1]$, which means by taking complements that $\Omega_n \xrightarrow[]{H} \emptyset$.

Finally, it is clear that $\Omega_n$ does not converge to any open set in the way of compacts. Suppose $\Omega_n \xrightarrow[]{K} \Omega$. If $\Omega=\emptyset$. Pick $K \subset [0,1]$. Then it would mean that $K\subset \Omega_n$ for $n$ large enough. But the “holes” in $\Omega_n$ become “denser” in $[0,1]$, which means that if we pick some compact containing an open set, that compact will not satisfy the given definition. On the contrary, if the sequence would converge to a nonempty open set, some compact within it will not belong to all $\Omega_n$.