Luzin's Theorem

Numbered 2.91 in Axler.

Rewrote in my own words and scanned the notes to markdown. A personal exercise.

Luzin Theorem [2.91]

• \(g:\mathbb{R}\to\mathbb{R}\) Borel measurable.

Then \(\forall \varepsilon>0\), \(\exists\) closed \(F\subseteq\mathbb{R}\) s.t. \(|\mathbb{R}\setminus F|<\varepsilon\) & \(g|_F\) continuous on \(F\).

Proof Idea

1. Prove Luzin for simple functions.
2. Apply Luzin to simple approximation.
3. Apply Egorov to get uniform continuity. Note Egorov need finite \((n,n+1)\).
4. Uniform limits preserve continuity.

Proof

Proof

Proof on simple functions

Let

\[ g=\sum_{k=1}^n a_k\chi_{D_k} \]

for distinct non zero \(a_k\in\mathbb{R}\) & disjoint Borel \(D_k\subseteq\mathbb{R}\).

By (2.71) Lebesgue equivalent definition \(\exists\) open \(G_k\), closed \(F_k\) \(\forall k\in\{1,\ldots,n\}\) s.t.

\[ F_k\subseteq D_k\subseteq G_k \]

where

\[ |D_k\setminus F_k|<\frac{\varepsilon}{2n} \quad\&\quad |G_k\setminus D_k|<\frac{\varepsilon}{2n}. \]

Then

\[ G_k\setminus F_k = (G_k\setminus D_k)\cup(D_k\setminus F_k) \]

has \(|G_k\setminus F_k|<\frac{\varepsilon}{n}\) \(\forall k\in\{1,\ldots,n\}\). Let

\[ F:= \left(\bigcup_{k=1}^n F_k\right) \cup \left(\bigcap_{k=1}^n \mathbb{R}\setminus G_k\right). \]

Then \(F\) closed subset of \(\mathbb{R}\) &

\[ \mathbb{R}\setminus F \subseteq \bigcup_{k=1}^n G_k\setminus F_k. \]

Latter because

\[ \begin{align*} \mathbb{R}\setminus F &= \left(\bigcap_{k=1}^n \mathbb{R}\setminus F_k\right) \cap \left(\bigcup_{k=1}^n G_k\right) \\ [\text{distributivity}] &= \bigcup_{k=1}^n \left( \left(\bigcap_{j=1}^n \mathbb{R}\setminus F_j\right) \cap G_k \right) \\ &\subseteq \bigcup_{k=1}^n \left((\mathbb{R}\setminus F_k)\cap G_k\right) \\ &= \bigcup_{k=1}^n G_k\setminus F_k. \end{align*} \]

Thus \(|\mathbb{R}\setminus F|<\varepsilon\).

\(F_k\subseteq D_k\Rightarrow g\) identically \(a_k\) on \(F_k\Rightarrow g|_{F_k}\) continuous.

\[ \bigcap_{k=1}^n \mathbb{R}\setminus G_k \subseteq \bigcap_{k=1}^n \mathbb{R}\setminus D_k \]

\(\Rightarrow g\) identically \(0\) on \(\bigcap_{k=1}^n\mathbb{R}\setminus G_k\Rightarrow g|_{\bigcap_{k=1}^n\mathbb{R}\setminus G_k}\) continuous.

Hence by Ex 9 (disjoint open continuity union is continuous) \(g|_F\) is continuous.

Apply Luzin to simple approximations

By simple approximation of measurable functions (2.89) \(\exists g_k:\mathbb{R}\to\mathbb{R}\) \(\forall k\in\mathbb{Z}^+\) that converges pointwise to \(g\) where \(g_k\) simple, Borel measurable.

By simple Luzin \(\exists\) closed \(C_k\subseteq\mathbb{R}\) s.t.

\[ |\mathbb{R}\setminus C_k|<\frac{\varepsilon}{2^{k+1}}, \qquad g_k|_{C_k}\text{ continuous} \]

\(\forall k\in\mathbb{Z}^+\). Let

\[ C=\bigcap_{k=1}^{\infty}C_k. \]

Thus \(C\) is closed & \(g_k|_C\) continuous \(\forall k\in\mathbb{Z}^+\) &

\[ \mathbb{R}\setminus C \subseteq \bigcup_{k=1}^{\infty}\mathbb{R}\setminus C_k \Rightarrow |\mathbb{R}\setminus C| \leq \frac{\varepsilon}{2}. \]

Apply Egorov to each finite \((m,m+1)\), \(n\in\mathbb{Z}\)

For each \(n\in\mathbb{Z}\) the sequence \(g_k|_{(m,m+1)}\) converges pointwise to \(g\).

Egorov gives Borel \(E_m\subseteq(m,m+1)\) s.t. \(g_k\) converges uniformly to \(g\) on \(E_m\) &

\[ |(m,m+1)\setminus E_m| < \frac{\varepsilon}{2^{|m|+3}} \qquad \forall m\in\mathbb{Z}. \]

Then \(g_k\) converges uniformly on \(C\cap E_m\) to \(g\). Since \(g_k|_C\) is continuous the uniform limit in \(C\cap E_m\), \(g|_{C\cap E_m}\), is continuous by 2.84.

Let

\[ D=\bigcup_{m\in\mathbb{Z}} C\cap E_m. \]

Then \(g|_D\) is continuous because \(C\cap E_m\) are subsets of disjoint open intervals & subsets of disjoint open intervals keep continuity.

Now just need to show \(\mathbb{R}\setminus D\) arbitrarily small & then take arbitrarily small closed approximate \(F\).

\[ \begin{align*} \mathbb{R}\setminus D &= \bigcap_{n\in\mathbb{Z}} \left( (\mathbb{R}\setminus C)\cup(\mathbb{R}\setminus E_n) \right) \\ &= (\mathbb{R}\setminus C) \cup \left( \bigcap_{m\in\mathbb{Z}}\mathbb{R}\setminus E_m \right) \qquad [\text{distributivity}]. \end{align*} \]

\[ \bigcap_{m\in\mathbb{Z}}\mathbb{R}\setminus E_m \subseteq \mathbb{Z} \cup \left( \bigcup_{m\in\mathbb{Z}}(m,m+1)\setminus E_m \right). \]

Hence,

\[ \mathbb{R}\setminus D \subseteq (\mathbb{R}\setminus C) \cup \mathbb{Z} \cup \left( \bigcup_{m\in\mathbb{Z}}(m,m+1)\setminus E_m \right). \]

\[ \Rightarrow |\mathbb{R}\setminus D| < \frac{\varepsilon}{2}+0+\frac{\varepsilon}{2} = \varepsilon. \]

Then by equivalent of Lebesgue 2.65 \(\exists\) closed \(F\subseteq D\) s.t.

\[ |D\setminus F| < \varepsilon-|\mathbb{R}\setminus D|. \]

Then,

\[ \mathbb{R}\setminus F = (\mathbb{R}\setminus D)\cup(D\setminus F) \]

and so

\[ \begin{align*} |\mathbb{R}\setminus F| &\leq |\mathbb{R}\setminus D|+|D\setminus F| \\ &< |\mathbb{R}\setminus D| +\varepsilon -|\mathbb{R}\setminus D| \\ &= \varepsilon. \end{align*} \]

\(g|_D\) continuous \(\Rightarrow g|_F\) continuous. \(\blacksquare\)