Exercise 1.3.23 (Lusin's theorem II)

Question

Let $f:\mathbb{R}^d	o\mathbb{C}$ be a
\begin{enumerate}[label=(oman*)]
\item locally absolutely integrable
\item Lebesgue measurable (almost everywhere finite)
\end{enumerate}    
Show that Lusin's theorem still holds for $f$.

Elu Thingol · Accepted Answer

\begin{enumerate}[label=(oman*)]
	\item We repeat the proof of Lusin's theorem, this time adjusting for locality. Pick an arbitrary $\epsilon>0$. For all $m\in\mathbb{N}$ consider the restricted function $f_m := f|_{B(0,m)}$. Then $(f_m)_{m\in\mathbb{N}}$ converges pointwise to $f$ (we will later use Egorov's theorem to translate this into the uniform convergence). Since each $f_m$ is absolutely integrable by theorem assumption, we can find a set $E_m$ such that $m(E_m) \leq \epsilon/2^{m+1}$ and $f_m|_{E_m^c}$ is continuous on $\mathbb{R}^d/E_m$ by Theorem 1.3.28. Let $E := \bigcup_{m=1}^{\infty} E_m$ with $m(E) \leq \epsilon/2$. Furthermore, since $(f_m)_{m\in\mathbb{N}}$ is a sequence of Lebesgue measurable functions converging pointwise to $f$, we can find a set $A\subset\mathbb{R}^d: m(A)\leq \epsilon/2$ such that $(f_m|_{E^c})_{m\in\mathbb{N}}$ converges uniformly to $f|_{E^c}$ on $\mathbb{R}^d/A$. Thus, set $E':=E \cup A$ with $m(E')\leq \epsilon$.  Then our sequence of continuous functions $(f_m|_{\mathbb{R}^d/E'})_{m\in\mathbb{N}}$ converges uniormly to $f$ by Egorov's theorem, and the uniform limit of continuous functions is again continuous.
	
	\item Now suppose that $f$ is Lebesgue measurable (and a.e. finite for the convergence to make sense). Define a sequence $(f_n)_{n\in\mathbb{N}}$ of Lebesgue measurable functions $f_n:= f\cdot 1_{\{x\in\mathbb{R}^d: |f(x)| \leq n\}} \cdot 1_{\{x\in\mathbb{R}^d: \|x\| \leq n\}}$. This implies an almost everywhere ``binary'' pointwise convergence of $(f_n)_{n\in\mathbb{N}}$ against $f$ (which we will convert to uniform convergence later). Each $f_n$ is a bounded Lebesgue measurable function; by definition we can find a sequence $(s_{n,m})_{m\in\mathbb{N}}$ of simple functions which converges to $f_n$ pointwise. Notice that $s_{n,m}$ are (eventually) bounded by the boundedness of their limit $f_n$ (that is where the $1_{\{x\in\mathbb{R}^d: |f(x)| \leq n\}}$ becomes useful); thus $\|s_{n,m}\|_{L^1} = \sum |c_i|m(I_i) < \infty$ are absolutely integrable by their simpleness. By Theorem 1.3.28 for each $n\in\mathbb{N}$ our simple functions $(s_{n,m})_{m\in\mathbb{N}}$ may be converted to continuous ones by exluding a set $E_n \subseteq \mathbb{R}^d: m(E_n) \leq \epsilon/3\cdot 1/2^{n}$ from the domain. Set $E=\bigcup_{n=1}^{\infty} E_n$ and for each $n\in\mathbb{N}$ we have $(s_{n,m}|_{E^c})_{m\in\mathbb{N}}$ are continuous. By Egorov's theorem (Theorem 1.3.26) we can find a set $A\subseteq \mathbb{R}^d: m(A) \leq \epsilon/3$ such that the convergence becomes uniform $(s_{n,m}|_{(E\cup A)^c})_{m\in\mathbb{N}} \stackrel{unf}{	o} f_n|_{(E \cup A)^c}$, which implies the continuity of each $f_n|_{(E \cup A)^c}$. Repeating this step, we can find a set $B\subseteq \mathbb{R}^d: m(B) \leq \epsilon/3$ such that the sequence $(f_n|_B)_{n\in\mathbb{N}}$ converges to $f|_B$ uniformly. Alas, set $K:= E \cup A \cup B$. We then have, outside of the set $K:m(K)\leq \epsilon$ the uniform convergence of the continuous functions $(f_n|_{K^c})_{n\in\mathbb{N}}$ to the $f|_{K^c}$, which then also must be continuous by Analysis II, Theorem 3.3.1.
\end{enumerate}

Exercise 1.3.23 (Lusin's theorem II)

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Exercise 1.3.23 (Lusin's theorem II)

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