Lusin's Theorem - Infinite Measure and not Real-Valued Extensions

[joypav]

Lusin's Theorem: Let $f$ be a real-valued measurable function on $E$. Then for each $\epsilon > 0$, there is a continuous function $g$ on $R$ and a closed set $F$ contained in $E$ for which $f=g$ on $F$ and $m(E - F)<\epsilon$.

I'm going through exercises in the book... almost finals time. The proof for Lusin's Theorem is in the book and I have made sure I understand it. However, I am having trouble with the extensions of the theorem. I have written a proof for when $E$ has infinite measure, but I'm not sure how to approach the extension when $f$ is not necessarily real-valued.

Prove the extension of Lusin's Theorem to the case that $E$ has infinite measure and the case where $f$ is not necessarily real-valued. (Exercise in Royden)

Proof for infinite measure:
Let $f$ be a real-valued measurable function on $E$.
Let,
$E_n = E \cap [n, n+1)$
Then for $m \neq n, E_n \cap E_m = \emptyset$
Each $E_n$ has finite measure. (From a homework problem I did previously in the semester)

Apply Lusin's Theorem.

$\exists F_n$ closed, $g_n : F_n \rightarrow R$ continuous such that $m(E - F_n) < \frac{\epsilon}{2^{n+1}}$, and $g_n = f$ on $F_n$.
Let,
$F = \cup_{n \in \Bbb{N}} F_n$ and $g(x) = \sum_{n \in \Bbb{N}} g_n(x) \chi_{F_n}(x)$
Then $g$ is continuous when restricted to $F$.

Consider the sequence $(x_n) \subset F$ such that $x_n \rightarrow x, x \in E$.
Since $x \in E_n$, some $n$,
$\implies \exists N \in \Bbb{N}$ such that $(x_n)_{n \geq N} \subset F_{n-1} \cup F_n$. ($F_{n-1} \cup F_n$ closed)
$\implies x \in F_{n-1} \cup F_n \subset F$
$\implies F$ closed.

Now, extend $g$ continuously to $R \implies g=f$ on $F$.
(I know I can extend $g$ by a previous exercise in this section.)
$m(E - F) = m(\cup_{n \in \Bbb{N}} E_n - F_n) = \sum_{n \in \Bbb{N}} m(E_n - F_n) < \epsilon$I would appreciate feedback on this proof and help with the not necessarily real-valued extensions!

 

[joypav]

I have not yet made progress on the second proof... (f not necessarily real-valued), but I asked my professor about it and he informed me that I am not supposed to be thinking about complex numbers. What this question is referring to is the extended real line.

 

[math771]

Thanks!

Your proof for the extension to infinite measure looks good. For the case where $f$ is not necessarily real-valued, we can still use a similar approach.

Let $f$ be a measurable function on $E$, which may take values in $\mathbb{R}$ or $\mathbb{C}$.
We can write $f$ as $f = f_1 + if_2$, where $f_1$ and $f_2$ are real-valued functions.
Apply Lusin's Theorem to $f_1$ and $f_2$ separately, with $\frac{\epsilon}{2}$ instead of $\epsilon$.
We get continuous functions $g_1$ and $g_2$ on $R$ and closed sets $F_1$ and $F_2$ contained in $E$ such that $f_1 = g_1$ on $F_1$ and $f_2 = g_2$ on $F_2$, and $m(E - F_1) < \frac{\epsilon}{2}$ and $m(E - F_2) < \frac{\epsilon}{2}$.

Define $g(x) = g_1(x) + ig_2(x)$ for $x \in F_1 \cap F_2$.
Then $g$ is continuous on $F_1 \cap F_2$.
Now, for $x \in E - (F_1 \cap F_2)$, we can choose a sequence $(x_n) \subset F_1 \cap F_2$ such that $x_n \rightarrow x$.
Then, $g(x_n) = g_1(x_n) + ig_2(x_n) \rightarrow f_1(x) + if_2(x) = f(x)$.
Thus, $g(x) = f(x)$ for $x \in E - (F_1 \cap F_2)$.

Finally, let $F = F_1 \cap F_2$.
Then $F$ is closed, and $g = f$ on $F$.
Also, $m(E - F) \leq m(E - F_1) + m(E - F_2) < \epsilon$.

I hope this helps!