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joypav

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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!