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joypav
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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!
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!