MHB Understanding Lusin's Theorem for $\mathbb{R}$ and Its Proof

[joypav]

Problem:
Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be measurable. Then there exists a sequence of continuous functions $(g_n)$ such that $limg_n(x)$ exists for all $x \in \mathbb{R}$ and $limg_n(x) = f(x)$ a.e. x.

Is this like Lusin's Theorem? Lusin's theorem for the real numbers? If so, how does this change the proof?

 

[joypav]

Here is my proof (posting for completeness but if you have corrections feel free to comment!).

Proof:
Apply the Approximation Theorem to $f$.
Let $(f_n)$ be a sequence of simple functions such that $f_n \rightarrow f$ point wise.
For every $n \in \Bbb{N}$, let
$F_n \subset E$ be closed such that $m(R - F_n) < \frac{\epsilon}{2^k}$ such that $f_n |_{F_n}$ is continuous.
(By Egorov's) $\implies \exists B \subset R$ such that $m(B) < \epsilon$ and $f_n \rightarrow f$ uniformly on $R - B$.
WLOG, assume $B$ is open.
Then $R - B$ is closed.
Let $F_0 = R - B$.

Claim: $f |_{F_\infty}$ is continuous.
A uniformly convergent sequence of continuous functions is continuous
$\implies f |_{F_\infty} = F_0 \cap \cap_{k}F_k$ is continuous.
Then,
$m(E - F_\infty) \leq \sum_{n=0}^{\infty}m(E - F_n) < 2\epsilon$
$\implies$ We've found $F \subset R$ closed such that $m(R - F) < \epsilon$ and $f |_F$ is continuous.

Now, consider our set F.
$\exists G_1, G_2,...$ open and pairwise disjoint such that $F^c = \cup_{i \in \Bbb{N}}G_i$.

Let $G_i = (a_i, b_i)$. Let,
$g(x) = $
$f(x), x \in F$
$(\frac{b_i - x}{b_i - a_i})f(a_i) + (\frac{x - a_i}{b_i - a_i})f(b_i), x \in (a_i, b_i)$

Then,
$f |_F$ continuous $\implies g |_F$ is continuous.
We extend $g$ so that it is also continuous on $F^c$ by the definition for $g$ given.
We conclude that,
$g : R \rightarrow R$ is continuous and $g |_F = f$.