Background Information:
Theorem 1.18 - If $E\in M_{\mu}$, then \begin{align*} \mu(E) &= \inf\{\mu(U):E\subset U, U \ \text{open}\}\\ &=\sup\{\mu(K):E\subset K, K \ \text{compact}\}\end{align*}
Theorem 2.26 - If $f\in L^1(\mu)$ and $\epsilon > 0$, then
a.) there is an integrable simple function $\phi = \sum_{1}^{n}a_j\chi_{E_j}$ such that $\int |f - \phi|d\mu < \epsilon$.
b.) If $\mu$ is a Lebesgue-Stieltjes measure on $\mathbb{R}$, the sets $E_j$ in the definition of $\phi$ can be taken to be finite unions of open intervals.
c.) Moreover, in situation b.), there is a continuous function $g$ that vanished outside a bounded interval such that $\int |f - g|d\mu < \epsilon$.
Corollary 2.32 - If $f_n\rightarrow f$ in $L^1$, there is a sub-sequence $\{f_{n_j}\}$ such that $f_{n_{j}}\rightarrow f$ a.e.
2.33 Egoroff's Theorem - Suppose that $\mu(X) < \infty$, and $f_1,f_2,\ldots$ and $f$ are measurable complex-valued functions on $X$ such that $f_n\rightarrow f$ a.e. Then for every $\epsilon > 0$ there exists a set $E\subset X$ such that $\mu(E) < \epsilon$ and $f_n\rightarrow f$ and $f_n\rightarrow f$ uniformly on $E^c$.
Question:
Lusin's Theorem - If $f:[a,b]\rightarrow \mathbb{C}$ is Lebesgue measurable and $\epsilon > 0$, there is a compact set $E\subset [a,b]$ such that $\mu(E^c) < \epsilon$ and $f|E$ is continuous.
Attempted proof - Let $f:[a,b]\rightarrow\mathbb{C}$ be Lebesgue measurable and $\epsilon > 0$. By theorem 2.26 we can build a sequence of continuous functions $\{g_n\}$ such that $$g_n\rightarrow f \ \text{in} \ L^1$$ Then by Corollary 2.32 there is a sub-sequence $\{f_{n_j}\}$ of $\{g_n\}$ such that $f_{n_j}\rightarrow f$ a.e. Now, by Egoroff's theorem there exists a set $E$ with $\mu(E) < \infty$ such that $g_n\rightarrow f$ uniformly on $X\setminus E$. Note since $\{f_{n_j}\}$ is a sub-sequence of $\{g_n\}$ then we have $f_{n_j}\rightarrow f$ uniformly on $X\setminus E$ as well. Now by theorem 1.18 we can take $E$ to be a compact subset of $[a,b]$ since $K$ is compact and $E\subset K$. Note that $X\setminus E$ is the space on which $g_n\rightarrow f$ uniformly so then $E^c$ must be part of that space where $g_n\rightarrow f$ uniformly and thus $\mu(E^c) < \epsilon$ by definition. Finally since $\{g_n\}$ are continuous functions and $g_n\rightarrow f$ uniformly on $X\setminus E$ then $f$ is continuous.
I am not sure this is exactly correct, any suggestions is greatly appreciated.