Lusin's Theorem for finite measure spaces Lusin's Theorem. Let $f$ be a real-valued measurable function on $E$. Then for each $\varepsilon&gt;0$,  there is a continuous function $g$ on $\mathbb{R}$ and a closed set $F$ contained in $E$ for which $f=g$ on $F$ and $m(E\sim F)&lt;\varepsilon$.


can anybody guide me the proof of the theorem or can advice me some book from where i can go through the proof.
Here is a proof of the Lusin's Theorem where $f:[a,b]\to \mathbb{C}$.

Lusin's Theorem: If $f:[a,b]\to \mathbb{C}$ is Lebesgye measurable and $\epsilon &gt; 0$, there is a compact set $E\subset [a,b]$ such that $\mu(E^c) &lt; \epsilon$ and $f|E$ is continuous. (Note I will use Egoroffs theorem, Theorem 2.26, theorem 1.18, and Corollary 2.32 in Folland's Real Analysis, I am sure there are the same theorems in your book)

Proof: Let $f:[a,b]\to \mathbb{C}$ be Lebesgue measurable and $\epsilon &gt;0$. By theorem 2.26 we can build a sequence of continuous functions $\{g_n\}$ such that $$g_n\to f \in L^1$$
Then by corollary 2.32 there is a subsequence $\{g_{n_j}\}$ of $\{g_n\}$ such that $g_{n_{j}}\to f$ a.e. Now by Egoroff's theorem, for any $\epsilon &gt; 0$ there exists a set $F\subset [a,b]$ with $\mu(F) &lt; \epsilon/2$ such that $g_{n_j}\to f$ uniformly on $F^{c}$.

Now by theorem 1.18, since $\mu([a,b]) &lt; \infty$, there is $E$ compact subset of $[a,b]$ such that $E\subset F^{c}$ and $$\mu(F^c) - \epsilon/2 &lt; \mu(E) \leq \mu(F^c)$$ So $F\subset E^c$ and we have 
\begin{align*}
\mu(E^c) &amp;= \mu(F) + \mu(E^c\setminus F)\\
&amp;= \mu(F) + \mu(E^c\cap F^c)\\
&amp;= \mu(F) + \mu(F^c\setminus E)\\
&amp;= \mu(F) + (\mu(F^c) - \mu(E))\\
&amp;\leq \epsilon/2 + \epsilon/2\\
&amp;= \epsilon
\end{align*}
Note since $E\subset F^c$ and $g_{n_j}\to f$ uniformly on $F^c$, we have that $g_{n_j}\to f$ uniformly on $E$. Since, for all $j$, $g_{n_j}$ is continuous, we have that $f$ is continuous on $E$, that is $f|E$ is continuous. 

Hope this helps.
Here is a proof of the Lusin's Theorem where $f:[a,b]\to \mathbb{C}$.

Lusin's Theorem: If $f:[a,b]\to \mathbb{C}$ is Lebesgye measurable and $\epsilon &gt; 0$, there is a compact set $E\subset [a,b]$ such that $\mu(E^c) &lt; \epsilon$ and $f|E$ is continuous. (Note I will use Egoroffs theorem, Theorem 2.26, theorem 1.18, and Corollary 2.32 in Folland's Real Analysis, I am sure there are the same theorems in your book)

Proof: Let $f:[a,b]\to \mathbb{C}$ be Lebesgue measurable and $\epsilon &gt;0$. By theorem 2.26 we can build a sequence of continuous functions $\{g_n\}$ such that $$g_n\to f \in L^1$$
Then by corollary 2.32 there is a subsequence $\{g_{n_j}\}$ of $\{g_n\}$ such that $g_{n_{j}}\to f$ a.e. Now by Egoroff's theorem, for any $\epsilon &gt; 0$ there exists a set $F\subset [a,b]$ with $\mu(F) &lt; \epsilon/2$ such that $g_{n_j}\to f$ uniformly on $F^{c}$.

Now by theorem 1.18, since $\mu([a,b]) &lt; \infty$, there is $E$ compact subset of $[a,b]$ such that $E\subset F^{c}$ and $$\mu(F^c) - \epsilon/2 &lt; \mu(E) \leq \mu(F^c)$$ So $F\subset E^c$ and we have 
\begin{align*}
\mu(E^c) &amp;= \mu(F) + \mu(E^c\setminus F)\\
&amp;= \mu(F) + \mu(E^c\cap F^c)\\
&amp;= \mu(F) + \mu(F^c\setminus E)\\
&amp;= \mu(F) + (\mu(F^c) - \mu(E))\\
&amp;\leq \epsilon/2 + \epsilon/2\\
&amp;= \epsilon
\end{align*}
Note since $E\subset F^c$ and $g_{n_j}\to f$ uniformly on $F^c$, we have that $g_{n_j}\to f$ uniformly on $E$. Since, for all $j$, $g_{n_j}$ is continuous, we have that $f$ is continuous on $E$, that is $f|E$ is continuous. 

Hope this helps.