Consider a lebesgue measurable function $f:S\rightarrow \mathbb{R}$, where $S$ is a Lebesgue measurable subset of $\mathbb{R}$. I would like to show that there must exist a subset $S_{0}$ of $S$ with $|S_{0}|>0$, $|.|$ denotes Lebesgue measure, such that $f\in L^{\infty}(S_{0})$.
I suggest arguing by contradiction: Suppose the contrary. Then $f$ is not essentially bounded on any subset $U$ of $S$ with $|U|>0$. Then $|f|=\infty$ almost everywhere.
Does this make sense ?
Here is the answer from (quoted) the link https://en.wikipedia.org/wiki/Lusin%27s_theorem
Let $ (X,\Sigma ,\mu )$ be a Radon measure space and Y be a second-countable topological space, let $f:X\rightarrow Y$ be a measurable function. Given $\epsilon > 0$, for every $ A\in \Sigma$ of finite measure there is a closed set $E$ with $\mu(A \setminus E) < \epsilon$ such that $f$ restricted to $E$ is continuous. If A is locally compact, we can choose E to be compact and even find a continuous function$f_{\varepsilon }:X\rightarrow Y$ with compact support that coincides with $f$ on $E$ and such that $ \sup _{x\in X}|f_{\varepsilon }(x)|\leq \sup _{x\in X}|f(x)|$.