Prove theorem similar to Egorov theorem

Question

I am trying to prove the following theorem, which reminds me of Egorov's theorem.

Let $f$ be a bounded function on $[ -\pi , \pi]$ and suppose $\epsilon >0$. Then there is a step function $g$ and a set $E$ such that $|f-g|<\epsilon$ on the complement of E and $\mu(E)<\epsilon$

Answer 1

I assume your function $f$ is supposed to be Lebesgue measurable, then I think I know how you can tackle the problem in 2 steps.

1. You can prove that if $f:[a,b]\longrightarrow \mathbb{R}$ is a measurable function then you can express it as the pointwise limit of a sequence of simple functions $(\psi_n)_n:\mathbb{R}\longrightarrow\mathbb{R}$ (I call a simple function one of the form $\psi=\sum\alpha_i\chi_{E_i}$ where $E_i$ are Lebesgue Measurable sets).

2. You can use that any Lebesgue Measurable set $E$ with $\mu(E)<\infty$ is "close to a finite union of intervals" (or more precisely: $\forall \epsilon>0,\exists I_1,\ldots,I_n$ intervals such that $\mu(E\Delta\bigcup_{i=1}^{n}I_i)<\epsilon$) to prove that you can also find a sequence $(\phi_n)_n$ of step functions that converges pointwisely to $f$ $\mu$-almost everywhere. Then using Egorov's theorem on this sequence gives you what you want.

There may be an easier way to do it, but these are the properties I'm familiar with right now.