I am struggling with proving the following:
Suppose that $\phi: [a, b] \rightarrow \mathbb{R}$ is a simple function such that $$\sup_{x\in [a,b]} |\phi(x)| \leq B.$$
Let $\epsilon > 0$. Prove that there exists a step function $\psi: [a,b] \rightarrow \mathbb{R}$ and a Lebesgue-measurable set $E \subset [a,b]$ such that $$\sup_{x\in[a,b]}|\psi(x)| \leq B, \quad m(E) < \epsilon,$$ $$\forall x \in E^c, \quad |\phi(x) - \psi(x)| < \epsilon.$$
I have already shown that given such a step function $\psi$, we can find a continuous function $g: [a,b] \rightarrow \mathbb{R}$ that approximates $\psi$ on a similar measurable set $E$. But now I am not sure how to approximate the simple function with a step function. Seemingly, this should be simple- a simple function is just a step function if it can be expressed as $$\psi = \sum_{i=1}^n a_i \chi_{U_i},$$ where $\chi$ is the indicator function, $a_i$ are the values in the finite set of values that $\psi$ takes on, and $U_i \subset [a, b]$ is a finite union of disjoint intervals that cover $[a, b]$. What am I missing? Any help is appreciated!
Lemma: Suppose $m(E)<\infty$. Then, for every $\delta>0$, there is a finite union of open intervals $U$ such that $m(U \backslash E)+m(E \backslash U)<\delta$
Let $\epsilon > 0$. For $\phi=\sum_{i=1}^N a_i \chi_{E_i}$, take $U_i$ such that $m(U_i \backslash E_i)+m(E_i \backslash U_i)<\epsilon/N$ for each $i$, as per lemma.
Take $\psi=\sum_{i=1}^N a_i \chi_{U_i}$. Then $\forall x \in U \cap E = (U \Delta E)^c, \quad|\phi(x)-\psi(x)|= 0 \quad$ (where $\Delta$ is the symmetric difference).
And $m(U \Delta E) = \sum_{i=1}^N [m(U_i \backslash E_i)+m(E_i \backslash U_i)]<N\epsilon/N$.
Thus $m(U \Delta E) < \epsilon$ and we are done $\quad \square$