The statement of the simple approximation lemma is given below:
My question is:
My professor added this remark After proving it "If $f\geq 0$, then $\varphi \geq 0.$", but I do not understand why, If $f\geq 0$ this leads to that $\varphi \geq 0,$ could anyone explain this for me, please?
What your professor meant is that in the case that $f\geq 0$, then you can additionally demand that $\varphi_\epsilon\geq 0$ in the statement of the theorem. To prove this, note that if you have $\varphi_\epsilon$ which is negative somewhere, you can simply define $\varphi_\epsilon'=\max(\varphi_\epsilon,0)$ and then $\varphi_\epsilon'\geq 0$ and it still satisfies $\varphi_\epsilon'\leq f\leq \psi_\epsilon$ and $0\leq \psi_\epsilon-\varphi_\epsilon'<\epsilon$ on $E$.