Suppose $f\in L^1(\mathbb{R}^n)$ has compact support. For any $\varepsilon>0,$ can we find some simple function $\varphi$ (namely $\varphi$ can be written as $\varphi=\sum_{i=1}^{m}a_i\chi_{E_i}$) so that $f\leq\varphi ,a.e.$ and $\left\|f-\varphi\right\|_{1}<\varepsilon$？
I think this is true beacuse first we can approximate $f$ by simple function $\psi$ so that $|\psi|\leq |f|$ and then try to adjust $\psi$ slightly so that it satisfies our requirement, but I failed in this step, can someone help me？
No. There exist (non-negative) $f$ which belong to $L^1$ but not $L^{\infty}$, so for any measurable function $\phi$ such that $f\leq \phi$ a.e, we must have $\phi\notin L^{\infty}$, and hence $\phi$ cannot be a simple function (they are bounded since by definition their image is a finite set).
As a concrete example, look at $f(t)=\frac{1}{\sqrt{t}}$ on $(0,1)$ and $0$ elsewhere. This is in $L^1$, but is not in $L^{\infty}$.
Also, as a side remark, it doesn't make sense to speak of the support of an $L^1$ function. Only continuous ones. The best you can say is there is a compact set $K$ and a measure zero set $E$ such that the support is contained in $K\cup E$ (and the above (equivalence class of) function satisfies this condition with $K=[0,1]$).