Let $sl_{2}$ be $2 \times 2$ traceless matrices over field $K$ of characteristic $0$. How can we define a $\mathbb{Z}$-grading on $sl_{2}$?
Let consider $h ,e ,f$ as follows respectively: \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} then the grading $L_{-1}=\mathbb{K}h$ , $L_{0}=\mathbb{K}h , \mathbb{K}f$ and $L_{i}=0$ for all $ i \notin \{-1,0,1\}$. How can it be considered as $\mathbb{Z}$-graded?
In order for $L=\bigoplus L_i$ to be $\mathbb{Z}$-graded, you need $[L_i,L_j]\subseteq L_{i+j}$, or in other words $[x,y]\in L_{i+j}$ whenever we have $x\in L_i$ and $y\in L_j$. In this case, we may define
$$ L_{-1}=Ke \qquad L_0=Kh \qquad L_1=Kf. $$
(And $L_i=0$ otherwise.) To check it is $\mathbb{Z}$-graded, it suffices to check the commutator conditions for the basis elements. For example, $e\in L_{-1}$ and $h\in L_0$ so we ought to have $[e,h]\in L_{-1}$. Is this true?
If you want, you can make a $3\times 3$ table of commutators $[x,y]$ of basis elements $x,y\in\{e,h,f\}$, and then check each entry is in the correct graded component in this way. A couple other observations you can make: all diagonal entries are $0$, so they are all in the correct graded component (why?), and if an off-diagonal entry is in the correct graded component then the correponding entry on the other side of the diagonal is also in the correct graded component (why?).