I want to check if my proof below is correct.
Because the dual of $L^1(\Bbb R )$ is $L^{\infty }(\Bbb R )$ it is enough to show that $L^1(\Bbb R )$ is not the dual of $L^{\infty }(\Bbb R )$.
Let $B:=\{\chi _{[q,q+1]}:q\in \Bbb Q \}$ and note that $B$ is a linearly independent subset of $L^{\infty }(\Bbb R)$. Now define a linear functional in $\operatorname{span}(B)$ by setting $\varphi (\chi _{[q,q+1]}):=\delta _{q,0}$ where the RHS is a Kronecker $\delta $ function.
Then its easy to see that $\|\varphi \|=1$, therefore by the Hahn-Banach theorem there is some bounded linear functional $\tilde \varphi \in (L^{\infty }(\Bbb R))'$ such that $\|\tilde \varphi \|=\|\varphi \|$ and $\tilde \varphi|_{\operatorname{span}(B)}=\varphi $.
Suppose that there is some $f\in L^1(\Bbb R)$ such that $\int f\chi _{[q,q+1]}\,\mathrm d \lambda =\delta _{q,0}$ for all $q\in \Bbb Q $. Then by the dominated convergence theorem we find that $$ \lim_{n\to \infty }\int f\chi _{[1/n,1+1/n]}\,\mathrm d \lambda =\int f\chi _{(0,1]}\,\mathrm d \lambda =\int f\chi _{[0,1]}\,\mathrm d \lambda=1 $$ what imply that there is some $n\in \Bbb N $ such that $\int f\chi _{[1/n,1+1/n]}\,\mathrm d \lambda>0$, thus such $f$ cannot exists.$\Box$