Let be $C_0(\mathbb{R})=\{f:\mathbb{R}\rightarrow \mathbb{K} \ | \ \forall \xi>0 \ \exists K_{\xi} \subset \mathbb{R} \ compact: \forall x \in \mathbb{R} \setminus K \ |f(x)|<\xi \}$ (or if you prefer $C_0(\mathbb{R})=\{f:\mathbb{R}\rightarrow \mathbb{K} \ | \lim_{|x|\rightarrow+\infty}|f(x)|=0\}$) and $L_1(\mathbb{R})=\{f:\mathbb{R}\rightarrow \mathbb{K}\ | \int_{\mathbb{R}}|f(x)|dx<+\infty\}$ where for $L_1(\mathbb{R})$ consider $f$ as an equivalence class $[f]=\{g:\mathbb{R}\rightarrow \mathbb{K} \ | \int_{\mathbb{R}}|f(x)-g(x)|dx =0\}$. Proove that $[C_0(\mathbb{R})]^*$ cannot be identified as $L_1(\mathbb{R})$, otherwise, prove that: $$\exists L\in [C_0(\mathbb{R})]^* \ \nexists f\in L_1(\mathbb{R}):Lg=\int_{\mathbb{R}}fgdx, \ \ g\in C_0(\mathbb{R})$$ $\mathbb{K}=\mathbb{R} \ or\ \mathbb{C}$
So as suggested from the exercise, I tried to prove that $Lg=g(0)$ has no $f\in L_1(\mathbb{R})$ as defined, and assuming as an absurd that this $f$ exists I prooved, by using the Lusin theorem on $f$ and the dominated convergence theorem that this would lead to: $$f(0)=\int_\mathbb{R}f^2dx$$ But I don't know if I could really know if I could use the dominated convergence theorem in this situation because the sequence of functions $f_n$ that the Luisin theorem gives me could not have an upperbound integrable function. The only function for sure that could be an upperbound integrable function for $f_n$ is $f$, but from the Lusin theorem I can conclude that: $$\sup_{x\in\mathbb{R}}|f_n|\leq\sup_{x\in\mathbb{R}}|f|$$ if $f$ is bounded, which I can't know for sure. So I'm stuck, any help?
But you know exactly what this $L$ is right? It is the Dirac measure centred at $0$, $\delta_0$. Its action on a continuous function cannot be identified with integration against some $f \in L^1(\mathbb{R})$ simply because $\delta_0$ is not absolutely continuous with respect to the Lebesgue measure.