Consider the measurable space $(\Bbb{R},\mathcal{B},\lambda)$, where $\mathcal{B}$ is the Borel algebra in $\Bbb{R}$ and $\lambda$ is Lebesgue in $\mathcal{B}.$ Let $$\mathcal{F}=\left\{f:\Bbb{R}\rightarrow\Bbb{R}:f\,\textrm{is measurable and integrable with respect with }\lambda\right\}, $$ and consider the subspace $\mathcal{C}\subset\mathcal{F}$ given by
$$\mathcal{C}=\left\{f:\Bbb{R}\rightarrow\Bbb{R}: f\,\textrm{is continuous}\right\}. $$ I want to prove that $\mathcal{C}$ is dense in $\mathcal{F}$ in the norm $L_{1}$, that is, in the induced metric $d(f,g)=||f-g||_{1}$.
I know that there is some demonstrations of this (and similar) facts. For example, a simple demonstration using the Weierstrass approximation theorem. However, my context does not use this theorem, and I'm thinking in a strategy to solve this. Since I'm in a class of measure-theory, maybe I can do a common strategy seen in questions like this.
My ideia is to prove that all $f\in\mathcal{F}$ can be approximated by measurable integrable functions.
1) Do this for $f$ a characteristic function, $f=\chi_{E};$
2) Use 1) and linearity to prove it for simple functions $f$;
3) Use 2) and Monotone Convergence Theorem for a general $f$ measurable integrable.
The steps 2) and 3) are pretty simple, so I'm stucked in the step 1).
Intuitively, I can see that, if $f=1$ in an interval $E$ and $f=0$ outside $E$, I can build straight lines that tighten at the ends of the interval. But what if $E$ is an arbitrary measurable set?
If you know some fact about mollifier, then number one goes through: Consider $\varphi_{\epsilon}\ast\chi_{E}$, this is continuous compactly supported, and we have $\varphi_{\epsilon}\ast\chi_{E}\rightarrow\chi_{E}$ in $L^{1}$.
Of course, one should start with bounded $E$, for $\chi_{|x|\leq N}\chi_{E}\uparrow\chi_{E}$, the simple functions can be taken as linear combination of bounded sets.