I would like to show the following statement:
For each $g\in L_1(\mathbf{R}^n,m)$ and each $\epsilon>0$, there exists a continuous function $f:\mathbf{R}^n\to\mathbf{R}$ such that $$ \int |f-g|\,dm <\epsilon. $$ I have proven the following statement which was told to be useful:
For each bounded Lebesgue measurable $A\subset\mathbf{R}^n$ and each $\epsilon>0$, there exists a continuous function $f_A:\mathbf{R}^n \to [0,1]$ with compact support such that $$ \int |f_A-\chi_A|\,dm <\epsilon. $$
Attempted Proof for the first statement: Let $g \in L_1(\mathbf{R}^n, m)$, $\epsilon > 0$ be given. In particular, there exists some $k \in \mathbf{R}$ such that $$ g = g|_{[-k, k]^n} + g|_{\mathbf{R}^n - [-k, k]^n}, $$ and that $$ \int_{\mathbf{R}^n - [-k, k]^p} |g| \, dm < \frac{\epsilon}{2}. $$ Now note that there exists simple functions $$ s(x)= \sum_{i = 1} ^m a_i \chi_{A_i}(x) $$ such that $\| g|_{[-k, k]^p} - s \|_{L_1} < \epsilon/2$ with $A_i$ bounded. Now note that by the statement proven, for all $i$, there exists continuous functions $f_{A_i}: \mathbf{R}^n \to [0, 1]$ such that $$ \int |f_{A_i} - \chi_{A_i}| \, dm < \epsilon. $$ Therefore, we have \begin{align*} \int |g - \sum_{i = 1} ^m a_i f_{A_i}| \, dm &= \int | g|_{[-k, k]^n} - \sum_{i = 1} ^m a_i f_{A_i}| \, dm + \int |g|_{\mathbf{R}^n - [-k, k]^n}| \, dm \\ &\leq | g|_{[-k, k]^n} - \sum_{i = 1} ^m a_i \chi_{A_i}| \, dm + \int |\sum_{i = 1} ^m a_i f_{A_i} - \sum_{i = 1}^m a_i \chi_{A_i}| \, dm + \int |g|_{\mathbf{R}^n - [-k, k]^n}| \\ &< \frac{\epsilon}{2} + \sum_{i = 1} ^m |a_i| \int |f_{A_i} - \chi_{A_i}| \, dm + \frac{\epsilon}{2} \\ &< \epsilon + \sum_{i = 1} ^m |a_i| \epsilon \\ &= \epsilon(1 + \sum_{i = 1} ^m |a_i|). \end{align*} Since $\epsilon > 0$ is arbitrary, we are done.
My Doubts: I think the last sentence where we reached the conclusion is faulty. In particular, I think our $a_i$ in the last part of the inequality actually depends on $\epsilon > 0$. Therefore, it is not very clear how this gives the desired result. However, I am not sure how to fix this.
Once the simple function $f$ is fixed, you can choose $f_{A_i}$ such that $$ \lvert a_i\rvert \int |f_{A_i} - \chi_{A_i}| \, dm < \epsilon/m. $$
Alternatively, you can define $g_N(x)=g(x)\mathbf{1}_{\lvert g(x)\rvert\leqslant N}$, take $N$ such that $\lVert g-g_N\rVert_1<\varepsilon/2$ and then use the statement with $f_A=g_N$.