Consider $\mathbb{R}=\bigcup_i [-i,+i]$.
1st ingredient:
For all $i\geq 1$, find continuous $g_i$ supported on $[-i,+i]$ s.t. $\int_{[-i,+i]} \left|f-g_i \right| < \frac{1}{i}$.
This is possible thanks to the proof in the link you provided.
2nd ingredient:
Now since $f:\mathbb{R}\rightarrow\mathbb{R}$ is Lebesgue-integrable, $\int_{\mathbb{R}-[-i,+i]}|f|\rightarrow0$ as $i\rightarrow\infty$.
Can you conclude the argument using those ingredients?
You showed that $l^*(E) = \inf l^*(\mathcal{O})$ (the infimum runs over all open sets $\mathcal{O}$ containing $E$).
Suppose $E$ is of finite exterior measure -then for any $n\in \mathbb{N}$ we have an open set $\mathcal{O_n}$ such that
$$l^*(\mathcal{O_n}) \leq l^*(E)+n^{-1}<\infty$$
And since $l^*(\mathcal{O_n}) = l^*(E)+l^*(\mathcal{O_n} \setminus E)$ we have $l^*(\mathcal{O_n} \setminus E) \leq n^{-1}$ (The equality is infact a restatement of the measurability of $E$, see Equivalent Definition of Measurable set)
Taking $\mathcal{O} = \bigcap_{n=1}^\infty \mathcal{O_n}$ we have come up with a $G_\delta$ set (countable intersection of open sets) which is obviously a borel set, with the property: $$l^*(\mathcal{O}\setminus E) \leq l^* (\mathcal{O_n}\setminus E) \leq n^{-1}$$ for all $n\in\mathbb{N}$ which is the first corollary.
If $E$ is of infinte exterior measure then we denote $E_n = E \cap B(0,n)$ ($E$'s intersection with the ball of radius $n$ centered at the origin). Each $E_n$ is bounded and so has finite exterior measure (Since it obviously is encompassed within $B(0,n)$) And we may extract $G_n$ sets such that $$l^*(G_n \setminus E_n)=0$$ Putting $G=\bigcup_{n=1}^\infty G_n$ we have $$l^*(G\setminus E) = l^*\left(\bigcup_{n=1}^\infty G_n \setminus \bigcup_{n=1}^\infty E_n \right)\leq \sum_{n=1}^\infty l^*(G_n \setminus E_n ) =0$$
This shows the first corollary.
For the second: For every measurable $E^C$ we obviously have a borel set $G\supset E^C$ such that $l^*(G\setminus E^C) =0$ (Take the set from the first claim) that shows that $G\setminus E^C$ is a lebesgue measurable set as it is a null set.
Then obviously $E^C= G \setminus (G\setminus E^C)= G \cap (G\cap E)^C$. Taking completements on both sides we have:$$ E = G^C \cup (G\setminus E^C)$$
$G^C$ is again a borel set (As a completement of borel set).
As a final comment on this excercise: One notices we used 2 main properties of the Lebesgue measure:
- It is outer regular meaning that for every measurable subset $E$ , for every $\varepsilon>0$ we have an open set $\mathcal{O}\supset E$ with the property $l^*(\mathcal{O} \setminus E) \leq \varepsilon$ which we had to prove.
- It is defined on borel sets, bounded sets have finite exterior-measure.
It turns out that the second implies the first and the first property is what we actually needed. In this proof I made a slight detour (as i wasn't sure which Lebesgue measurability criterion you are using).
Best Answer
You know from Show that there exists a continuous function $f$ such that $\int |\chi_A-f| d\lambda\lt \epsilon$ that for every $\epsilon \gt0$ and every indicator function $\chi_A$ there exists a continuous function $f$ such that $$||\chi_A-f||_1\lt \varepsilon.$$
Step 1: Assume $g$ is a simple function, i.e. it can be written as a linear combination of indicator functions: $$ g = \sum_{i=1}^n \chi_{A_i} .$$ From the above result, we have that for each $i=1,\dots,n$ exists a continuous function $f_i$ with $$||\chi_{A_i}-f_i||_1 \lt \frac{\varepsilon}{n}.$$ Thus, the function $f:=\sum_{i=1}^n f_i$ is continuous and we have $$ || g - f ||_1 \leq \sum_{i=1}^n || \chi_{A_i} - f_i || < \varepsilon.$$ In short: For every simple function $g$ and every $\varepsilon >0$ exists a continous function $f$ with $$ || g - f ||_1 < \varepsilon.$$
Step 2: If $ g $ is integrable, then there exists a sequence of simple functions $g_n$ with $||g_n - g||_1 \to 0$.
From step 1 we have that for any $n \in \mathbb{N}$ a continuous function $f_n$ exists with $|| f_n - g_n || < \frac{1}{n}$. Thus $$ ||f_n -g ||_1 \leq ||f_n - g_n ||_1 + ||g_n -g ||_1 < \frac{1}{n} + ||g_n -g ||_1 \to 0.$$
In short: For every integrable function $g$ and every $\varepsilon >0$ exists a continuous function $f$ with $$ || g - f ||_1 < \varepsilon.$$