I claim that the function $g$ is continuous at every point where $g(x) = \infty$. Here is one way to see this. Recall that a function $h:\mathbb{R}\to {\mathbb R}\cup\{-\infty,\infty\}$ is lower semicontinuous provided
$$
h(x) \le \liminf_{y \to x} h(y)
$$
for every $x\in \mathbb R$. Put $h_n(x) = 2^{-n}f(x-r_n)$. Then it is easy to check that $h_n$ is lower semicontinuous.
Now it can be shown that any sum of non-negative lower semicontinuous functions is lower semicontinuous. Thus $g$ is lower semicontinuous and so
$$
g(x) \le \liminf_{y\to x} g(y)
$$
for all $x\in \mathbb R$. Now suppose that $g(x) = \infty$. Then we automatically get
$$
\limsup_{y\to x}g(y) \le g(x).
$$
Hence we have
$$
\limsup_{y\to x} g(y) \le g(x) \le \liminf_{y\to x} g(y)
$$
and so $g$ is continuous at $x$.
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
Hint: For any interval, $(a,b)$, and any integer $M$, show that $\{x:g(x)>M\}\cap (a,b)$ has nonzero measure. Removing a null set does not change the measure of this set; in particular it will still be nonempty.
It will help to find rational number $r_n$ which lies in $(a,b)$ and consider the "spike" $2^{-n}f(x-r_n)$. As long as the spike is larger than $M$, than $g$ will be as well. You should be able to find a nontrivial interval where the spike is large.