Show that every Lebesgue integrable function can be approximated in norm and almost everywhere by a sequence of continuous functions.
I'm not sure where to even start with this. The question doesn't really make sense to me, so if someone could start by explaining that, I'd be very thankful.
Best Answer
Suppose $f \in L^1(\mathbb{R})$. For every $\epsilon > 0$ there exists a simple function $g=\sum_{n=1}^{N}a_n\chi_{E_n}$ such that $\|f-g\|_{L^1} < \epsilon/3$. I'll assume you can prove that using a standard approximation.
So you can obtain a proof of the result that you want by showing that a characteristic function $\chi_{E}$ of a set $E$ of finite measure can be approximated to any desired accuracy by a continuous function.
Let $E$ be a set of finite Lebesgue measure, and let $\epsilon > 0$ be given. Then there is a finite or countable collection of open intervals $I_n=(a_n,b_n)$ such that $E\subset\bigcup_n I_n$ and $$ 0 < m|E| < \sum_{n}m|I_n| < m|E| +\epsilon/3. $$ Choose $\epsilon_n$ such that $\sum_n \epsilon_n < \epsilon/3$, and define a continuous function $f_n$ to be $1$ on $[a_n,b_n]$, to be $0$ on $\mathbb{R}\setminus[a_n-\epsilon_n,b_n+\epsilon_n]$ and to be linear in between. Then $$ \int|f_n-\chi_{[a_n,b_n]}|dx = \int f_n-\chi_{[a_n,b_n]} \le \epsilon_n. $$ Therefore, $$ 0 \le \chi_{E} \le \sum_n \chi_{[a_n,b_n]} \le \sum_n f_n, $$ and \begin{align} \int |\sum_n f_n - \chi_{E}|dx & = \int \sum_n f_n-\chi_{E} dx \\ & = \int \sum_n (f_n-\chi_{[a_n,b_n]})+(\sum_{n}\chi_{[a_n,b_n]}-\chi_{E})dx \\ & = \sum_n \int (f_n-\chi_{[a_n,b_n]})+\sum_n m|I_n|-m|E \\ & \le \sum_n \epsilon_n + \frac{\epsilon}{3} < \frac{2\epsilon}{3}. \end{align} Using the above, you can then truncate the sum $\sum_n f_n$ to a finite sum $\sum_{n=1}^{N}f_n$ such that $\|\sum_{n=1}^{N}f_n - \chi_{E}\| < \epsilon$. The sum $F=\sum_{n=1}^{N}f_n$ is continuous, which gives the existence of a continuous function $F$ such that $\|F-\chi_{E}\|< \epsilon$ if $E$ is a set of finite measure, which completes the proof.