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Here is my attempt of proving $cl(C[a,b])=L^p[a,b]$. I would be immensely thankful to anybody reading it, pointing out the mistakes or improving it. Since I just began studying functional analysis, it's important to me that my understanding is right. Thanks!


Let $\Sigma$ be Borel $\sigma$-Algebra of $[a,b]$.
We can assume $cl(span\{ \chi_A|A \in \Sigma\})=L^p$.

Claim 1. $span\{\chi_O|O \subseteq [a,b] \text{ open} \}$ is dense in $span\{ \chi_A|A \in \Sigma\}$.

Proof: Let $x=\sum_{i=1}^n a_i \chi_{A_i}\in span\{ \chi_A|A \in \Sigma\}, \epsilon>0$ be arbitrarily chosen.
Define $y:=\sum^n_{i=1}a_i \chi_{O_i}$ where $O_i$ are open sets such that $$ \lambda(O_i\setminus A_i)\leq \max_i\lambda(O_i\setminus A_i) < \frac{\epsilon}{n\cdot \max_i |a_i|}$$ This construction is possible since Lebesgue measure is outer regular. Indeed \begin{align} \Vert x-y \Vert_p^p&= \Vert \sum_{i=1}^n a_i (\chi_{A_i}- \chi_{O_i}) \Vert_p^p\\ &\leq \max_i |a_i| \sum_{i=1}^n \Vert \chi_{A_i}- \chi_{O_i} \Vert_p^p\\ &\leq n\cdot \max_i |a_i| \cdot \max_j \Vert \chi_{A_j}- \chi_{O_j} \Vert_p^p\\ &\leq n\cdot \max_i |a_i| \cdot \max_j \lambda(O_j\setminus A_j) <\epsilon \end{align} q.e.d.

Claim 2. $span\{\chi_O|O \subseteq [a,b] \text{ open} \}$ is dense in $L^p$.

Proof: By claim 1 and idempotency of closure we have
$$ cl(span\{\chi_O|O \subseteq [a,b] \text{ open} \})=cl(span\{ \chi_A|A \in \Sigma\})=L^p$$ q.e.d.

Claim 3. $span\{\chi_I| I\subset [a,b] \text{ open interval}\}$ is dense in $span\{\chi_O|O \subset [a,b] \text{ open} \}$ and hence in $L^p$.

Proof: Let $x=\sum_{i=1}^m b_i \chi_{O_i}\in span\{\chi_O|O \subset [a,b] \text{ open} \}, \epsilon>0$ arbitrary. Each $O_i$ can be written as the disjoint union of open intervals $I_{ij}$ with $j$ ranging over not necessarily finite index set $J_i$. Now for each $O_i$ we pick enough number $n(i)$ of those intervals such that $$\lambda(O_i \setminus \cup_{j=1}^{n(i)} I_{ij})\leq \max_{i} \lambda(O_i \setminus \cup_{j=1}^{n(i)} I_{ij})\leq \frac{\epsilon}{m\cdot \max_k |b_k|}$$ Define $y:=\sum_{i=1}^m a_i \sum^{n(i)}_{j=1} \chi_{I_{ij}} \in span\{\chi_I| I\subset [a,b] \text{ open interval}\}$, since it's finite linear combination. Furhermore

\begin{align} \Vert x-y \Vert_p^p&=\Vert \sum_{i=1}^m b_i (\chi_{O_i}-\sum^{n(i)}_{j=1} \chi_{I_{ij}}) \Vert_p^p\\ &\leq \max_k |b_k|\cdot m \cdot \max_{i} \lambda(O_i \setminus \cup_{j=1}^{n(i)} I_{ij})\leq \epsilon \end{align} q.e.d.

Claim 4. $C[a,b]$ is dense in $span\{\chi_I| I\subset [a,b] \text{ open interval}\}$ and hence in $L^p$.

Claim 4. $span\{\chi_I| I\subset [a,b] \text{ open interval}\} \subseteq cl(C[a,b])$ and hence dense in $L^p$.

Proof: Let $x=\sum_{i=1}^n c_i \chi_{I_i}$, $\epsilon>0$. We need to show $B_\epsilon(x)\cap C[a,b]$ is not empty.
For it we construct such function $f$ that tightly enough (in sense of $\Vert f-x \Vert_p^p <\epsilon$) and continuously encapsulates graph of $x$. (I am not sure how to explicitly define it.)

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    $\begingroup$ Claim 4 does not make sense. $C[a,b]$ is not contained in this span. Instead, you can show that the span is contained in the closure of $C[a,b]$. $\endgroup$ – tomasz Sep 3 '18 at 0:15
  • $\begingroup$ @tomasz You are right, I don't know what I was thinking... Thanks for taking a look! $\endgroup$ – user3342072 Sep 3 '18 at 1:02
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$C[a,b]$ is not dense in $L^{\infty}[a,b]$. So I am assuming $1 \le p < \infty$. If $C[a,b]$ is not dense in $L^{p}[a,b]$, then there exists $\Phi \in (L^{p})^*$ such that $\Phi=0$ on $L^{p}[a,b]$, but $\Phi \ne 0$. However, $\Phi(f)=\int_{a}^{b}f(t)\phi(t)dt$ for all $f\in L^p$ and some $\phi \in L^{q}=(L^{p})^*$. Because all characteristic functions $\chi_{E}$, where $E$ is a measurable subset of $[a,b]$, are in $L^p$, then $\int_{E}f(t)dt=0$ for all such $E$. From this it follows that $f=0$.

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