I came across this proof for a similar question but I am trying to solve it differently so I'm making another post.
I aim to prove the following statement:
${\bigcap}_{\alpha}$ {$R_{\alpha}([a,b])$ $\alpha$ is non-decreasing} = $C([a,b])$
The proof technique I'm trying to use is C([a,b]) $\subseteq$ $R_{\alpha}([a,b])$ and $R_{\alpha}([a,b])$ $\subseteq$ C([a,b]) $\Longrightarrow$ C([a,b]) $=$ $R_{\alpha}([a,b])$. If this statement were to hold, then surely $\bigcap_{\alpha}$ $R_{\alpha}([a,b])$ = C([a,b]) would hold.
I have shown that C([a,b]) $\subseteq$ $R_{\alpha}([a,b])$ for any non-decreasing $\alpha$ however, I'm having difficulty showing the converse. Any help would be appreciated.
My proof is below:
Suppose $f$ $\in$ $R_{\alpha}([a,b])$. I aim to show that $f$ $\in$ C([a,b]) i.e $f$ is continuous on the interval [a,b]. Proceed by contradiction. Suppose $f$ $\notin$ C([a,b]). Therefore, $f$ is discontinuous at some point $c$ $\in$ $[a,b]$. From here I want to find a contradiction with the Riemann condition i.e I want to show $U_{\alpha}(f:P)$ $-$ $L_{\alpha}(f:P)$ > $0$. I have been trying to think about a specific non-decreasing ${\alpha}^{\ast}$ that this would hold. Also, if I were to find such an ${\alpha}^{\ast}$ would it effect my overall proof technique?
The specific $\alpha$ you need is defined by $\alpha (x)=0$ for $x <c$ and $1$ for $x\geq c$.