Problem: Suppose $f\in R[a,b]$ then for any $[c,d]\subseteq [a,b]$ we have that $f\in R[c,d]$.
Let $\epsilon>0$. As f is integrable, there exists a partition $D$ of $[a,b]$, given by
$D=$ $\{$ $a,x_1,x_2,x_3....,x_n=b$ $\}$
Consider $D'= D \cup$ $\{$ $c,d$ $\}$. Then $D'$ is a refinement of $D$. Let $D''=D'\cap [c,d]$. Then $D''$ is a partition for $[c,d]$ and is a subset of $D$, therefore $U(f,D'')-L(f,D'')\leq U(f,D')-L(f,D')\leq U(f,D)-L(f,D) < \epsilon$. The first inequality holds as $D''\subseteq D'$, a dissection of $[a,b]$ containing c and d.
Is the proof correct?
Is there a simpler proof?
You can simply argue by using Lebesgue's Integrability Criterion. If $f \in R[a,b]$, then $f$ is atleast continuous almost everywhere in $[a,b]$, so $f$ will be continuous atleast almost everywhere in any interval $[c,d] \subseteq [a,b]$. So clearly $f \in R[c,d]$.