If a function $f$ is integrable (in the Riemann sense) on a closed interval $I$, is it true that $f$ is integrable on every closed subinterval of $I$?
Intuitively, if there was a subinterval such that $f$ was not integrable over it, using the $\epsilon$ criterion for integrability on that interval we get the Riemann sum is $> \epsilon$ for that interval, so the Riemann sum on the whole interval is $> \epsilon$ which would yield a contradiction.
You can argue directly from the necessary and sufficient condition for Riemann integrability. If it is integrable on $I$ its set of discontinuities has measure zero. Therefore so does the set of discontinuities in the subinterval. Hence, it will be Riemann integrable there too.
Yes. Take $\varepsilon>0$. There is a partition $P$ of $I$ such that$$\overline{\Sigma}(f,P)-\underline\Sigma(f,P)<\varepsilon.\tag1$$If $[a,b]\subset P$ and if $P'=\{a,b\}\cup(P\cap[a,b])$, then $\overline{\Sigma}(f|_{[a,b]},P')\leqslant\overline{\Sigma}(f,P)$ and $\underline{\Sigma}(f|_{[a,b]},P')\geqslant\underline{\Sigma}(f,P)$. It follows then from $(1)$ that$$\overline{\Sigma}(f|_{[a,b]},P')-\underline{\Sigma}(f|_{[a,b]},P')<\varepsilon.$$
