Does (Riemann) integrability of a function on an interval imply its integrability on every subinterval? For example, if $f$ is integrable on $[0,3]$, is it also integrable on $[1,2]$? I tried thinking of a counterexample but couldn't, since I've only learned what implies integrability but not what integrability implies.
First note that if $a \leq a'$ and $b' \leq b$ and $g: [a,b] \to \mathbb R$ is $0$ outside of $[a',b']$, then $\int_a^b g(x)\, dx = \int_{a'}^{b'} g(x)\,dx$. This is easily shown using the definition of the Darboux (or Riemann, if you prefer that definition) integral using partitions (tagged partitions for the Riemann definition).

The product of Riemann&ndash;Darboux integrable functions are Riemann&ndash;Darboux integrable, so in particular you have (in your concrete example) that $\int_1^2 f(x)\, dx = \int_0^3 f(x) \chi_{[1,2]}(x)\, dx$, where $\chi_{[1,2]}$ denotes the characteristic function of the interval $[1,2]$, i.e. 

$\displaystyle\qquad \chi_A(x) = \begin{cases} 1 &amp; \text{if } x \in A \\ 0 &amp; \text{if } x \notin A\end{cases}$
If $f$ is integrable on $[0,3]$ then there is a $F:[0,3]\to \Bbb R$ :$F'=f$ on  $[0,3]$.The restriction of $F$ on $[1,2]$ let's name it $G$ means that $G'=f$ on $[1,2]$. So if $f$ is integrable on $[0,3]$ is integrable on every connected subset(&lt;=>here is equal to every subinterval)
 of $[0,3]$