Well, going with the hint you could argue that since f(x) is continuous in $\displaystyle [\theta,1]$, for any $\displaystyle \theta>0$, then f is uniformly continuous there, so $\displaystyle \forall\,\, \epsilon > 0\,\,\exists\,\,\delta>0\,\,s.t.\,\,|x-y|<\delta\,\Longrightarrow |f(x)-f(x)|<\epsilon$ , and then: taking a partition P with n points of $\displaystyle [\theta,1]$ s.t. $\displaystyle \max \{x_i-x_{i-1} /\,[x_{i-1},x_i ]\in P\}<\frac{\delta}{n\epsilon}$, we get $\displaystyle U(f,P)-L(f,P)<\epsilon$.
Of course, we still need to deal with $\displaystyle [0,\theta)$, but this is not problem: the difference $\displaystyle \theta-0=\theta.$ can be made as little as wanted.
Now, not going with the hint is much simpler and shorter, but perhaps you guys haven't yet studied this: f is Riemann integrable there since it is bounded and the set of points of discontinuities of f in [0,1] (which is one single point) has (Borel-Lebesgue) measure zero.
Tonio