I have been given the following problem:
Let $f$ be a bounded function on $[0,1]$. Suppose that $f$ is integrable on each closed interval $[a,b]$ of $(0,1)$. Prove that $f$ is intergrable on $[0,1]$.
Now, I was thinking of starting by defining $E=[0,1-\varepsilon]$, which would mean that $f$ is integrable over $E$. But I am not really sure where to go with this idea. Some guidance would be great!
This answer shows that $ f $ is Darboux integrable.
It suffices to show that for each $ \epsilon > 0$, there is a partition P of $ [0, 1] $ such that the upper and the lower Riemann sums (denoted by $ S(P) $ and $ s(P) $) differ by $ \epsilon $ at most.
Following your idea, since $ f $ is integrable on $ [0, 1 - \delta] $ for any $ 0 < \delta \leq 1 $, there is a partition $ Q $ of this interval such that $ S(Q) - s(Q) < \delta/2 $. Then, $ P = Q \cup \{1\} $ is the desired partition for an appropriate choice of $ \delta $, as we have that
$$ S(P) - s(P) = S(Q) + \delta \sup_{x\in [1 - \delta, 1]} f(x) - s(Q) - \delta \inf_{x\in [1 - \delta, 1]} f(x) $$
and now $ f $ is bounded, so that we have $ m \leq f(x) \leq M $ in the interval for some constants $ m, M $ and
$$ < \frac{\delta}{2} + (M - m) \delta = \frac{1 + 2M - 2m}{2} \delta$$
and picking $ \delta = 2\epsilon/(1 + 2M - 2m) $ gives that $ S(P) - s(P) < \epsilon $.