# let f be a bounded function on [0,1] and integrable on $[\delta, 1 ]$, for every $0 < \delta < 1$. Prove that f is integrable

let f be a bounded function on [0,1] and integrable on $$[\delta, 1 ]$$, for every $$0 < \delta < 1$$. Prove that f is integrable.

Could anyone give me a hint for proving this?

EDIT

Will I use this corollary?

• Are we talking about Riemann or Lebesgue integration ? – nicomezi Nov 30 '18 at 6:35
• All kinds of answers can be given depending what results we can and we cannot use. – Kavi Rama Murthy Nov 30 '18 at 6:37
• we are taking about darboux integrable @nicomezi – hopefully Nov 30 '18 at 8:37
• we are taking about darboux integrable @KaviRamaMurthy – hopefully Nov 30 '18 at 8:38
• I am sorry for being unclear – hopefully Nov 30 '18 at 8:38

## 2 Answers

Within Riemann Integration Theory:

Let the bound of $$f$$ be $$B/2$$, i.e. $$|f| \leqslant B/2$$ on $$[0,1]$$. Given $$\varepsilon >0$$, choose $$\delta \in (0, \varepsilon / (2B))$$. Since $$f$$ is integrable on $$[\delta, 1]$$, there is a partition $$P$$ of $$[\delta, 1]$$ s.t. $$U(f,P,[\delta, 1]) - L(f, P, [\delta, 1]) <\epsilon /2$$. Then $$P' = P \cup \{0\}$$ is a partition of $$[0,1]$$. On $$[0, \delta]$$, since $$\vert f \vert \leqslant B/2$$, $$-B/2 \leqslant f \leqslant B/2$$, so $$\sup_{[0, \delta]} - \inf _{[0, \delta]} \leqslant B/2 - (-B/2) = B$$. Thus $$U(f,P') - L(f, P') \leqslant (\delta - 0)(\sup_{[0, \delta]} f - \inf_{[0, \delta]} f) + U(f, P, [\delta, 1]) - L(f, P, [\delta, 1]) \leqslant B\delta + \frac \varepsilon 2 < B \cdot \frac \varepsilon {2B} + \frac \varepsilon 2 = \varepsilon.$$ Hence $$f$$ is integrable on $$[0,1]$$.

Define $$h_n ( u) =|f(u)|$$ for $$u\in (n^{-1} ,1)$$ and $$h_n (u) =0$$ otherwise. Then the sequence $$h_n$$ is a sequence of measurable functions and hence it's limit which is equal to $$|f|$$ is measurable. Now since $$|f|$$ is bounded measurable function defined on finite measure set $$[0,1]$$ it is integrable on this set but this implies a integrability of the function $$f.$$

• Without measure theory just advanced calculus course – hopefully Nov 30 '18 at 8:53