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?


Will I use this corollary?

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  • $\begingroup$ Are we talking about Riemann or Lebesgue integration ? $\endgroup$ – nicomezi Nov 30 '18 at 6:35
  • 1
    $\begingroup$ All kinds of answers can be given depending what results we can and we cannot use. $\endgroup$ – Kavi Rama Murthy Nov 30 '18 at 6:37
  • $\begingroup$ we are taking about darboux integrable @nicomezi $\endgroup$ – hopefully Nov 30 '18 at 8:37
  • $\begingroup$ we are taking about darboux integrable @KaviRamaMurthy $\endgroup$ – hopefully Nov 30 '18 at 8:38
  • $\begingroup$ I am sorry for being unclear $\endgroup$ – hopefully Nov 30 '18 at 8:38

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 $f.$

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

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