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If $f(x)$ is Riemann integrable on $[a, b]$ and [$\forall x \in [a, b] \ f(x) \ne 0 \land \frac{1}{f(x)}$ is bounded]
then prove $\frac{1}{f(x)}$ is Riemann integrable on $[a, b]$.

I saw Prove the reciprocal of a function is integrable if it is bounded but I don't understand if: \begin{eqnarray} f(x) = \begin{cases} \frac{1}{2} & ( x = \frac{1}{4n} ) \\ 1 & ( x =\frac{1}{4n+1} ) \\ -\frac{1}{2} & ( x =\frac{1}{4n+2} ) \\ -1 & ( x =\frac{1}{4n+3} ) \\ 1 & ( other ) \end{cases} ( n \in \mathbb{N} ) \end{eqnarray} f(x) is integrable on $[0, 1]$.
Let $P=\{0=x_0,x_1,...,x_n=1\}$
Let $M_i,m_i,G_i,g_i$ same to above page.
then $M_1=1,m_1=-1,G_1=2,g_1=-2$.
Why does not satisfy [$G_1-g_1 = \frac{M_1 - m_1}{|M_1 m_1|}$]?


I have proved this probrem in a different way. Thanks everyone!

  • This equation generally fails, but you could reach this by using finer partitions. In your example, you may take a finer partition to avoid the situation that the sup of $f$ and the inf of $g$ [or the conjugate pair] are reached at different $x$. e.g. take $x_1 = 2/(8n+3)$ and $x_2 = 1/(4n-1)$. His method is essentially correct. – xbh Dec 22 '17 at 06:00
  • I have prove integrability in a different way. Thank you. –  Dec 22 '17 at 08:50

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