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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 am thinking:
Since $f(x)$ and $\frac{1}{f(x)}$ are bounded, $\exists M \gt 0 s.t. |f(x)| \lt M \land \frac{1}{|f(x)|} \lt M$
Therefore $\frac{1}{M} \leq |f(x)| \leq M$.
$\left| \frac{1}{f(x)}-\frac{1}{f(y)} \right|= \left| \frac{f(y)-f(x)}{f(x)f(y)} \right| \leq M^2 |f(y)-f(x)| \ \ (x, y \in [a, b])$
but I don't know what to do.


  • See this question: https://math.stackexchange.com/questions/2537952/proving-that-the-reciprocal-of-a-riemann-integrable-function-is-also-riemann-int – Christian Blatter Dec 21 '17 at 15:34
  • See also https://math.stackexchange.com/questions/1103545/prove-the-reciprocal-of-a-function-is-integrable-if-it-is-bounded – Robert Z Dec 21 '17 at 15:35

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