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I have been working on this problem for a while, but I cannot seem to get it right each time I try to attempt it. Can anyone help?

This is the problem:

Suppose $f:[a,b] \to R$ is Riemann integrable, and suppose $g = \frac{1}{f}$ is well defined and bounded on $[a,b]$. Prove that $g$ is Riemann integrable on $[a,b]$.

This is what I have worked on so far:

Let $\phi(x) = \frac{1}{x}$. Note that $\phi$ is continuous for all $x \neq 0$. Since $f \in R[a,b]$, and $f$ is bounded, then $\phi \circ f$. Let $g= \phi \circ f$. Then there exists $M > 0$ such that $|f(x)| \le M$ for all $x \in [a,b] \implies f(x) \in [-M,M]$. $g$ is bounded so there exists $m > 0$ such that $|g(x)| \le m$ for all $x \in [a,b]$ which results into $|\frac{1}{f}| \le m \implies |f(x)| \ge \frac{1}{m}$ for all $x \in [a,b]$. Let $S$ be the union of two compact intervals as $S=\{y \in R \mid \frac{1}{m} \le |y| \le M \}$. $\phi$ is continuous because $0 \not\in S$. So $\phi: S \to R$ is continuous with $S$ compact. Using this Theorem which states "Suppose $f:[a,b] \to S$, $f \in R(x)$ on $[a,b]$, and $\phi: S \to R$ is continuous with $S$ compact. Then $\phi \circ f \in R(x)$ on $[a,b]$", we know that the composition of a continuous function with a compact domain and an integrable function with be integrable.

From here on out, I am struggling to find/prove a lemma to conclude this proof with $\phi \circ f \in R[a,b]$. Any help will be appreciated.

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Try modifying the function $\phi$ so that it is continuous on $[-M,M]$ and still satisfies $\phi\circ f={1\over f}$. –  David Mitra Apr 5 '12 at 21:35

3 Answers 3

It suffices to prove that given $\epsilon > 0$, we can find a partition $P = \{t_{1}, ..., t_{n}\}$ of $[a, b]$ such that $\sum \omega'_{i}(t_{i} - t_{i - 1}) < \epsilon$, where $\omega'_{i} = \sup g - \inf g$, being taken over the interval $[t_{i - 1}, t_{i}]$. Denote by $\omega_{i} = \sup f - \inf f$ over the same interval. We know that there must be $k$ such that $0 < k < |f(x)|$. Take a parition $P$ such that $\sum \omega_{i}(t_{i} - t_{i - 1}) < \epsilon \cdot k^{2}$. For any $x, y$ in the ith interval we have

$\left| \displaystyle\frac{1}{f(y)} - \frac{1}{f(x)} \right| = \displaystyle \frac{|f(x) - f(y)|}{|f(y)f(x)|} \leq \frac{\omega_i}{k^{2}}$.

It follows that $\omega'_{i} \leq \frac{\omega_{i}}{k^{2}}$ and thus $\sum \omega'_{i}(t_{i} - t_{i - 1}) < \epsilon$; therefore $g$ is integrable.

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First a function is Riemann integrable iff it is bounded and continuous almost anywhere on the domain (proof). Now you only have to show the the composition $\phi \circ f$ is bounded and continuous almost everywhere and you showed that. You are done.

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Define $\phi:[-M,M]\rightarrow\Bbb R$ via

$$ \phi(x)=\cases{ 1/x, & $ |x|\ge m$ \cr x/m^2, & $ |x|<m$ }. $$

Then $\phi$ is continuous on $[-M,M]$ and, since $M\ge|f(x)|\ge m$ for all $x$, we have $\phi\circ f={1\over f}$. Now you may apply your Theorem.

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