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How would I go about proving that if $fg$ is Riemann integrable, given that $g$ is continuous, nonzero, and bounded (so $g$ Riemann integrable), how would I go about showing that $f$ is Riemann integrable?

I was thinking about taking the upper and lower sums, but after I get there I am stuck. This question might be ill formed as well (and might not even be true - I just thought of the proposition).

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That is not true.

Take $f(x)=1/x$ and $g(x)=x$ and integrate it on $[0,1]$ for example

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    $\begingroup$ Note that if $g$ is also bounded away from $0$ (i.e. there is a positive $\epsilon > 0$ such that $|g(x)| >\epsilon$, then $f$ is integrable. This is because $1/g$ is bounded and continuous and therefore Riemann integrable, which makes $fg \cdot 1/g$ Riemann integrable. $\endgroup$ – Arthur Apr 28 '16 at 7:41
  • $\begingroup$ @Arthur Upvoted, you beat me to the answer! $\endgroup$ – yoyostein Apr 28 '16 at 7:44
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I think you may need another criterion for $g$ for it to be true: $g\geq c$ for some $c>0$, i.e. $g$ has a positive lower bound. Or more generally, as Arthur pointed out, $|g(x)|>\epsilon$ for some $\epsilon>0$.

Lebesgue's criterion: A function $f:[a,b]\to\mathbb{R}$ is Riemann integrable iff it is bounded and continuous almost everywhere.

So, $fg$ is bounded and continuous almost everywhere. $f=\frac{fg}{g}$ (well-defined since $g\geq c$ is nonzero) is also bounded and continuous almost everywhere.

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