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I am trying to check the converses of a few theorems.

I know that that if $g$ is integrable then $|g|$ is integrable. However, if $|g|$ is Riemann Integrable, then is $g$ Rieman integrable?

I know that if $g$ is integrable then $g^2$ is integrable. However, is the converse true?

I have a hunch that they aren't true, but am failing to device the counter examples.

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up vote 8 down vote accepted

Let $f(x)=1$ when $x$ is rational, $-1$ when $x$ is irrational. Interval say $[0,1]$.

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Notice that this example extends to Lebesgue integration by taking the characteristic of a non-measurable set. – JSchlather Jan 31 '13 at 2:06
@JacobSchlather: Thanks for the answer. By non-measurable set, do you mean uncountable sets? Also, I was wondering whether you had anything to say about the other question about : know that if $g$ is integrable then $g^2$ is integrable. However, is the converse true? If no, why? – user43901 Jan 31 '13 at 2:11
The example above takes care of your other question, since for the $f$ used in my answer, and Peter L. Clark's, has the property that $f^2=1$. So if you take that as $g$, then $g^2$ is Riemann integrable but $g$ is not. – André Nicolas Jan 31 '13 at 2:14
@user43901 That's not what I meant, although non-measurable sets are necessarily uncountable. If you haven't studied measure theory yet, I wouldn't worry about it. I was making a remark at the utility of the counterexample. – JSchlather Jan 31 '13 at 2:17
Non-measurable and uncountable are quite different. It so happens that every countable set is Lebesgue measurable. But plenty of uncountable ones are also, like any interval (open or closed), any countable union of intervals, many others. – André Nicolas Jan 31 '13 at 2:18

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