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There is a theorem in our textbook that says, "Let $f$ be a bounded function on a set of finite measure $E$. Then $f$ is Lebesgue integrable over $E$ if and only if it is measurable."

So I was wondering about an example of a function that was Lebesgue integrable but not measurable. I tried to search for some examples online but couldn't really find anything useful...

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    $\begingroup$ Could you expand a bit on exactly what you want? The Lebesgue integral only makes sense to define for a measurable function. $\endgroup$ – JSchlather Mar 14 '13 at 17:41
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    $\begingroup$ It's the other direction: a function can be measurable without being Lebesgue integrable. $\endgroup$ – Qiaochu Yuan Mar 14 '13 at 17:41
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    $\begingroup$ Try $f(x) = 1/x$ on $E = (-1,0) \cup (0,1)$ with Lebesgue measure. $\endgroup$ – Nate Eldredge Mar 14 '13 at 18:20
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    $\begingroup$ @NateEldredge Why include $(-1,0)$? Just curious. $\endgroup$ – Julien Mar 14 '13 at 20:30
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    $\begingroup$ @julien: Some people might consider $1/x$ to be integrable on $(0,1)$, since the Lebesgue integral exists and has the value $+\infty$. The most common definition of "integrable" excludes this case, but I wanted something a little stronger. $\endgroup$ – Nate Eldredge Mar 14 '13 at 23:05
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The function $1/x$ on $\mathbb{R}$ (defined arbitrarily at $0$) is measurable but it is not Lebesgue integrable. In general, a function is Lebesgue integrable if and only if both the positive part and the negative part of the function has finite Lebesgue integral, which is not true for $1/x$.

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    $\begingroup$ How come the positive and negative parts of $1/x$ do not have finite Lebesgue integral? I thought you could only say that the Riemann integral is infinite. Is it just because the Lebesgue and Riemann integrals are the same whenever the Riemann one exists? $\endgroup$ – Glassjawed Nov 22 '14 at 6:57
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Following the construction of the lebesgue integral that is in the book "Real Analysis" (Royden), a function need to be measurable if it want to be lebesgue integrable.

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  • $\begingroup$ Yes, that was cleared out in the comments. Did you read them? $\endgroup$ – Pedro Tamaroff Aug 24 '13 at 18:44
  • $\begingroup$ oh i didn't, sorry. $\endgroup$ – fernanfio Aug 28 '13 at 5:21
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The Royden's book has this theorem because it doesn't assume the "measurability" when constructing Lebesgue integral and integrability. However, eventually measurability is still proven.

See the hidden motivation for this approach at: A function that is Lebesgue integrable but not measurable (not absurd obviously)

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