A function that is measurable but not Lebesgue integrable.

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...

• Could you expand a bit on exactly what you want? The Lebesgue integral only makes sense to define for a measurable function. – JSchlather Mar 14 '13 at 17:41
• It's the other direction: a function can be measurable without being Lebesgue integrable. – Qiaochu Yuan Mar 14 '13 at 17:41
• Try $f(x) = 1/x$ on $E = (-1,0) \cup (0,1)$ with Lebesgue measure. – Nate Eldredge Mar 14 '13 at 18:20
• @NateEldredge Why include $(-1,0)$? Just curious. – Julien Mar 14 '13 at 20:30
• @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. – Nate Eldredge Mar 14 '13 at 23:05

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 $1/x$.
• 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? – Glassjawed Nov 22 '14 at 6:57