# A function that is bounded and measurable but not Lebesgue integrable

Could you give me concrete examples about

"a function that is bounded and measurable but not Lebesgue integrable".

Royden's textbook "Real analysis" says a bounded measurable function is said to be integrable if its lower Lebesgue integrale is equal to its upper Lebesgue integral.

(I know if the domain is of finite measure, then a bounded function is Lebesgue integrable iff it is measurable, so my desired example need to be on a domain of infinite measure.)

• Hmm... if the domain is of finite measure, then a function is Lebesgue integrable iff it is measurable? Take $g(x) = \dfrac{1}{x}$ on the domain $(0, 1]$. $g$ is continuous on this domain and thus measurable. Also this domain has finite measure. But $\int \limits_{(0,1]} |g(x)| \,dm = \infty$, right? So $g$ isn't integrable and seems to contradict your claim. What's wrong with my example? – layman Jun 21 '15 at 14:19
• @user46944 Presumably the word "bounded" should have been in the parenthetical as well. – Ian Jun 21 '15 at 14:23
• @Ian Good point, thanks for the comment. – layman Jun 21 '15 at 14:25
• @Ian: thank you. I just correct the proposition in the parenthesis. – Thang Jun 21 '15 at 14:27

Let $f : \Bbb R \to \Bbb R$ be defined as:

$$f(x) = \begin{cases} 1 & x \in [0, \infty) \\ 0 & \text{else} \end{cases}.$$ Clearly, $f$ is measurable since $f = \chi_{[0, \infty)}$ (and $[0, \infty)$ is a Lebesgue measurable set, so its characteristic function is measurable).

Also clearly $f$ is bounded. But $\int \limits_{\Bbb R} |f| \,dm = \infty$.

• Unfortunately there is some issue with terminology; sometimes "integrable" means "the absolute value of the function has finite integral", or in other words "the function is in $L^1$". Other times it means that at least one of the positive and negative parts of the function have finite integral, in which case the integral can be defined but might be infinite. It would be good to have the definition spelled out. – Ian Jun 21 '15 at 14:22
• @Ian Ah, I see. So the OP could have meant Lebesgue integrable in the way you describe. You're right, it should be made clear. I usually think of finite integral of absolute value when I think of integrable. – layman Jun 21 '15 at 14:24
• I've never actually seen the other definition of "integrable". Which is not to say that Ian lied, of course, there are many things I haven't seen. In the expositions I know "integrable" means the integral of the absolute value is finite, and the funny bit is not any actual ambiguity in the formal definitions, the funny bit is just that "has an integral" does not imply "integrable". – David C. Ullrich Jun 21 '15 at 17:34
• @DavidC.Ullrich Yeah, I know what you mean. Actually, if I remember correctly, in my analysis class we defined for a non-negative measurable function that it is integrable if its integral is finite. But then I think we defined for an arbitrary measurable function that it is integrable if the integral of at least one of its positive or negative parts is finite. Talk about confusing terminology. – layman Jun 21 '15 at 17:41
• @Svetoslav You've actually asked two different questions, I think inadvertently. Saying "$\int f<\infty$" is not the same as "$\int f$ is finite". Assuming $\int f<\infty$ does not imply $f$ is integrable; could be that $\int f^+<\infty$, $\int f^-=\infty$, so $\int f=-\infty<\infty$ and $\int |f|=\infty$. On the other hand if $\int f$ is finite then the integrals of $f^+$ and $f^-$ must both be finite, so $\int|f|<\infty$. – David C. Ullrich Jan 18 '16 at 13:47

This happens exactly when the integral of the positive part and the integral of the negative part are both infinite. One nice example is

$$\int_1^\infty \frac{\sin(x)}{x} dx$$

which exists in the improper Riemann sense and not in the Lebesgue sense. A more extreme example where this is easier to prove would be

$$\int_0^\infty \sin(x) dx.$$