10
$\begingroup$

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.)

$\endgroup$
4
  • 2
    $\begingroup$ 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? $\endgroup$
    – layman
    Jun 21, 2015 at 14:19
  • 1
    $\begingroup$ @user46944 Presumably the word "bounded" should have been in the parenthetical as well. $\endgroup$
    – Ian
    Jun 21, 2015 at 14:23
  • 1
    $\begingroup$ @Ian Good point, thanks for the comment. $\endgroup$
    – layman
    Jun 21, 2015 at 14:25
  • $\begingroup$ @Ian: thank you. I just correct the proposition in the parenthesis. $\endgroup$
    – Thang
    Jun 21, 2015 at 14:27

3 Answers 3

Reset to default
10
$\begingroup$

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 $\int \limits_{\Bbb R} |f| \,dm = \infty$.

$\endgroup$
9
  • 4
    $\begingroup$ 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. $\endgroup$
    – Ian
    Jun 21, 2015 at 14:22
  • 1
    $\begingroup$ @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. $\endgroup$
    – layman
    Jun 21, 2015 at 14:24
  • $\begingroup$ 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". $\endgroup$
    – David C. Ullrich
    Jun 21, 2015 at 17:34
  • 1
    $\begingroup$ @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 $\int|f|<\infty$. $\endgroup$
    – David C. Ullrich
    Jan 18, 2016 at 13:47
  • 1
    $\begingroup$ @DavidC.Ullrich , you are right, I have written it sloppily. By $\int{f}<\infty$ I meant $\int{f}$ is finite. So, you say, if we know that $\int{f}$ is finite (computed by $\int{f}:=\int{f^+}-\int{f^-}$ ) then necessarily $\int{|f|}<\infty$. This is because necessarily $\int{f^+},\int{f^-}$ are both finite. $\endgroup$
    – Svetoslav
    Jan 18, 2016 at 14:37
9
$\begingroup$

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

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

$\endgroup$
0
$\begingroup$

Analytic - Lebesgue Integrable:

Any non-zero polynomial $p(x)=\sum_{i=0}^n r_i x^i$ where $r_i\in \mathbb{R}$, is not only measurable, but it is analytic! However, by Jensen's inequiality and the continuity of (Lebesgue) measure we have that $$ \int_{x \in \mathbb{R}} |p(x)| dx \geq \left| \int_{x \in \mathbb{R}} p(x) dx \right| \geq \left| \int_{x \in [0,\infty)} p(x) dx \right| \geq \lim_{t \uparrow \infty} \left| \int_{x \in [0,t]} p(x) dx \right| \geq \lim_{x \mapsto \infty} \left| \sum_{i=0}^{n} \frac{r_i t^{i+1}}{i+1} \right| =\infty, $$ where the right-hand equality holds since $\min_{i=0,\dots,n}t{|r_i|}>0$ by hypotesis of $p$ being non-zero.

Analytic + Bounded - Lebesgue Integrable:

If you want bounded, then simply consider the case where $r_0\neq 0$ and $r_i=0$.

Bonus: Class $C^k$ + Bounded - Lebesgue Integrable:

Since every analytic function is $k$-dimension continuisouly differentiable then for every $k\in \mathbb{N}$ there exists a $k$-times differentiable function which is not Lebesgue integrable. In particular, since every differentiable function is continuous and continuous function is measurable, then we get what you asked for (+bonuses).

Bonus II: Excessively many examples

Let $g$ be such that, $M\leq |g(x)|\geq \delta>0$. Then $$ \int_{x \in \mathbb{R}} |g(x)p(x)| dx \geq \delta \int_{x \in \mathbb{R}} |p(x)| dx = \infty. $$

Note: By arguing componentwise you can extend this to any Bochner space between separable Hilbert spaces (since there is only one up to isometric linear isomorphism).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .