4
$\begingroup$

I am looking for an Integrable function $f : [0,1] \to \mathbb{R}$ which is not essentially bounded on any subinterval of $[0,1]$.

Thank you in advance.

$\endgroup$
  • $\begingroup$ That does not work: Your $f$ is zero almost everywhere, so it is essentially bounded. $\endgroup$ – user61527 Dec 29 '13 at 2:32
  • $\begingroup$ Which kind of integral? $\endgroup$ – hmakholm left over Monica Dec 29 '13 at 2:35
  • $\begingroup$ @T.Bongers Bad Overlook !!! You are right... So that "cannot" happen !! I deleted mine. $\endgroup$ – the8thone Dec 29 '13 at 2:38
  • $\begingroup$ @HenningMakholm Lebesgue Integral $\endgroup$ – the8thone Dec 29 '13 at 2:39
  • $\begingroup$ @T.Bongers It seems impossible to find an example, being unbounded on a set of strictly positive measure (within any subinterval), and at the same time being integrable !!! but seemingly there is such a function $\endgroup$ – the8thone Dec 29 '13 at 2:47
6
$\begingroup$

Let $$ g(x) = \begin{cases} -\log x & x\in(0,1] \\ 0 & x \notin (0,1] \end{cases}$$ Then $\int g = 1$, and $g$ is not essentially bounded.

Let $q_1, q_2, q_3, \ldots$ be an enumeration of the rationals in $(0,1)$, and consider $$ f(x) = \sup_{n\in\mathbb N} g(2^n(x-q_n)) $$

Then $f(x)$ is nowhere essentially bounded, yet it must be integrable with integral at most $\sum_{n=1}^\infty 2^{-n} = 1$.

(It is imaginable that the supremum in the definition of $f$ doesn't exist at some points $x$, but that has to be a measure zero set, so just declare $f(x)$ to be $0$ there).

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks ! it should be $\sup g(2^n|x-q_n|)$ , right ? $\endgroup$ – the8thone Dec 29 '13 at 2:52
  • $\begingroup$ It seems to me by multiplying $|x-q_n|$ by $2^n$, you are pushing that number outside of the support of $g(x)$. How about multiplying by $2^{-n}$? What am I missing here ... $\endgroup$ – the8thone Dec 29 '13 at 3:02
  • 1
    $\begingroup$ @Roozbeh-unity (first comment): Doesn't matter. Even with $g(2^n(x-q_n))$, $f$ will have an interval of large values to the right of any rational. So to prove that $f$ is not essentially bounded in $(a,b)$ just choose some rational in the interior of the interval, and there will be sufficiently many high values to the right of it. $\endgroup$ – hmakholm left over Monica Dec 29 '13 at 3:03
  • 1
    $\begingroup$ @Roozbeh-unity: Multiplying by $2^n$ makes the peak narrower, which is essential for the total width (and area) of the peaks to have a finite sum. Otherwise $f$ wouldn't be integrable. The final $f$ consists of a lot of very narrow peaks -- for every $C$, $\{x\mid f(x)>C\}$ is dense in $(0,1)$, but has measure between $\frac12 e^{-C}$ and $e^{-C}$. $\endgroup$ – hmakholm left over Monica Dec 29 '13 at 3:07
  • $\begingroup$ Thank you so much ! although I have to think more to get it completely $\endgroup$ – the8thone Dec 29 '13 at 3:12
1
$\begingroup$

I will give a second solution. The key idea is Borel-Cantelli Lemma.

Let $\{q_n\}$ be the enumeration of rational numbers, and we work with $$I_n = [q_n, q_n+1/4^n).$$ Since $\sum_n |I_n| <\infty$, from Borel-Cantelli lemma, for almost every $x\in \mathbb{R}$, $x$ is only in finitely many $I_n$. Now define $$g(x) = \sup \{2^n : x\in I_n\},$$ this sup is taken from a finite index set for a.e. $x$, thus $g < \infty$ a.e.

To show $g$ is not essentially bounded, for each non empty open interval $(a,b)$, there is an infinite set $A\subset \mathbb{N}$ such that for $n\in A$ we have $I_n \subset (a,b)$. For each $M \in \mathbb{R}$, we can find $n^*\in A$ (because $A$ is infinite) such that $2^{n^*} \geq M$ this implies $g\geq 2^{n^*}\geq M$ on the set $I_{n^*} \subset (a,b)$.

To show $g$ is integrable, observe that $$\int_\mathbb{R} g \leq \sum_{n=1}^\infty 2^n |I_n| =1.$$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.