# An Integrable function which is not essentially bounded on any subinterval of $[0,1]$

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

• That does not work: Your $f$ is zero almost everywhere, so it is essentially bounded. – user61527 Dec 29 '13 at 2:32
• Which kind of integral? – hmakholm left over Monica Dec 29 '13 at 2:35
• @T.Bongers Bad Overlook !!! You are right... So that "cannot" happen !! I deleted mine. – the8thone Dec 29 '13 at 2:38
• @HenningMakholm Lebesgue Integral – the8thone Dec 29 '13 at 2:39
• @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 – the8thone Dec 29 '13 at 2:47

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).

• Thanks ! it should be $\sup g(2^n|x-q_n|)$ , right ? – the8thone Dec 29 '13 at 2:52
• 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 ... – the8thone Dec 29 '13 at 3:02
• @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. – hmakholm left over Monica Dec 29 '13 at 3:03
• @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 $e^{-C}$. – hmakholm left over Monica Dec 29 '13 at 3:07
• Thank you so much ! although I have to think more to get it completely – the8thone Dec 29 '13 at 3:12

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 $$\int_\mathbb{R} g \leq \sum_{n=1}^\infty 2^n |I_n| =1.$$