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 $[0,1]$.
Thank you in advance.
 A: 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).
A: 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 
$$\int_\mathbb{R} g \leq \sum_{n=1}^\infty 2^n |I_n| =1.$$
