let $\Omega:=\left( a,b\right) \subset\mathbb{R}$, and suppose
$f:\Omega\rightarrow\left[ 0,\infty\right) $ is a bounded (Lebesgue)
measurable function, with $f\not \equiv 0$ almost everywhere. It is *not* true
in general that there exists some interval $I\subset\Omega$ such that
*essinf*$\_{I}f>0$. 


My question is: is the assertion true adding the hypothesis
$|($supp $f)\setminus\Omega\_{+}|=0$ ? Here, supp $f$ is the support of $f$,
$\Omega\_{+}$ is the largest open subset of $\Omega$ where $f>0$ almost
everywhere, and $\left\vert Z\right\vert $ denotes the Lebesgue measure of
$Z\subset\mathbb{R}$.


Thanks in advance.


Uriel



On the essential infimum (over subdomains) of nonnegative measurable functions
One can construct a measurable $f$ with $f > 0$ almost everywhere yet $\operatorname{essinf}\_I f = 0$ for every interval. In particular we have $\operatorname{supp} f = \Omega\_+ = (a,b)$.


Enumerate as $(q\_n, r\_n)$ the subintervals of $(a,b)$ having rational endpoints. For each $n$, let $I\_n$ be a subinterval of $(q\_n, r\_n)$ having length less than $1/2^n$. Set $f\_n = 1/n$ on $I\_n$ and $1$ elsewhere.


Now let $f = \inf\_n f\_n$ which is certainly measurable. Let $I$ be any subinterval of $(a,b)$. Then $I$ contains infinitely many different subintervals having rational endpoints (note I do not require them to be disjoint), so for any $N$ there is $n \ge N$ such that $I\_n \subset (q\_n, r\_n) \subset I$. As such, we have $f \le f\_n = 1/n \le 1/N$ on $I\_n \subset I$, where $I\_n$ has positive measure. Hence $\operatorname{essinf}\_I f \le 1/N$. But $N$ was arbitrary so $\operatorname{essinf}\_I f = 0$.


On the other hand, if $f(x) = 0$ then $x$ must have been in infinitely many of the intervals $I\_n$. Since $\sum m(I\_n) \le \sum 1/2^n < \infty$, the Borel-Cantelli lemma says that the set of such $x$ has measure zero. Thus $f > 0$ almost everywhere.


