Assume that a non-negative function $f \in L^{\infty}(\Omega)$ for some open set $\Omega \subseteq \mathbb{R}^n$ is given. Does there exist an open set $\mathcal{O} \subseteq \Omega$ and a constant $K>0$ such that $f \geq K$ is valid almost everywhere on $\mathcal{O}$?
I think this is the case, but unfortunately have not yet found a good argument for it. So far I have two approaches:
Since $f$ is bounded, one can find a sequence of simple functions $f_n : \Omega \to [0,\infty)$ such that $f_n \leq f_{n+1}$ and $f_n \to f$ uniformly on $\Omega$ apply. In particular, it follows that $f_n \leq f$ holds. Suppose that every $f_n$ has the representation \begin{equation*} f_n = \sum_{k=0}^{N_n} c_k^n \chi_{E_k^n} \end{equation*} with $c_0^n = 0$. Here one can exclude sets $E_k^n \subseteq \Omega$ with Lebesgue measure $0$. If one can now show that there is an $n \in \mathbb{N}$ and a $k \in \{1, \dots, N_n\}$ such that $\mathcal{O} \subseteq E_k^n$ for an open set $\mathcal{O} \subseteq \Omega$, then one would have shown the assertion.
We choose a set $A \subseteq \Omega$ with $\lambda(A) > 0$ such that $f > 0$ holds everywhere on $A$. This can be achieved by using another representative of $f$. If we now define for each $n \in \mathbb{N}$ the set \begin{equation*} M_n = \left \{ x \in A \: \Big| \: 2^{-n-1} \| f \|_{L^{\infty}(\Omega)} < f(x) \leq 2^{-n} \| f \|_{L^{\infty}(\Omega)} \right\}, \end{equation*} at least one set must have Lebesgue measure greater than $0$. Now my idea would be to use the regularity of the Lebesgue measure to show that an open set $\mathcal{O}$ with $\mathcal{O} \subseteq M_n$ exists.
I am grateful for any hints.
