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Let $A\subset\mathbb{}\mathbb{R}^N$ and $u$ a measurable function on $A$ bounded by above in every compact set contained in $A$. Suppose $x\in A$ and define:

$$\liminf_{y\rightarrow x}u(y)=\lim_{r\rightarrow 0}\inf\{u(y):\ y\in B(x,r)\cap (A\setminus\{x\})\}$$


$$ess\liminf_{y\rightarrow x}u(y)=\lim_{r\rightarrow 0}ess\inf_{B(x,r)}u$$

Is true that:

$$\liminf_{y\rightarrow x}u(y)\leq ess\liminf_{y\rightarrow x}u(y)$$

UPD: I think the answer is more easy than o thougth, because by definition $u(y)\geq \inf u$ almost everywhere, hence i\m gonna change the question. I just want to know a example where the inequality is strict.

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up vote 1 down vote accepted

Let $A=\mathbb{R}^n$ and let $N=\mathbb{Q}^n$. Then $N$ is measure zero and dense subset of $A$. Let $u=0$ on $N$ and $u=1$ on $A\setminus N$. Then for every nonempty open set $G$, $\inf_{y\in G} u(y)=0$ but $\mathrm{essinf}_{y\in G} u(y)=1$. Therefore, for any $x\in A$,

$$\liminf_{y\to x}u(y)=0<1=\mathrm{ess}\liminf_{y\to x}u(y).$$

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