# $\limsup$ bounded almost everywhere

Consider $z \in \mathbb{R}^n$ and a sequence $\{ z_i \}_{i=1}^{\infty}$ such that $z_i \rightarrow z$.

Let $\phi: \mathbb{R}^n \times X \rightarrow \mathbb{R}_{\geq 0}$. $X$ is unbounded.

I wonder if the proposition

$$\limsup_{i \rightarrow \infty} \phi(z_i, x) \text{ bounded almost everywhere on } X$$

is equivalent to

$$\exists \delta > 0 \text{ such that: } \sup_{\epsilon \in \delta \mathbb{B}} \phi(z + \epsilon, x) \text{ bounded almost everywhere on } X$$

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Are there any conditions on $\phi$? –  in_wolframAlpha_we_trust Apr 2 '12 at 7:27
$\forall z \in \mathbb{R}^n$ $x \mapsto \phi(z,x)$ is locally bounded and measurable; also $\forall x \in X$ $z \mapsto \phi(z,x)$ is locally bounded. –  Adam Apr 2 '12 at 15:17