Let $E$ be a bounded measurable set of $\mathbb{R}^n$. If every countinuous function $f:E\to\mathbb{R}$ is also uniformly continuous, then $E$ is compact.
Since I do not know topological property of $f(E)$, it seems that I must use some measure property to show $E$ is closed. However, measure and topology seems not quite relevant.
Thank you!
Suppose $x_n \in E, x_n \to x$ and $x \notin E$. Let $f(y)=\frac 1 {\|y-x\|}$. Then $f$ is continuous, hence uniformly continuous by hypothesis. Since $\|x_n-x_m\| \to 0$ it follows from uniform continuity that $\frac 1 {\|x_n-x\|}-\frac 1 {\|x_m-x\|} \to 0$. Since Cauchy sequences in $\mathbb R^{n}$ are bounded it follows that $\frac 1 {\|x_n-x\|}$ is bounded. But this sequence tends to $\infty$. This contradiction shows that $E$ is closed. Since it is bounded, it is compact.