# Hausdorff outer measure counterexample

Let $A$ be a $k$-dimensional rectangle in $\mathbb{R}^k$.

Then $\displaystyle H_p(A)=\sup_{n\in N}H_{p,1/n}(A) \geq \inf_{n \in N}H_{p,1/n}(A)$

How can I find an example (A) such that $H_p(A) > \inf_{n \in N}H_{p,1/n}(A)$

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I changed "sup" and "inf" in this posting to "\sup" and "\inf", and set one line in "displaystyle". When either in "display" mode or with "displaystyle", this causes subscripts to appear directly under "sup" and "inf", thus: $\displaystyle\sup_{n\in N}$. It also prevents "sup" and "inf" from being italicized (as if they were each a product of three variables) and results in proper spacing before and after "sup" and "inf". And it is conventionally correct usage. –  Michael Hardy May 10 '12 at 12:34

In $\mathbb R^1$, consider rectangle $A = [0,1]$ and $p = 1/2$. Then $H_p(A) = \infty$ but $H_{(p,1)}(A)=1$.
It is only a few very special cases when $H_{(p,\epsilon)}(A)$ does not vary with $\epsilon$. Case $p=1$ in $\mathbb R^1$, for example.