Is this statement true?

Any set $A \in \mathbb{R}^k$ with $\lambda^k(A) > 0$ contains a bounded set $K \subseteq A$ with $\lambda(K) > 0$, where $\lambda^k$ denotes the $k$-dimensional Lebesgue measure.


1 Answer 1


Let $B_n = A \cap [-n,n]^k$ where $A$ is measurable. Then $(B_n)$ is an increasing sequence of bounded measurable sets contained in $A$, and $A = \cup_{n\geq 0} B_n$, hence $\lambda^k(B_n) \to \lambda^k(A)$.

  • $\begingroup$ great answer thanks ! $\endgroup$ Feb 13, 2015 at 17:13

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