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Post Made Community Wiki by Scott Morrison♦
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Here's another answer in a similar spirit to several othersgeneralization of Pete's answer. If $A$ and $B$ are open subsets of $\mathbb{R}^n$, show that some measure space with $A \subseteq B$ and $\mu(A) \ge \mu(B)$ where $\mu$ is Lebesgue measure. Of course many variations on this are possible, an obvious one being that if you drop the assumption of openness mu(B)$, then $A$ and $B$ are only equal up to a set of measure 0, which is of course good enough for many some purposes. |
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Here's another answer in a similar spirit to several others. If $A$ and $B$ are open subsets of $\mathbb{R}^n$, show that $A \subseteq B$ and $\mu(A) \ge \mu(B)$ where $\mu$ is Lebesgue measure. Of course many variations on this are possible, an obvious one being that if you drop the assumption of openness then $A$ and $B$ are only equal up to a set of measure 0, which is of course good enough for many purposes. |
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