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Show that if $X \subset [a,b]$ is a set that does not have a null measure, then there exists $\epsilon \gt 0$ such that for every partition $P$ of $[a, b]$, the sum of the lengths of $P$ intervals that contain $X$ points within it is greater than $\epsilon$.

my attempt

By definition we say that $X \subseteq \mathbb R$ has null measure if given $\epsilon \gt 0$ there is a countable collection $\{I_j\}_{j \in J}$ of open ranges such that $$X \subseteq \bigcup_{j \in J} I_j$$ $$\sum_{j \in J} \mid I_j \mid \lt \epsilon$$

If there is no null measure then either $$X \subseteq \bigcup_{j \in J} I_j$$ is false or $$\sum_{j \in J} \mid I_j \mid \lt \epsilon$$ is false or the two are false.

If $\sum_{j \in J} \mid I_j \mid \lt \epsilon$ is false then it's done

If $X \subseteq \bigcup_{j \in J} I_j$ is false then X is not contained in the union and therefore the sum of the union of the points of X are non-countable and greater than $\epsilon$.

I don't think my attempt was very good, so any help is welcome.

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  • $\begingroup$ Every partition of $[a,b]$ covers (by definition) the interval $[a,b]$. So, it also cover its subsets. Then your first part can't happend. $\endgroup$
    – YCB
    Commented Jun 23, 2021 at 14:23

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