# Show that if $X \subset [a,b]$ is a set that does not have a null measure, then there exists $\epsilon \gt 0$....

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.

• 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.
– YCB
Commented Jun 23, 2021 at 14:23