I saw this claim:
if $v(E)$ = 0, then E is a null-set. The converse statement is wrong.
where $v(E)$ is the Jordan measure of $E$,
I am not sure if this right, because I know $\int_E1_E=0 $ for $E$ null set and for any bounded function.
2026-04-02 21:37:35.1775165855
The volume of null-set
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