Property of Lebesgue measure in $mathbb{R}^2$

Yes, it is true. Proving that $overline{Isetminus E}subseteq I$ is trivial.

For proving $Isubseteqoverline{Isetminus E}$ let $(x,y)in I$ and assume that $(x,y)notinoverline{Isetminus E}$.

Then some open set $U$ must exist with $(x,y)in U$ and $Ucap(Isetminus E)=varnothing$ or equivalently $Ucap Isubseteq E$.

But $Ucap I$ has positive Lebesgue measure.

So this contradicts that $E$ is a set with Lebesgue measure zero and we conclude that our assumption must be wrong.

That means that $(x,y)in I$ implies that $(x,y)inoverline{Isetminus E}$ and we are ready.

If the complement of $E$ is not dense in $I$, then $E$ contains some open rectangle, so it cannot be of measure zero.