I'm working on a proof for the following lemma,
If $E \subset \mathbb{R}$ is a measurable set such that there exists $t \in (0,1)$ for which
$$ |E\cap I| \leq t|I| $$
for every interval $I \subseteq \mathbb{R}$, $E$ has measure zero.
My attempt is as follows: first off, we can generalize the inequality for arbitrary open sets since (in $\mathbb{R}$) they are countable union of disjoint intervals. If $G = \cup_{n\geq0}I_n$ is an open set,
$$ |E\cap G| = |E\cap \bigcup_{n\geq0}I_n| = |\bigcup_{n\geq0}E \cap I_n| \leq \sum_{n\geq0}|E\cap I_n| \leq t\sum_{n\geq0}|I_n| = t|G| $$
with this last equality given by the fact that the intervals are disjoint. Now, given $\epsilon > 0$, there exists $G$ open with $E \subseteq G$ and
$$ |G| = |G|_e < |E|_e + \epsilon = |E| + \epsilon $$
and therefore,
$$ |E| = |E\cap G| \leq t|G| < t|E| + t\epsilon $$
for each $\epsilon$. Taking limits on both sides when $\epsilon \to 0$, we get
$$ |E| \leq t|E| $$
which is only true if $|E|$ = 0, since if not then the previous inequality would imply $|E|<|E|$, completing the proof.
This seems like a sound argument to me, but I'm just beginning to study measure theory. Am I missing something?
As pointed out in the comments, the argument fails if $|E|$ is not of finite measure, but can be fixed as follows: for each $n\in\mathbb{N}$, we define $E_n := E \cap (-n,n)$ which is of finite measure. Now if $I$ is an interval and $n\in\mathbb{N}$,
$$ |E_n \cap I| \leq |E\cap I| \leq t|I| $$
and so by the original argument, $|E_n| = 0 \ (\forall n \in \mathbb{N})$. Finally, we prove the original claim by noting that
$$ 0 \leq |E| = |\bigcup_{n\geq0} E_n| \leq \sum_{n\geq0}|E_n| = 0 $$