Proving a particular set in $[0,1]$ has measure 1 I saw this question while looking through some real analysis PhD qualifier exams.

Let $E$ be a Lebesgue measurable subset of $[0,1]$. Let $c$ be a positive constant such that for all $0\leq a,b\leq1$, we have $$m(E \cap [a,b]) \geq c(b-a).$$ Prove that $m(E)=1$.

My approach is to show that the complement $F := [0,1] \backslash E$ has measure zero. By countable additivity of the Lebesgue measure, we have
$$m(E \cap [a,b])+m(F \cap [a,b]) = b-a \implies (1-c)(b-a) \geq m(F\cap [a,b]).$$
Let $\{x_k\}$ enumerate the rational numbers in $\mathbb{Q}$, and for a fixed $\epsilon > 0$ and each $k$ we consider the interval $$I_k = [x_k - \frac{\epsilon}{(1-c)2^{k+1}}, x_k+\frac{\epsilon}{(1-c)2^{k+1}}].$$
Then
$$m(F) = m(F \cap [0,1]) = \sum ^\infty _{k=1} m(F \cap I_k)\leq (1-c)\sum _{k=1}^\infty \frac{\epsilon}{(1-c)2^k} = \epsilon.$$
Here I made the assumption that the intervals $I_k$ would cover $[0,1]$, but I'm not sure if this is actually true.
EDIT. Okay the assumption is definitely not true, since the intervals are getting smaller and thus we cannot cover all irrational numbers in $[0,1]$. How do we work around it?
 A: This is an easy consequence of Lebesgue Density Theorem.
If $m(E) <1$ then there would be points of $E^{c}$ at which $E$ has Lebesgue density  $0$ but the hypothesis implies that the Lebesgue density is at least $c$.
A: Lebesgue differentiation theorem is powerful enough to give a neat and short proof. However, a proof using weaker result is also available:
For each partition $\Pi = \{ I_1, \ldots, I_n \}$ of $[0, 1]$ into non-overlapping subintervals $I_1, \ldots, I_n$, define the operator $T_{\Pi} : L^1[0,1] \to L^1[0, 1]$ by
$$ T_{\Pi} f(x) = \sum_{k=1}^{n} \biggl( \frac{1}{m(I_k)}\int_{I_k} f(t) \, \mathrm{d}t \biggr) \mathbf{1}_{I_k}(x). $$
Let $\|\Pi\| = \max\{m(I_k) : k = 1, \ldots, n\}$ be the mesh size of $\Pi$. Then we have:

Lemma. Let $(\Pi_n)$ be a sequence of partitions of $[0, 1]$ such that $\|\Pi_n\| \to 0$ as $n\to\infty$. Then for each $f \in L^1[0, 1]$,
$$ T_{\Pi_n} f \to f \quad \text{in $L^1$ as $n\to\infty$}$$

So, if $(\Pi_n)$ is as in the lemma, then by passing to a subsequence if necessary, we find that $(T_{\Pi_n}\mathbf{1}_E)$ converges to $\mathbf{1}_E$ a.e. However, the assumption tells that $T_{\Pi_n}\mathbf{1}_E \geq c > 0$ everywhere. So by passing to the limit, $\mathbf{1}_E \geq c$ a.e. and hence $\mathbf{1}_E = 1$ a.e. Therefore $m(E) = 1$.

Proof of Lemma. We easily find that
$$ \| T_{\Pi} f\|_{L^1} = \sum_{k=1}^{n} \biggl| \int_{I_k} f(t) \, \mathrm{d}t \biggr| \leq \|f\|_{L^1}. $$
Moreover, if $g$ is any continuous function on $[0, 1]$, and if $(\Pi_n)$ is a sequence of partitions on $[0, 1]$ such that the mesh size converges to $0$ as $n\to\infty$, then each $T_{\Pi_n}g$ is bounded by $\sup|g|$, and $T_{\Pi_n} g \to g$ pointwise as $n\to\infty$. So, for any $f \in L^1[0,1]$ and $g \in C[0,1]$, the dominated convergence theorem shows that
\begin{align*}
&\limsup_{n\to\infty} \| T_{\Pi_n} f - f \|_{L^1} \\
&\leq \limsup_{n\to\infty} \left( \| T_{\Pi_n}(f-g) \|_{L^1} + \| T_{\Pi_n} g - g \|_{L^1} + \| f-g \|_{L^1} \right) \\
&\leq 2\| f-g \|_{L^1}.
\end{align*}
Since $C[0, 1]$ is a dense subspace of $L^1[0, 1]$ and $g \in C[0,1]$ is arbitrary, the above bound can be made arbitrarily small. Therefore the desired conclusion follows. $\square$
A: Take any $\delta >0.$ Let $D$ be a closed set with $D\subset E$ and $m(D)>m(E)-\delta.$ Let $V$ be a family of open intervals with $\bigcup V\supset D$ and $m([0,1]\cap (\,\bigcup V\,))<m(D)+\delta.$
Now $D$ is compact so there exists a finite $W\subset V$ with $\bigcup W\supset D.$  So $m([0,1]\cap (\,\bigcup W\,))\le m([0,1]\cap (\,\bigcup V\,))<m(D)+\delta\le m(E)+\delta.$
Let $T=[0,1]\setminus \bigcup W.$ Then $m(([0,1]\setminus D)\cap E)\ge m(T\cap E).$ But the complement in $[0,1]$ of a finite union of open intervals is the union of a finite set of pair-wise disjoint intervals (including degenerate 1-point intervals), so $m(T\cap E)\ge c\cdot m(T).$  Therefore $$m(E)=m(D)+m(([0,1]\setminus D)\cap E)\ge$$ $$\ge m(D)+m(T\cap E)\ge$$ $$\ge m(E)-\delta+c\cdot m(T)=$$ $$=m(E)-\delta+c(1-m([0,1]\cap (\,\bigcup W\,))\,)\ge$$ $$\ge m(E)-\delta+c(1-m(E)-\delta))$$ which simplifies to $$\delta\ge c(1-m(E)-\delta).$$ This is not possible for every $\delta>0$ unless $m(E)=1.$
