Proving a particular set in $[0,1]$ has measure 1

Solution 1:

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 $m(E)=1.$

Solution 2:

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. $\square$