A Half Partition of Borel Sets will Give Half the Borel Measure

Question

Here is the problem:

Let $n \in \mathbf{N}$ and define $A_n := \bigcup_{k \in \mathbf{Z}} [\frac{2k}{2^n}, \frac{2k + 1}{2^n})$. If $E \in \mathcal{B}(\mathbf{R})$, show the limit $$ \lim_{n \to \infty} \lambda(A_n \cap E) = \frac{\lambda(E)}{2}. $$ I have successfully proven that for all $E \in \mathcal{B}(\mathbf{R})$ that is bounded. The problem that I am facing right now is to extend from the bounded case to unbounded case.

My idea was to split $E = \bigcup_{m = 1} ^\infty E \cap B(0, m) =: \bigcup_{m = 1} ^\infty E_m$, with $B(0, m)$ being the ball with radius $m$. This would give us by the continuity of measure that $$ \lim_{n \to \infty} \lambda(A_n \cap E) = \lim_{n \to \infty} \lim_{m \to \infty} \lambda(A_n \cap E_m). $$ Now ideally we would really like to switch the order of limit here. However, I can not justify if we could do so or not. Is there another way to approach the unbounded case? Any idea would be appreciated.

Answer 1

I will assume that OP has already established the claim for bounded Borel subsets. Then we will prove the claim for arbitrary Borel $E \subseteq \mathbb{R}$.

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1. First, assume $\lambda(E) < \infty$. We fix a partition $(E_j)_{j\in J}$ of $E$ into bounded Borel sets. (For example, we can set $E_j = E \cap [j, j+1)$ and $J = \mathbb{Z}$.) Note that $ \sum_{j\in J} \lambda(E_j) = \lambda(E) < \infty$. Then

$$ 0 \leq \lambda(E_j \cap A_n) \leq \lambda(E_j) \qquad\text{and}\qquad \lim_{n\to\infty} \lambda(E_j \cap A_n) = \frac{1}{2}\lambda(E_j), $$

and so, the dominated convergence theorem is applicable. Hence,

\begin{align*} \lim_{n\to\infty} \lambda(E \cap A_n) &= \lim_{n\to\infty} \sum_{j\in J} \lambda(E_j \cap A_n) \\ &= \sum_{j\in J} \lim_{n\to\infty} \lambda(E_j \cap A_n) = \sum_{j\in J} \frac{1}{2} \lambda(E_j) = \frac{1}{2} \lambda(E). \end{align*}

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2. Now suppose $\lambda(E) = \infty$. Then for any $r > 0$,

$$ \liminf_{n\to\infty} \lambda(E \cap A_n) \geq \liminf_{n\to\infty} \lambda(E \cap [-r, r] \cap A_n) = \frac{1}{2}\lambda(E \cap [-r, r]). $$

Since $r$ is arbitrary, letting $r \to \infty$ shows that

$$ \liminf_{n\to\infty} \lambda(E \cap A_n) \geq \infty, $$

which immediately implies the desired claim for unbounded Borel sets.

Answer 2

Consider $E \subset \mathbb R$ with $\lambda(E) < \infty$. Given $\epsilon$, there is $R$ such that $\lambda(E \backslash [-R,R]) < \epsilon$. Then $$ \liminf_{n \to \infty} \lambda(A_n \cap E) \ge \lim_{n \to \infty} \lambda(A_n \cap E \cap [-R,R]) = \frac{1}{2} \lambda(E \cap [-R,R]) > \frac{1}{2} (\lambda(E) - \epsilon)$$ while $$ \limsup_{n \to \infty} \lambda(A_n \cap E) \le \lim_{n \to \infty} \lambda(A_n \cap E \cap [-R,R]) + \lambda(E \backslash [-R,R]) = \frac{1}{2} \lambda(E \cap [-R,R]) + \lambda(E \backslash [-R,R]) \le \frac{1}{2} \lambda(E) + \epsilon/2$$

Take $\epsilon \to 0$ and we get $\lim_{n \to \infty} \lambda(A_n \cap E) = \frac{1}{2} \lambda(E)$.

On the other hand, if $\lambda(E) = \infty$, then for any $M$ there is $R$ so that $\lambda(E \cap [-R,R]) > M$. We get $$\liminf_{n \to \infty} \lambda(A_n \cap E) \ge \lim_{n \to \infty} \lambda(A_n \cap E \cap [-R,R]) = \frac{1}{2} \lambda(E \cap [-R,R]) > M/2$$ Takng $M \to \infty$, we see that $\lim_{n \to \infty} \lambda(A_n \cap E) = \infty$.