I'm trying to solve following problem:

Let $A \subseteq \mathbb{R}$. Suppose that for any $\epsilon > 0$, there are Lebesgue measurable sets $B, C$ so that $B \subseteq A \subseteq C$ and $m(C \cap B^c) < \epsilon$. Show that $A$ is Lebesgue measurable.

Here $m$ is the Lebesgue measure on $\mathbb{R}$. A set $X \subseteq \mathbb{R}$ is Lebesgue measurable if $$m^*(E) = m^*(E \cap X) + m^*(E \cap X^c)$$ for every $E \subseteq \mathbb{R}$, where $m^*$ is the outer measure given by $$m^*(E) = \inf\{\space \sum_{n=1}^{\infty} \text{length}(I_n) : (I_n)_{n \in \mathbb{N}} \space \text{a cover of}\space E \space\text{by open intervals}\}.$$

I can show that $m^*(E) \leq m^*(E \cap A) + m^*(E \cap A^c)$, by sub-additivity. Can anyone help me with the other direction?


Use the completeness of the Lebesgue measure.

Let $\epsilon = \frac{1}{n}$, and create a sequence of measureable sets $B_n \subset A \subset C_n$ such that $m(C_n\setminus B_n) < \frac{1}{n}$. The sets $B = \cup_n B_n$ and $C = \cap_n C_n$ are measurable, and $B\subset A \subset C$, with $m(C\setminus B) = 0$.

We have $A = (A\cap B) \cup (A \setminus B) = B \cup (A \setminus B)$. However, $(A \setminus B) \subset (C \setminus B)$, and any subset of a set of measure zero is Lebesgue measurable, hence $(A \setminus B)$ is measurable and has measure zero. Hence $A = B \cup (A \setminus B)$ is measurable and $mA = mB$.

Note: To show that any subset of a set of measure zero is measurable, suppose $Z \subset N$, where $mN = 0$. By inclusion, we have $m^* Z = 0$. Now let $E$ be an arbitrary set, then since $N$ is measurable, we have $m^* E = m^* (N \cap E) + m^* (E\setminus N) = m^* (E\setminus N)$.

Since $Z \subset N$, we have $(E \setminus N) \subset (E \setminus Z) \subset E$ and so we have $m^* (E \setminus Z) = m^* E$. Since $m^* (Z \cap E) \le m^* Z = 0$, we have $m^* E = m^* (Z \cap E) + m^* (E\setminus Z)$, hence $Z$ is measurable. It follows that $mZ = m^* Z = 0$.

  • $\begingroup$ Thanks! To show that m(C\B) = 0, I used that C\B = intersection of all the (C_n \ B_n), so for any N, 0 =< m(C\B) =< m(C_N \ B_N) -> 0 as N goes to infinity. Is this OK? $\endgroup$ – Ryan Mar 22 '13 at 20:48
  • $\begingroup$ Yes, you have $m(C\setminus B) <\frac{1}{n}$ for all $n$, hence it is zero. $\endgroup$ – copper.hat Mar 22 '13 at 20:51
  • $\begingroup$ Brilliant. Thank you again. $\endgroup$ – Ryan Mar 22 '13 at 20:57
  • $\begingroup$ You are very welcome! $\endgroup$ – copper.hat Mar 22 '13 at 21:01

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