Finding a "big" bounded subset of a subset of $\mathbb R$

Question

I'm working in the following exercise.

First, we define the outer Lebesgue measure, $m(A)=\inf\{\sum_n l(I_n): (I_n)\text{ is a sequence of open intervals and} A \subset \bigcup_nI_n \}$, where $l$ denotes the lenght of the interval.

Suppose $A \subset \mathbb R$ and suppose $m(A)<\infty$. Show that for every $\epsilon>0$ there exists a bounded set $A_1 \subset A$ such that $m(A_1^c\cap A)<\epsilon$.

Progress: I was thinking about defining the decreascing sequence of sets $J_n=(-\infty, -n) \cup (n , \infty)$ and showing that one of them must have $\mu(A \cap J_n)<\epsilon$, then the remaining would follow easily, but I'm not managing to do it.

Answer 1

If $A$ is measurable Let $\epsilon>0$

Let $A = \bigcup_{k\in \mathbb{N}} A_k$, where $A_k$ is bounded. Let $B_i = \bigcup_{k=1}^i A_k$ and $B^*_i = B_i \setminus B_{i-1}$ (with $B^*_1=B_1=A_1$). We readily see that $\mu(A) = \sum_{k=1}^\infty \mu(B^*_k)$

Since $\mu(A)<\infty$, $\sum_{k=1}^\infty \mu(B^*_k) < \infty$ and then there exists $N$ such that $\sum_{k=N}^\infty \mu(B^*_k) < \epsilon$.

But then $B_N$ is bounded and $\mu((B_N)^C \cap A) = \mu(A \setminus B_N) = \mu( \bigcup_{k=N}^\infty B^*_k) < \epsilon$

If $A$ is not measurable, let $M \supset A$ such that $M$ is measurable. There is a bounded set $B\subset M$ such that $m(M\setminus B) <\epsilon$ but then $m(A\setminus (B\cap A)) = m(A\setminus B) \leq m(M\setminus B) <\epsilon$ and $B\cap A \subset A$ is bounded.

Answer 2

First, suppose $A$ is measurable.

For each $n \in \omega$, let $I_n=(-(n+1), -n]\cup (n, n+1]$.

For each $n \in \omega$, let $J_n=(-\infty, -n]\cup (n, \infty)$.

Let $\epsilon>0$.

Since each $I_n$ is measurable and since $\bigcup_{n \in \omega} I_n=\mathbb R$, we have:

$m(A)=m(A \cap \bigcup_n I_n)=m( \bigcup_n A \cap I_n)=\sum_{n=0}^\infty m(A \cap I_n)<\infty$ Therefore:

$0=\lim_{k\rightarrow\infty}\sum_{n=k}^\infty m(A \cap I_n)=\lim_{k\rightarrow\infty}m(\bigcup_{n=k}^\infty A \cap I_n)=\lim_{k\rightarrow\infty}m(A \cap\bigcup_{n=k}^\infty I_n)=\lim_{k\rightarrow\infty}m(A \cap J_n)$

So there exists $n \in \omega$ such that $m(J_n \cap A)<\epsilon$. Pick $n$

Put $A_1=J_n^c\cap A$. Then $A_1 \subset A$. Since $A_1^c \cap A=(J_n\cup A^c)\cap A=J_n \cap A$, we have $m(A^c_1\cap A)<\epsilon$ and the proof is complete.

Now suppose $A$ is arbitrary. There exists a measurable set $B$ such that $A \subset B$ and $m(A)=m(B)$. Let $\epsilon>0$ be given. There exists a bounded $B_1$ such that $m(B_1^c \cap B)<\epsilon$ and $B_1 \subset B$. Let $A_1=B_1\cap A$. Notice that $A_1$ is bounded, $A_1 \subset A$ and $A_1^c\cap A=B_1^c \cap A$, therefore $m(A_1^c \cap A)= m(B_1^c\cap A)\leq m(B_1^c\cap B)<\epsilon$.