Prove that for each $\epsilon>0$, there exist a compact $K\subset E$ such that $m(E\cap K^c)<\epsilon$

Question

Let $E\subset\mathbb R$ be Lebesgue measurable with $m(E)<\infty$. Prove that for each $\epsilon>0$, there exist a compact $K\subset E$ such that $m(E\cap K^c)<\epsilon.$

Attempt

By contradiction, suppose that for all compact $K\subset E$, $m(E\cap K^c)\ge\epsilon.$

Notice that $K$ is Lebesgue measurable, thus $m(E)=m(E\cap K)+m(E\cap K^c)\ge(E\cap K)+\epsilon$

But from here is not clear how to get a contradiction.

Could someone help please?

Answer 1

This is a slog through a bit of set manipulations, and the definition of (outer measure):

Let $\epsilon>0.$ First suppose that $E$ is bounded. Then, there is a closed cube $Q$ such that

$\tag 1m(Q\cap E)=m(E).$

$\ Q\setminus E\ $ is measurable, so there is an open set $V\supseteq Q\setminus E$ such that

$\tag 2 m(V)<m(Q\setminus E)+\epsilon.$

Now, $Q\setminus V$ is compact, (it is closed and bounded), and since, from $(2)$, we have

$\tag3 m(Q\setminus E)>m(V)-\epsilon,$

we get

$\tag4 m(Q\setminus E)>m(V)-\epsilon\Rightarrow m(Q)>m(V)+m(Q\cap E)-\epsilon\Rightarrow m(Q\setminus V)>m(Q\cap E)-\epsilon.$

It now follows from $(4)$ and $(1)$ that

$\tag 6 m(Q\setminus V)>m(Q\cap E)-\epsilon=m(E)-\epsilon.$

Thus, $\tag 7 m(E\setminus (Q\setminus V))<\epsilon.$

If $E$ is not bounded, then we choose cubes $Q_i\subseteq Q_{i+1}$ increasing to $\mathbb R^n$ and apply the foregoing to $E_i= E\cap Q_i$ to find compact sets $K_i\subset E_i\subseteq E$ with the advertised property. Now choose $N$ so that

$\tag 8m(E) \le m(E_N ) + \epsilon/2.$

This is possible because $\bigcup E_i=E$ and $E_i\subseteq E_{i+1}$, so that $\lim m(E_i)=m(E).$ But we also have a compact set $K\subseteq E_N$ with

$\tag 9 m(E_N )\le m(K) + \epsilon/2.$

This is possible since $E_N$ is bounded.

It now follows from $(8)$ and $(9)$ that

$\tag {10} m(E)\le m(K)+\epsilon,$

as desired.

Answer 2

Consider $\mathrm{B}_n$ to be the closure of the ball $\mathrm{B}(0; n),$ which is compact. Observe that $\mathrm{E}_n = \mathrm{E} \cap \mathrm{B}_n$ converges to $\mathrm{E},$ hence its measure also converges. This shows $m(\mathrm{E}) = \lim_n m(\mathrm{E}_n)$ and since $\mathrm{E}_n \subset \mathrm{E},$ it turns out that $0 = \lim m(\mathrm{E} \setminus \mathrm{E}_n) = \lim m(\mathrm{E} \cap \mathrm{B}_n^\complement).$ Q.E.D.

Answer 3

The required claim easily follows from the regularity of the Lebesgue measure. For instance, by Proposition 6.2 from here if $n\ge 1$ is an integer, $m$ is the Lebesgue measure on $\Bbb R^n$, and $E$ be any Lebesgue measurable subset of $\Bbb R^n$ then $$m(E) = \sup\{m(K) : K\mbox{ is a compact subset of }\Bbb R^n\mbox{ with }K\subset E\}.$$

Now, given $\epsilon>0$, there exists a compact subset $K$ of $E$ such that $m(K)>m(E)-\epsilon$. Since $E=(E\cap K)\sqcup (E\cap K^c)=K\sqcup (E\cap K^c)$, we have $m(E)=m(K)+m(E\cap K^c)$, so $m(E\cap K^c)= m(E)-m(K)<\epsilon$.

Answer 4

Note that $E=\bigcup\limits_{n=1}^{\infty}E \cap [-n,n]$. Hence, by the continuity of measure, $m(E)=\lim\limits_{n \rightarrow \infty}m(E \cap [-n,n])$. This implies that given an $\epsilon>0$ we can find a $n_{\epsilon}\in\mathbb{N}$ such that $m(E)-m(E \cap [-n_{\epsilon}, n_{\epsilon}])<\epsilon$. By the additivity of measure the LHS equals $m(E\cap [-n_{\epsilon}, n_{\epsilon}]^C)$, so since $[-n_{\epsilon}, n_{\epsilon}]$ is compact we are done.