Approximation of open subsets of $\Bbb R^2$ by compact sets. I am having difficulty to prove one of the regularity properties of Lebesgue measure. Here it is $:$
Let $(\Bbb R^2, \mathcal L_{\Bbb R^2}, \lambda_{\Bbb R^2})$ be the Lebesgue measure space on $\Bbb R^2.$ Then prove that $\lambda_{\Bbb R^2} (U) = \sup \left \{\lambda_{\Bbb R^2}(K)\ |\ K\ \text {is compact},\ K \subseteq U \right \},$  for any open subset $U \subseteq \Bbb R^2.$
How do I proceed? Any help will be highly appreciated.
Thanks in advance.
 A: I proceed along the same lines as Greg Martin pointed out in his comment above. Here's my answer.
Let $S : = \left \{\lambda_{\Bbb R^2} (K)\ |\ K\ \text {is compact},\ K \subseteq U \right \}.$ We need to prove that $\lambda_{\Bbb R^2} (U) = \sup S.$
Since $U$ is open, $U$ is a neighbourhood of each of it's points. So for any $x \in U$ there exists an open ball $B_x$ surrounding $x$ such that $B_x \subseteq U.$ Take any slightly small closed ball $B_x'$ inside $B_x$ for each $x \in U.$ So $B_x'$ is a compact neighbourhood of $x$ sitting inside $U,$ for each $x \in U.$ Since $\Bbb R^2$ is second countable it has a countable open base $\mathcal B = \left \{B_n \right \}_{n=1}^{\infty}.$ So for each $x \in U,$ there exists $B_{m(x)} \in \mathcal B$ such that $x \in B_{m(x)} \subseteq B_x'.$ Now consider the collection $\mathcal B' := \left \{B_{m(x)}\ |\ x \in U \right \}.$ Then we have $U = \bigcup\limits_{x \in U} B_{m(x)}.$ Since $\mathcal B' \subseteq \mathcal B$ and $\mathcal B$ is countable it follows that $\mathcal B'$ is also countable. So we can index the elements of $\mathcal B'$ by natural numbers say $\mathcal B' = \left \{B_{n_r} \right \}_{r=1}^{\infty}$ (This collection may as well be finite; so for each $r \in \Bbb N,$ the corresponding open ball $B_{n_r}$ in the collection $\mathcal B'$ might not be distinct). So we have $U = \bigcup\limits_{r=1}^{\infty} B_{n_r}.$
Now by the construction of $\mathcal B'$ it follows that for any $r \in \Bbb N,$ there exists $x_r \in U$ such that $B_{n_r} \subseteq B_{x_r}' \subseteq U.$ Since $U = \bigcup\limits_{r=1}^{\infty} B_{n_r},$ it follows that $U = \bigcup\limits_{r=1}^{\infty} B_{x_r}'.$ Let $K_n : = \bigcup\limits_{r=1}^{n} B_{x_r}'.$ Then $\{K_n \}_{n=1}^{\infty}$ is a sequence of compact subsets of $U,$ with $K_n \subseteq K_{n+1},$ for all $n \geq 1$ and moreover $\bigcup\limits_{n=1}^{\infty} K_n = \bigcup\limits_{r=1}^{\infty} B_{x_r}' = U.$ So $\{K_n \}_{n=1}^{\infty}$ is sequence of compact subsets of $U$ such that $K_n\ \bigg \uparrow\ U.$  Since $\lambda_{\Bbb R^2}$ is a measure on $\mathcal L_{\Bbb R^2}$ so it is countably additive and hence it is continuous from below. Therefore $\lim\limits_{n \to \infty} \lambda_{\Bbb R^2} (K_n) = \lambda_{\Bbb R^2} (U).$ But each $K_n$ is a compact subset of $U$ and hence $$\lambda_{\Bbb R^2} (K_n) \leq \sup S,\ \text {for all}\ n \geq 1.$$ Therefore $$\lim\limits_{n \to \infty} \lambda_{\Bbb R^2} (K_n) \leq \sup S.$$ But this implies that $$\lambda_{\Bbb R^2} (U) \leq \sup S.\ \ \ \ \ \ \ (1)$$
On the other hand for any compact set $K,$ with $K \subseteq U$ we have by monotonicity of $\lambda_{\Bbb R^2},$ $\lambda_{\Bbb R^2} (U)$ is an upper bound of $S.$ Hence
$$\lambda_{\Bbb R^2} (U) \geq \sup S.\ \ \ \ \ \ \ \ (2)$$ Combining $(1)$ and $(2)$ we have $$\lambda_{\Bbb R^2} (U) = \sup S$$ as required.
QED
