Borel Measures: Regularity Given the complex plane $\mathbb{C}$.
Consider a Borel measure:
$$\mu:\mathcal{B}(\mathbb{C})\to\mathbb{R}_+:\quad\mu(\mathbb{C})<\infty$$
Then it is regular:
$$\mu(A)=\sup_{A\supseteq K}\mu(K)=\inf_{A\subseteq U}\mu(U)$$
How can I prove this?
This thread refers to: Regular Measure
 A: One way is to use Dynkins $\pi$-$\lambda$ theorem. Set
$$
G :=\{A\in\mathcal{B}(\Bbb{C})\mid \mu(A)=\sup_{K\subset A}\mu(K)=\inf_{U \supset A}\mu(U)\}. 
$$
Note that $G$ contains the $\pi$ system of all open sets (since each open set in $\Bbb{C}$ is sigma compact), which also generates the Borel sigma algebra. 
Hence, it suffices to show that $G$ is a $\lambda$ system, I.e. it is closed under countable disjoint unions and complements. Using a $\epsilon/2^n$ argument (and the fact that $\mu$ is a finite measure), this is not too hard to do. 
EDIT: Ok, let us first show closure under countable disjoint unions:
Let $A_n \in G$ be pairwise disjoint with $A := \biguplus A_n$. Let $\varepsilon > 0$. For each $n$, there is $U_n \supset A_n$ open with $\mu(U_n) < \mu(A_n) + \varepsilon / 2^n$. Then $U := \bigcup_n U_n$ is open with $U \supset A$ and
$$
\mu(A) \leq \mu(U) \leq \sum_n \mu(U_n) \leq \sum_n \mu(A_n) + \varepsilon = \mu(A) + \varepsilon.
$$
Similarly, there is $N \in \Bbb{N}$ with $\mu(A) \leq \varepsilon/2 + \sum_{n=1}^N \mu(A_n)$ (here we use sigma additivity and finiteness of $\mu(A)$). Further, for each $n \in \{1, \dots, N\}$, there is $K_n \subset A_n$ with $\mu(A_n) \leq \varepsilon / 2N + \mu(K_n)$. Thus, $K := \bigcup_{n=1}^N K_n \subset A$ is compact with
$$
\mu(A) \leq \frac{\varepsilon}{2} + \sum_{n=1}^N \mu(A_n) \leq \frac{\varepsilon}{2} + \sum_{n=1}^N \bigg[\mu(K_n) + \frac{\varepsilon}{2N}\bigg] = \mu(K) + \varepsilon.
$$
Here, the last step used disjointness of the $K_n$, because of $K_n \subset A_n$ and disjointness of the $A_n$.
Since $\varepsilon >0 $ was arbitrary, we get $A \in G$.
Finally, let us show closure under complements. Let $A \in G$ and $\varepsilon > 0$. Then there are $K \subset A \subset U$ with $\mu(U) - \varepsilon <\mu(A) < \mu(K) + \varepsilon$. Hence (because $\mu(X)$ is finite, where in your case $X = \Bbb{C}$), we get
\begin{eqnarray*}
\mu(K^c) - \varepsilon   &=&\mu(X) - [\mu(K) + \varepsilon] \\ &<& \mu(X) - \mu(A) \\ &=& <\mu(A^c) \\ &=& \mu(X) - \mu(A) \\ &<& \mu(X) - [\mu(U) - \varepsilon] \\ &=& \mu(U^c) + \varepsilon.
\end{eqnarray*}
Finally, note that $U^c \subset A^c \subset K^c$ with $U^c$ closed and $K^c$ open.
It remains to note that $X (= \Bbb{C})$ is $\sigma$-compact, so that also the closed set $U^c$ is $\sigma$-compact. This allows us to find a sequence of compacta $K_n \subset U^c \subset A^c$ with $\mu(K_n) \to \mu(U^c)$, which (together with the above estimate) easily implies $A^c \in G$.
