Approximation theorem from measure theory Let $a$ be an algebra, $\mu_0$ a pre-measure on it, and $\mu$ be a measure on the generated $\sigma$-algebra. Let $E \in \sigma (a) $, such that $\mu (E) <\infty $. Show that $\forall \epsilon>0$ there exists $A_{\epsilon} \in a$ such that $\mu (E \Delta A_{\epsilon}) <\epsilon $.
Is there a direct way of using the Caratheodory Extension Theorem to solve this? Also, I think we require that  $\mu_0$ is $\sigma$-finite over $a$.
 A: Note that the measure $\mu$ constructed in Caratheodory's extension theorem is given by
$$
\mu(E) = \inf \{\sum_n \mu_0(A_n) \mid A_n \in a \text{ and } E \subset \bigcup_n A_n \}.
$$
Hence, for $\varepsilon > 0$, there is a family $(A_n)_n$ of sets $A_n \in a$ with $E\subset \bigcup A_n$ and
$$
\mu(E) \leq \sum_n \mu_0 (A_n) \leq \mu(E) + \varepsilon /2.
$$
For $A := \bigcup_n A_n$, this implies $E \subset A$ and hence
$$
\mu(E) \leq \mu(A) \leq \sum_n \mu_0 (A_n) \leq \mu(E) + \varepsilon /2 < \infty.
$$
Finally, the sequence $B_n := \bigcup_{k=1}^n A_k$ is increasing with $\bigcup_n B_n = A$, so that continuity of the measure $\mu$ from below yields
$$
\mu(B_n) \to \mu(A).
$$
Hence, there is $n_0$ with $\mu(B_{n_0}) \leq \mu(A) \leq \mu(B_{n_0}) + \varepsilon/2$.
Finally, we conclude
\begin{eqnarray*}
\mu(B_{n_0} \Delta E) & \leq & \mu(B_{n_0} \Delta A) + \mu(A \Delta E) \\
& \overset{(\ast)}{=} & \mu(A \setminus B_{n_0}) + \mu(A \setminus E) \\
&=& \mu(A) - \mu(B_{n_0}) + \mu(A) - \mu(E) \leq \varepsilon.
\end{eqnarray*}
This proves the claim, because $a$ is an algebra, so that $B_{n_0} \in A$.
At $(\ast)$, we made use of $B_{n_0} \subset A$ and of $E \subset A$.
You should check the validity of the first step, which is essentially the triangle inequality for the (pseudo)-metric
$$
d(A,B) = \mu(A \Delta B).
$$
One way to do this (if you already know the concept of integration w.r.t. measures) is that
$$
\chi_{A \Delta B} = |\chi_A - \chi_B|,
$$
so that
$$
\mu(A \Delta B) = \int |\chi_A - \chi_B| d\mu \leq \int |\chi_A - \chi_E| d\mu + \int |\chi_E - \chi_B| d\mu = \mu(A \Delta E) + \mu(E \Delta B).
$$
Of course, one can also prove this without using the integral.
EDIT: The above proves the claim for the measure $\mu$ constructed in the Caratheodory theorem. If $\mu_0$ is $\sigma$-finite w.r.t. $a$, then this holds for any extension, because the extension will be unique.
