necessary conditions of measure approximation theorem Measure approximation theorem (I can't really remember its exact name) states that let $A$ be an algebra, $\mu$ a measure on $\sigma(A)$ and $\mu$ is $\sigma$-finite on $A$. Let $E\in \sigma(A)$ such that $\mu(E) < \infty$. Then for all $\epsilon > 0$, there exists a $A_{\epsilon} \in A$ such that $\mu(E \bigtriangleup A_\epsilon) < \epsilon$. Here $\bigtriangleup$ means symmetric difference.
My question is can we give counterexamples to show that 
1) $\mu$ is $\sigma$-finite on $A$
2) $\mu(E) < \infty$
are necessary. i.e. it's not the case that for all $\epsilon$ there exists such an $A_\epsilon$ when 
1) $\mu$ is $\sigma$-finite on $A$ but $\mu(E) = \infty$
(2) $\mu(E) < \infty$ but $\mu$ is not $\sigma$-finite on $A$
I have trouble coming up with examples to prove the necessity of those two conditions. Can someone help me? 
 A: For both problems, we can take $\mu$ as a suitable measure on $X = [0,\infty)$ equipped with the Borel-$\sigma$-algebra.
This $\sigma$-algebra is generated by the algebra(!) of finite disjoint unions of sets intervals of the form $[a,b)$, where $b = \infty$ is allowed.
Ad 1) Take $\mu$ as Lebesgue measure on $X$ and $E = \bigcup_n [2n, 2n+1]$. Take $\varepsilon = 1$. For $B \in A$ there are only the following two cases (why?).


*

*We have $B \supset[m, \infty)$ for some $m \in \Bbb{N}$. This yields


$$
\mu(E \Delta B) \geq \mu(B \setminus E) \geq \mu(\bigcup_{\ell \geq \ell_0} (2\ell + 1, 2\ell) ) = \infty > \varepsilon.
$$
for a suitable $\ell_0$


*We have $\mu(B) < \infty$, which easily yields


$$
\mu(B \Delta E) \geq \mu(E \setminus B) = \infty > \varepsilon.
$$
Ad 2) Take $\mu$ as the counting measure on $X$ and $E = \{1,2,3\}$, as well as $\varepsilon = 1$. For $B \in A$, we either have $B = \emptyset$ or $\mu(B) = \infty$ (why?).
For $B = \emptyset$, we see $\mu(E \Delta B) = \mu(E) = 3 > \varepsilon$.
Otherwise, $\mu(B) = \infty$ and hence $\mu(E \Delta B) \geq \mu(B \setminus E) = \infty > \varepsilon$.
