Consequence of Riesz Representation Theorem from Rudin RCA 
It's Riesz Representation Theorem from Rudin's book. In the following chapter I met the following example:

It's obvious that $\sigma$-compact set has the  $\sigma$-finite measure.
But how to prove strictly if $E\in \mathfrak{M}$ and $E$ has $\sigma$-finite measure then $E$ is inner regular? 
Rudin wrote that it's easy. But I tried and thought about this and no results. 
How to prove that for any $\alpha$ with $\alpha <\mu(E)$ exists compact set $K\subset E$ such that $\mu(K)>\alpha$? If we show this then problem will be solved.
 A: Let $E\in\mathfrak{M}$. If $\mu(E)<\infty$ then $E$ is inner regular by $(d)$. Suppose $\mu(E)=\infty$ and $E$ $\sigma$-finite.
To show: $E$ is inner regular.
My idea
Since $E$ $\sigma$-finite, there exists a countable family $\{A_n\}_{n\in\mathbb{N}}\subset \mathfrak{M}$ such that $E=\bigcup_{n\in\mathbb{N}}A_n$ and $\mu(A_n)<\infty$ for all $n\in \mathbb{N}$. We can assume that all $A_n$ are disjoint, indeed it is enough to consider: $C_1:=A_1$, $C_2:=A_2\cap A_1^c$, $\dots$, $C_n = A_n\cap (\bigcup_{k=1}^{n-1}A_{k})^c$. Note that each $C_n$ is in $\mathfrak{M}$ and has finite measure since it is subset of $A_n$.
In order to say that $E$ is inner regular we have to find for each $\alpha\in\mathbb{R}_{>0}$ a compact subset $K\subset E$ such that $\mu(K)>\alpha$.
Let $\alpha\in\mathbb{R}_{>0}$ and $\epsilon>0$ small enough. Since for each $n\in\mathbb{N}$ $\mu(A_n)<\infty$ and $A_n\in\mathfrak{M}$, we can find by $(d)$ a compact subset $K_n\subset A_n$ such that $\mu(K_n)>\mu(A_n)-\epsilon$. Moreover, since we assume all $A_n$ disjoint, all $K_n$ are disjoint.
Clearly a finite union of compact is compact and $\bigcup_{n=1}^{m}K_n\subset E$ for all $m\in \mathbb{N}$.
Define for $m\in\mathbb{N}$ $B_m:=\bigcup_{n=1}^{m}A_n$, which is an increasing sequence of measurable sets for which $\bigcup_{m\in\mathbb{N}}B_m=\bigcup_{n\in\mathbb{N}}A_n$. Thus
\begin{equation}
\infty = \mu(E)=\mu(\bigcup_{n\in\mathbb{N}}A_n)=\mu(\bigcup_{m\in\mathbb{N}}B_m)=\lim_{m\to \infty}\mu(B_m)=\lim_{m\to \infty}\mu(\bigcup_{n=1}^{m}A_n).
\end{equation}
Therefore, it is possible to find $M\in\mathbb{N}$ such that
\begin{equation}
\mu(\bigcup_{n=1}^{M}A_n)>\alpha +1.
\end{equation}
Take $\epsilon = \frac{1}{M}$ and use that all $K_n$ are disjoint and also all $A_n$, we obtain
\begin{equation}
\mu(\bigcup_{n=1}^{M}K_n)=\sum_{n=1}^{M}\mu(K_n)>\sum_{n=1}^{M}[\mu(A_n)-\epsilon]=[\sum_{n=1}^{M}\mu(A_n)]-M\epsilon=\mu(\bigcup_{n=1}^M A_n)-1>\alpha.
\end{equation}
Thus, $K:=\bigcup_{n=1}^{M}K_n\subset E$ is the desired compact set.
A: Since $E$ is $\sigma$-finite, we can write $E = \cup E_n,$ where $E_1 \subset E_2 \subset E_3 \cdots,$ and $\mu(E_n) <\infty$ for each $n.$ By the result already known, for each $n$ we can choose a compact $K_n \subset E_n$ such that $\mu(E_n\setminus K_n) < 1/n.$ We then have
$$\mu(E) = \lim_{n\to \infty} \mu(E_n) = \lim_{n\to \infty} \left (\mu(K_n) +\mu(E_n\setminus K_n) \right).$$
Because $\mu(E_n\setminus K_n) \to 0,$ we get $\mu(E) = \lim \mu(K_n)$ as desired.
