Folland Chapter 7 Exercise 8 Suppose that $\mu$ is a Radon measure on X, If $\phi \in L^1(\mu)$ and $\phi \geq 0$, then prove that $\nu(E) = \int_E \phi d\mu$ is a Radon measure. (Hint: Use Corollary 3.6)
Corollary 3.6 says that if $f \in L^1(\mu)$, for every $\epsilon >0$, there exists $\delta > 0$ such that $|\int_E f d\mu| < \epsilon$ whenever $\mu(E) < \delta$.
Clearly $\nu$ is a measure and it is finite on compact set, for outer regularity:
Let $\epsilon > 0$ since $\mu$ is already a radon measure, for any E, we can find $U$ open and $E \subset U$ such that $\mu(U)-\mu(E) < \delta$, then $\nu(U)-\nu(E) < \epsilon$ By corollary 3.6, but since $v(U) > v(E)$ for all such $U$, 
$v(E) = \inf \{v(U): $ $U$ open, $E \subset U \}$.
Inner regularity can be proved similarly.
I am doubtful that this is the right solution, since this "solution" did not use anything from the section that contains the exercise, but instead only on very basic definition of radon measures. If anyone can see what went wrong here I would appreciate it if you can let me know.
thank you
 A: I said this seems correct, but the answer is no. This argument is correct just when $X$ is $\sigma$-compact. We know that compact sets have finite measure, but how about Borel set?
If we want to use hint, then we may approximate Borel sets using $\sigma$-compact sets. Fortunately, we have Proposition 7.5.
Suppose $E$ is a Borel set. Define $\{E_n\}_{n \in \mathbb{N}}$ as follows.
\begin{align}
 E_n := E \cap \phi^{-1}((n^{-1}, \infty))
\end{align}
Note that $E_n$ is finite(and hence, $\sigma$-finite) because $\phi$ is in $L^1$.
By proposition 7.5, for every $n \in \mathbb{N}$, there exists compact set $K_n$ such that $K_n \subseteq E_n$ and $\mu(E_n -K_n) < \varepsilon/2$.
By absolute continuity(Theorem 3.5, the hint),
for all $\varepsilon >0$, exists $\delta >0$ such that $\nu(F) < \varepsilon/2$ whenever $\mu(F) < \delta$.
Since $\nu$ is finite measure, for every $\varepsilon >0$, there exists $N \in \mathbb{N}$ such that
\begin{align}
 \nu(\bigcup_{n=N+1}^\infty K_n) < \varepsilon/2
\end{align}
By combining results, we get
\begin{align}
 \nu(E_n - \bigcup_{n=1}^N K_n) \le \nu(E_n - \bigcup_{n=1}^\infty K_n) + \nu(\bigcup_{n=N+1}^\infty K_n) < \nu(E_n - K_n) + \varepsilon/2 < \varepsilon.
\end{align}
Finally, by continuity of $\nu$ and monotonicity of $\{E_n\}$, we finally get
\begin{align}
 \nu(E - \bigcup_{n=1}^N K_n) = \lim_{n \to \infty} \nu(E_n - \bigcup_{n=1}^N K_n) < \varepsilon
\end{align}
Since $\varepsilon > 0$ is arbitrary and $N$ only depends on $\varepsilon$, we have proven the inner regularity on the Borel set. (Notice that this is stronger than what we want.)
EDIT: The proof for the outer regularity also faces same problem.
We can use inner regularity on the Borel set.
Suppose $E$ be a Borel set. Then, $E^c$ be another Borel set. Then, we can find a compact set $K$ such that $K \subseteq E^c$ and $\nu(E^c - K) < \varepsilon$.
Since $X$ is LCH, compact set is closed and so, $K^c$ is an open set. $E \subseteq K^c$ and
$$
\nu(K^c) = \nu(X) - \nu(K) < \nu(X) - \nu(E^c) + \varepsilon = \nu(E) + \varepsilon.
$$
It means that the outer regularity holds for $E$.
I refer the proof of Jonathan Conder's and just add some details
