Question on Radon measures from Folland's Real Analysis Greetings my mathematical friends.
I am taking a summer class on measures and the theory of real analysis, and I was given the following question from Folland's Real Analysis Second Edition Chapter 7 on Radon measures. It is question #9 which is:

The corollary they tell us to use in part a. is:
Given a Radon measur $ \mu $ and a non-negative lower semi continuous function f then we have:

Of course by lower semi continuous (LSC) we mean a function f such that the following set is open for all values of a $ \{ x | f(x)>a \} $ and by upper semi continuous (USC) the same for the set a $ \{ x | f(x)<a \} $
I like to think I am a good enough student but I haven't got the faintest clue as to how to do this which is frustrating, I have tried many times to look at the question and the hints in parts a and b but cannot really find a pattern. Could you please help me out?
Thanks friends
 A: a) I leave it to you to show that $\phi \chi_U$ is lower semi continuous.
The hint they gave you is actually a part of Proposition 7.14. Corollary 7.13  actually reads
$$
\int f \, d\mu = \sup \{\int g \,d\mu \,\mid\, 0\leq g \leq f \text{and } g\in C_c\}
$$
if $f$ is lower semicontinuous.
With this information, first try part (a) again. If you don's suceed, read on.

We get
$$\require{action}
\toggle{  \text{click for more hint}  }{
\begin{eqnarray*}
 &  & \nu\left(U\right)\\
 & \overset{\left(\text{why?}\right)}{=} & \int\phi\chi_{U}\,{\rm d}\mu\\
 & = & \sup\left\{ \int g\,{\rm d}\mu\,\middle|\,0\leq g\leq\phi\chi_{U}\text{ and }g\in C_{c}\right\} \\
 & \overset{h\phi\in C_{c}\Longleftrightarrow h\in C_{c}}{=} & \sup\left\{ \int h\phi\,{\rm d}\mu\,\middle|\,0\leq h\leq\chi_{U}\text{ and }h\in C_{c}\right\} \\
 & \overset{h\phi\in C_{c},\text{ definition of }\nu'}{=} & \sup\left\{ \int h\,{\rm d}\nu'\,\middle|\,0\leq h\leq\chi_{U}\text{ and }h\in C_{c}\right\} \\
 & \overset{\text{Corollary 7.13 again}}{=} & \int\chi_{U}\,{\rm d}\nu'=\nu'\left(U\right).
\end{eqnarray*}
}
\endtoggle$$
Here, we used in the last line that $\chi_{U}$ is lower semicontinuous and that $\nu'$ is Radon.
b) Let $E\subset X$ be Borel. If $\nu\left(E\right)=\infty$, it is
clear that $\nu$ is outer regular on $E$, so that we can assume
$\nu\left(E\right)<\infty$ in what follows. Now (with notation as
in the hint)
$$
E=\biguplus_{k\in\mathbb{Z}}\left(E\cap V_{k}\right).
$$
Let $\varepsilon>0$ be arbitrary. We will show that for each $k\in\mathbb{Z}$,
there is an open set $U_{k}\supset E\cap V_{k}$ with $\nu\left(U_{k}\setminus\left(E\cap V_{k}\right)\right)<\frac{\varepsilon}{2^{\left|k\right|}}$.
This will then imply the claim (why?).
What follows are hints for the construction of the $U_k$.

 Note that by replacing $U_{k}$ by $U_{k}\cap V_{k}$, we can assume $U_{k}\subset V_{k}$.

But then we have
$$\require{action}
\toggle
{\text{click for more hint}}
{
2^{k}\cdot\mu\left(U_{k}\setminus\left(E\cap V_{k}\right)\right)\leq\nu\left(U_{k}\setminus\left(E\cap V_{k}\right)\right)=\int_{U_{k}\setminus\left(E\cap V_{k}\right)}\phi\,{\rm d}\mu\leq2^{k+2}\cdot\mu\left(U_{k}\setminus\left(E\cap V_{k}\right)\right)\qquad\left(\ast\right),
}
\endtoggle
$$
so that it suffices to get $\mu\left(U_{k}\setminus\left(E\cap V_{k}\right)\right)$
small.

 To this end, note that an argument similary to $\left(\ast\right)$ shows that $\mu\left(E\cap V_{k}\right)<\infty$.

I will leave the remaining details to you.
c) First try this yourself.

 The measures $\nu$ and $\nu'$ agree on open sets (by part (a)) and are both outer regular (by part (b)). Hence?

$ $

 You have $\nu'(E) = \inf \{\nu'(U) \mid U \supset E \text{ open}\} = \inf \{\nu(U) \mid \dots \} = \nu(E)$.

