Several questions about Riesz–Markov–Kakutani representation theorem This is a list of questions about Riesz–Markov–Kakutani representation theorem
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1)If $f\in L^1(\mu)$, is it true that $\phi(f)=\int_Xfd\mu$, where $\mu$ is given by the theorem? 
I am quite sure it is correct, since $C_c(X)$ is dense in $L^1(\mu)$. And we can extract a subsequence that pointwise converges to $f$.
2)If $X$ is not locally compact, does the theorem still stands? If not, does there exist a non-regular measure such that $\phi(f)=\int_Xfd\mu$?
The property of locally compact is a must to use Urysohn lemma in the proof. The reason why Urysohn lemma (or partition unity) is widely used is because the way we construct the measure. So I am thinking if we can construct the same measure in another way that we can avoid Urysohn lemma.
3) If $X$ is not Hausdorff, what will happen as stated in 2)?
4)If $X$ is normal and not locally compact?
 A: Well, as a start, if you drop local compactness, you lose uniqueness.
Suppose $X$ is Hausdorff and not locally compact.  Then there exists a point $x \in X$ such that no open neighborhood of $U$ is contained in a compact set.  Now if $f \in C_c(X)$ and $f(y) \ne 0$, then the set $U = f^{-1}(\mathbb{R} \setminus \{0\})$ is an open neighborhood of $y$ which is contained in the support of $f$, which by assumption is compact.  As such, we must have $f(x)=0$ for every $f \in C_c(X)$.  Thus if we take $\phi=0$ to be the zero functional, $\mu=0$ the zero measure, and $\nu = \delta_x$ a point mass at $x$, we have for every $f \in C_c(X)$ that $\phi(f) = \int f\,d\mu = \int f\,d\nu$.
For instance, if you take $X = \mathbb{Q}$, then every compact set has empty interior, so no point has an open neighborhood which is contained in a compact set.  Hence $C_c(\mathbb{Q})$ consists only of 0.  There are lots of Borel measures on $\mathbb{Q}$, but there is only one linear functional on $C_c(\mathbb{Q})$ (the zero functional), and it's induced by every measure.
Note that $\mathbb{Q}$ is metrizable, hence normal, and every measure on $\mathbb{Q}$ is regular.  So that part is not the issue.
A similar example is to take $X$ to be an infinite-dimensional Banach space.  Again we get $C_c(X) = 0$ by Riesz's lemma.
You can also lose uniqueness if you drop Hausdorff.  Suppose $X$ is the line with two origins, call them $0$ and $0'$.  It's easy to show that for every continuous $f : X \to \mathbb{R}$ we have $f(0) = f(0')$, and moreover the sets $\{0\}$ and $\{0'\}$ are Borel.  So the point masses $\delta_0$ and $\delta_{0'}$ are two different measures which induce the same linear functional on $C_c(X)$.
I am not sure what happens to existence in either case.
A: For your first question, $C_c(X)$ is a subspace of $L^1(\mu)$, and $\phi$ is dominated by the $L^1(\mu)$ norm on $C_c(X)$. By the Hahn-Banach theorems, $\phi$ extends to some continuous linear functional $\Lambda$ on $L^1(\mu)$. For any $f\in L^1(\mu)$, we can find a sequence $\{f_n\}$ in $C_c(X)$ such that $\|f_n-f\|_1 \rightarrow 0$ as $n\rightarrow \infty$. By continuity of $\Lambda$,
$$
\Lambda f = \lim_{n\rightarrow \infty} \int_Xf_nd\mu.
$$
Also,
$$
\lim_{n\rightarrow \infty} \bigg|\int_X(f_n-f)d\mu\bigg| \leq \lim_{n\rightarrow \infty} \int_X|f_n-f|d\mu = 0
$$
implies
$$
\int_Xfd\mu = \lim_{n\rightarrow \infty} \int_{X}f_nd\mu,
$$
hence
$$
\Lambda f = \int_Xfd\mu.
$$
This in fact shows that $\Lambda$ is the unique continuous extension of $\phi$ to $L^1(\mu)$. In other words, $C_c(X)$ and $L^1(\mu)$ have the same dual space.
Note that we never used pointwise convergence, nor do we need it. A direct application of Fatou's lemma, however, does show that $f_n(x)\rightarrow f(x)$ as $n\rightarrow \infty$ for a.e. $x \in X$.
