Let $\Gamma$ be an abelian amenable group acting on a compact space $X$. Show that there is an invariant Radon measure $\mu$ on $X$.
Proof: Fix $x_0 \in X$. For $f \in C(X)$ define $g_f(t)=f(t.x_0)$. Then $g_f \in l^{\infty}(\Gamma)$. Now define $\psi:C(X) \to \mathbb{C}$ by $\psi(f)=\mu(g_f)$, where $\mu$ is the left invariant state on $l^{\infty}(\Gamma)$, the existence of which is given by the amenability of $\Gamma$. It is easy to see that $\psi \in C(X)^*$. By Riesz Representation Theorem, there exists $\nu$ on $X$ such that $$\psi(f)=\mu(g_f)=\int_X f d\nu$$
We already know that for $s \in \Gamma$, (Note: For $s \in \Gamma,s.\nu(E)=\nu(s^{-1}E)$)
$$\int_X f(sx)d\nu(x)=\int_Xf(x)d(s.\nu(x))$$
If we define $T:X \to X$ by $T(x)=sx$, then the above equation tells us that $$\int_X f\circ T(x)d\nu(x)=\int_Xf(x)d(s.\nu(x))$$
Now $$g_{f\circ T}(t)=f\circ T(t.x_0)=f(st.x_0)=s^{-1}.g_f(t)$$ gives us that $$\int_X fd(s\nu)=\mu(g_{f \circ T})=\mu(s^{-1}.g_f)=\mu(g_f)=\int_Xf d\nu, \forall f \in C(X)$$ which gives us that $s.\nu=\nu$ for all $s \in \Gamma$.
My Question: What goes wrong in my proof when $X$ being compact is replaced by $X$ being Locally Compact?. I was thinking that I can do the same argument with the modification of $C_0(X)$ in place of $C(X)$. But it can't be true since I know that when $\mathbb{Z}$ acts on $\mathbb{R}$ by translation, then the action doesn't induce any invariant Radon measure on $\mathbb{R}$.
You can see what goes wrong in your example of $\mathbb Z$ acting on $X=\mathbb R$ by translation. Or even simpler, take $X=\mathbb Z$.
The state $\mu:\ell^\infty(\mathbb Z)\to\mathbb C$ is basically a Banach limit (though two-sided). You can show that if $g:\mathbb Z\to\mathbb C$ tends to $0$ at infinity, then $\mu(g)=0$, and in particular the measure $\psi:C_0(X)\to \mathbb C$ you get is just the zero measure. In other words the problem is that $C_0(X)$ isn't unital.
You could extend to look at all bounded continuous functions, $\psi:C_b(X)\to \mathbb C$, but then the Riesz representation theorem fails and the functional you get isn't inner regular; it's more like a Banach limit.
Edit: here is a proof that a continuous shift-invariant linear function $g:c_0(\mathbb Z)\to\mathbb C$ must be zero (where $c_0$ means functions tending to zero, with sup norm). Let $e_i\in c_0(\mathbb Z)$ denote the function with $e_i(i)=1$ and $e_i(j)=0$ for $j\neq i$. Using linearity and shift invariance, $g(e_1)$ equals $g(\tfrac 1 N (e_1+\dots+e_N))$ for any $N$. Taking the limit as $N\to\infty$ gives $g(e_1)=g(0)=0$. By linearity and shift invariance, $g$ is zero on functions of finite support, but those are dense in $c_0(\mathbb Z)$.