I meet this problem:
$\Lambda(f)$ is a nonnegative bounded linear functional on $C[0,\infty)$. Assume $\Lambda(1) = 1$. Then $\Lambda$ has a representation $\Lambda(f) = \int_{R^+} f \mathrm{d}\mu$ if and only if $\Lambda$ satisfies $f_n \downarrow 0 \Rightarrow \Lambda(f_n) \rightarrow 0$, i.e. the monotone convergence theorem holds. Why don't we need this condition in the compact case?
One direction is just the monotone convergence theorem. I have difficulty dealing with the other side. When learning measure theory, I learned that a set function $\mu$ over a ring $R \subset \mathcal{P}(X)$ which is nonnegative, finitely additive, takes $0$ at $\varnothing$ and $\mu(X)<\infty$ is a measure iff $\mu$ is continuous at $\varnothing$. This seems similar to this problem, but I don't know how to proceed. Thank you for any help!
I haven't worked out the details, but I'll give you an idea. The spirit of the Riesz-Markov theorem is that you want to define $$ \mu(X)=\Lambda(1_X). $$ The problem is that $\Lambda$ is not defined on non-continuous functions. But I think that you can use your condition to extend $\Lambda$ to the simple functions.
Edit: some (scarce) details of what I meant in the comment. Easier answer after that.
If $V$ is an open set, then it is a disjoint countable union of intervals; this allows us to construct continuous functions $\{f_n\}$ such that $f_n\nearrow 1_V$. The condition on $\Lambda$ guarantees that $\{\Lambda(f_n)\}$ is Cauchy. So define $$ \mu(V)=\lim\Lambda(f_n). $$ This is well-defined, again by the property of $\Lambda$.
Given a Borel set $E\subset\mathbb R$, define $$ \mu(E)=\inf\{\mu(V):\ E\subset V,\ E\ \text{ open } \}. $$ And you can continue by mimicking the proof of the RRT.
Easier: since $C_c[0,\infty)\subset C[0,\infty)$, the RRT applies to give us a regular Borel measure $\mu$ with $\lambda f=\int_{[0,\infty)} f\,d\mu$ for all $f\in C_c[0,\infty)$. Given $f\in C[0,\infty)$ positive, let $g_n$ be continuous with $g_n\geq0$, $g_n=1$ on $[0,n]$ and $g_n=0$ on $[n+1,\infty)$. Then $fg_n\nearrow f$. As $f-fg_n\searrow0$, we have $\Lambda f=\lim_n\Lambda(fg_n)$. Then, by Monotone Convergence, $$ \Lambda f=\lim_n\Lambda(fg_n)=\lim_n\int_{[0,\infty)}fg_n\,d\mu=\int_{[0,\infty)}f\,d\mu. $$
Now any $f\in C[0,\infty)$ is a linear combination of positive functions, so the equality holds for all $f$.