Claim: Suppose that $X$ is a compact Hausdroff space and $\nu$ is a finite probability measure on $X$. The map $j: C(X) \to L^{\infty}(X,\nu)$ is injective iff $\nu$ has full support.
Note: Support($\nu$) is all those points $x \in X$ for which every neighborhood $N_x$ around $x$ has positive measure. It is equivalent to saying that $\nu(G)>0$ for all nonempty open sets $G$. Suppose that $\nu(G)$ is positive for every non empty open set. Then $\nu(N_x)$ is positive for every neighborhood $N_x$ around $x$. On the other hand suppose that Support($\nu$)=$X$. Let $G$ be a nonempty open set. Let $x \in G$. Since $G$ is a neighborhood of $x$, $\nu(G) > 0$.
Proof: Suppose that $\nu$ has full support and $j(f)=0$. Then $\|f\|_{\infty}=0$. Let $\epsilon >0$. Then there exists $a > 0$ such that $a < \|f\|_{\infty}+\epsilon=\epsilon$ and $\nu\left(\{x \in X: |f(x)| > a\}\right)=0$. Thus, $\nu\left(\{x \in X: |f(x)|>\epsilon\}\right)=0$ since the latter set is contained in the former. Since $f$ is continuous, $\{x \in X: |f(x)| > \epsilon\}$ is open. Since $\nu$ has full support, the set $\{x \in X: |f(x)| > \epsilon\}$ is empty. Thus $|f(x)| \le \epsilon$ for all $x \in X$. Since $\epsilon$ is arbitrary, we must have that $|f(x)|=0$ for all $x \in X$. Thus $f=0$.
On the other hand suppose that $\nu$ doesn't have full support. Then there exists a nonempty open set $G$ in $X$ such that $\nu(G)=0$. Fix $g \in G$. By Uryhson's Lemma, there exists $f \in C(X)$ such that $f(g)=1$ and $f|_{G^c}=0$. Let $\epsilon >0$. Since $\{x \in X: |f(x)| > \epsilon\} \subset G$, we have $\nu\left(\{x \in X: |f(x)| > \epsilon\}\right) \le \nu(G)=0$. Thus, $\epsilon \in \{a \ge 0: \nu\left(\{x \in X: |f(x)| > a\}\right)=0 \}$. Since, $\|f\|_{\infty}=\inf\{a \ge 0: \nu\left(\{x \in X: |f(x)| > a\}\right)=0 \}$, we have $\|f\|_{\infty} \le \epsilon$. Since $\epsilon > 0$ was arbitrary, we have $\|f\|_{\infty}=0$. But $f \ne 0$.
This looks right to me. I just wanted to have it double checked. Thanks for the help!!
The proof looks fine to me. It might be a little simpler if you start with a lemma that for a general measurable function $\|f\|_\infty = 0$ if and only if $f =0$ almost everywhere.