Let $X$ be a paracompact Hausdorff space and let $C_c(X)$ be the set of all continuous real-valued functions on $X$ with compact support. By a positive integral on $X$ I will mean a linear function $I: C_c(M) \to \mathbb{R}$ such that $$f \geq 0 \implies I(f) \geq 0. $$ A strictly positive integral on $X$ will be a positive integral $I$ on $X$ for which if $f \geq 0$ and $I(f) = 0$, then $f = 0$.
Let us assume that, for every $x \in X$, there is an open set $U_x$ containing $x$ such that there exists a strictly positive integral on $U_x$, denoted by $I_x : C_c(U_x) \to \mathbb{R}$. How can we show that there exists a strictly positive integral on $X$?
Of course, I immediately thought about partitions of unity. Since $X$ is paracompact and since the set $\{ U_x\}_{x \in X}$ is an open cover of $X$ (where $U_x$ is defined as above), there exists a partition of unity subordinated to this open cover, let us denote it by $\{\psi_x \ \mid \ x \in X \}$. We can also assume that $psi_x(y) \in [0,1]$, for every $y \in X$ and for every $x \in X$.
It is not difficult to see that for every $x \in X$, since $\text{supp} (\psi_x) \subset U_x$, we have that $I_{\psi_x}: f \mapsto I_x(\psi_x \cdot f)$ is a strictly positive integral.
Thus, I believe that the strictly positive integral on $M$ should be $$I = \sum_{x \in X} I_{\psi_x}. $$ However, how would we prove that this is continuous and linear, since I don't think that the sum above has only finitely many non-zero terms.