Definition 1 Let $\mathcal{L}$ be a non-empty collection of real-valued functions on a set $X$. Then $\mathcal{L}$ is a real vector space iff for all $f,g\in\mathcal{L}$ and $c\in\mathbb{R},cf+g\in\mathcal{L}$. Let $f\lor g:=\mathrm{max}(f,g),f\land g:=\mathrm{min}(f,g)$. A vector space $\mathcal{L}$ of functions is called a vector lattice iff for all $f$ and $g$ in $\mathcal{L}$, $f\lor g\in\mathcal{L}$. Then also $f\land g\equiv-(-f\lor-g)\in\mathcal{L}$. The vector lattice $\mathcal{L}$ will be called a Stone vector lattice iff for all $f\in\mathcal{L}, f\land1\in\mathcal{L}$.
Definition 2 Given a set $X$ and a vector lattice $\mathcal{L}$ of real functions on $X$, a pre-integral is a function $I$ from $\mathcal{L}$ into $\mathbb{R}$ such that:
(a) $I$ is linear: $I(cf+g)=cI(f)+I(g)$ for all $c\in\mathbb{R}$ and $f,g\in\mathcal{L}$.
(b) $I$ is nonegative, in the sense that whenever $f\in\mathcal{L}$ and $f\geq0$ (everywhere on $X$), then $I(f)\geq0$.
(c) $I(f_n)\downarrow0$ whenever $f_n\in\mathcal{L}$ and $f_n(x)\downarrow0$ for all $x$.
Theorem (Stone-Daniell) Let $I$ be a pre-integral on a Stone vector lattice $\mathcal{L}$. Then there is a measure $\mu$ on $X$ such that $I(f)=\int fd\mu$ for all $f\in\mathcal{L}$. The measure $\mu$ is uniquely determined on the smallest $\sigma$-ring $\mathcal{B}$ for which all functions in $\mathcal{L}$ are measurable.
Problem Let $f\geq0$ be a function on a set $X$ and $\mathcal{L}:=\{cf:c\in\mathbb{R}\}$. Then $\mathcal{L}$ is a vector lattice but not necessarily a Stone vector lattice. If $I$ is a pre-integral on $\mathcal{L}$, is it always true that $I(f)=\int fd\mu$ for some measure $\mu$?
My efforts
I cannot apply the Stone-Daniell Theorem since $\mathcal{L}$ is not necessarily a Stone vector lattice. Then I turned back to the construction of $\mu$ in the proof of the Stone-Daniell Theorem and found that $f\land1$ was used at the very beginning of the construction.
$\mathcal{L}$ is spanned only by $f$. If $I(f)$ is prescribed, then all $I(cf)=cI(f)$ is prescribed. So we only need to consider $f$.
For now let's take $X=\mathbb{R}$ and $\lambda$ be the Lebesgue measure. If $\int fd\lambda=M<\infty$, we can just take $\mu=I(f)\lambda/M$. If $\int fd\lambda=\infty$, e.g., $f(x)=e^x$, then I don't know how to continue.
I also think of using the following theorem:
Theorem (Generated $\sigma$-algebra) Let $(\Omega',\mathcal{A}')$ be a measurable space and let $\Omega$ be a nonempty set. Let $X:\Omega\rightarrow\Omega'$ be a map. The preimage $X^{-1}(\mathcal{A}'):=\{X^{-1}(A'):A'\in\mathcal{A}'\}$ is the smallest $\sigma$-algebra with respect to which $X$ is measurable. We say that $\sigma(X):=X^{-1}(\mathcal{A}')$ is the $\sigma$-algebra on $\Omega$ that is generated by $X$.
And I face the similar problem of $f(x)=e^x$ again.
This answer is provided by Dr. Xue Liu, the author of A new proof for the Daniel-Stone theorem for random probability measures.
If $f\equiv0$, then $I(f)=\int fd\mu=0$. Any measure $\mu$ works.
If $f\not\equiv0$, then there exists $z\in X$ with $f(z)>0$. We define $\mu:=\delta_{z}\times I(f)/f(z)$, where $\delta_{z}$ is the Dirac measure of $\{z\}$. Then $\int f d\mu=I(f)$.