Finite-dimensional vector lattice is a direct sum of one-dimensional vector lattices Prerequisites
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 $x$.
Lemma 1 For $k=1,\ldots,n$, let $X_k$ be disjoint sets and let $\mathcal{F}_k$ be a vector lattice of functions on $X_k$. Let $X=\bigcup_{1\leq k\leq n}X_k$. Let $\mathcal{F}$ be the set of all functions $f=(f_1,\ldots,f_n)$ on $X$ such that for each $k=1,\ldots,n,f_k\in\mathcal{F}_k$ and  $f=f_k$ on $X_k$.
(a) $\mathcal{F}$ is a vector lattice.
(b) If for each $k,I_k$ is a pre-integral on $\mathcal{F}_k$, for $f\in\mathcal{F}$ let $I(f):=\sum_{1\leq k\leq n}I_k(f_k)$. Then $I$ is a pre-integral on $\mathcal{F}$. (Then $(X,\mathcal{F},I)$ is called the direct sum of the $(X_k,\mathcal{F}_k,I_k)$.)
Lemma 2 Let $g_1,\ldots g_k$ be linearly independent real-valued functions on a set $X$, that is, if for real constants $c_j,\sum^k_{j=1}c_jg_j\equiv0$, then $c_1=c_2=\cdots=0$. Then for some $x_1,\ldots,x_k$ in $X,g_1,\ldots g_j$ are linearly independent on $\{x_1,\ldots,x_j\}$ for each $j=1,\ldots,k$.
Lemma 3 Let $\mathcal{F}$ be a vector lattice of functions on a set $\{p,q,r\}$ of three points such that for some $b$ and $c,f(p)\equiv bf(q)+cf(r)$ for all $f\in\mathcal{F}$. Then either $\mathcal{F}$ is one-dimensional, or $b\geq0,c\geq0$, and $bc=0$.
Proof Each $f\in\mathcal{F}$ can be written as a $3\times1$ vector
\begin{equation}
f=\begin{bmatrix}
bf(q)+cf(r)\\
f(q)\\
f(r)
\end{bmatrix}.
\end{equation}
Suppose there exists $f\in\mathcal{F}$ with $f(q)>0$ and $f(r)>0$. Then
\begin{equation}
f\land -f=\begin{bmatrix}
[bf(q)+cf(r)]\land -[bf(q)+cf(r)]\\
-f(q)\\
-f(r)
\end{bmatrix}.
\end{equation}
Since $f\land -f\in\mathcal{F}$, we must have $bf(q)+cf(r)\geq-[bf(q)+cf(r)]$, i.e., $bf(q)+cf(r)\geq0$. Therefore, $b$ and $c$ cannot be both negative. If one of them is nonpositive, the other must be nonnegative. If $f(q)=f(r)$, then $b+c\geq0$.
If $\mathcal{F}$ is one-dimensional, let $f$ be the basis. Then $f$ must satisfy $f(p)=bf(q)+cf(r)$. This relation is automatically satisfied by all the other functions in $\mathcal{F}$ and there is no constraint on $b$ and $c$.
Now consider the case that $\mathcal{F}$ is not one-dimensional. Then there exists a nonzero $g\in\mathcal{F}$ such that $g$ is not a multiple of nonzero $f$. Then
\begin{equation}
f\land g=\begin{bmatrix}
[bf(q)+cf(r)]\land [bg(q)+cg(r)]\\
f(q)\land g(q)\\
f(r)\land g(r)
\end{bmatrix}
\end{equation}
also belongs to $\mathcal{F}$. Note that $f(q)$ and $f(r)$ cannot be both zero. Similarly $g(q)$ and $g(r)$ cannot be both zero. Also $f(q)$ and $g(q)$ cannot be both zero. Otherwise $f$ is a multiple of $g$. Similarly $f(r)$ and $g(r)$ cannot be both zero.
(1) $f(q)=0$ and $g(r)=0$.
\begin{equation}
f\land g=\begin{bmatrix}
cf(r)\land bg(q) \\
0\land g(q)\\
f(r)\land 0
\end{bmatrix}.
\end{equation}
The above equation must hold when $f(r)$ and $g(q)$ take any nonzero values since any multiple of $f$ or $g$ belongs to $\mathcal{F}$. Let's take $g(q)>0$ and $f(r)>0$. Then
\begin{equation}
f\land g=\begin{bmatrix}
cf(r)\land bg(q) \\
0\\
0
\end{bmatrix}.
\end{equation}
Since $f\land g\in\mathcal{F}$, we must have $cf(r)\land bg(q)=b0+c0=0$. Either $c=0$ or $b=0$. Let's take $c=0$. Then for any $g(q)$ and $f(r)$,
\begin{equation}
f\land g=\begin{bmatrix}
0\land bg(q) \\
0\land g(q)\\
f(r)\land 0
\end{bmatrix}.
\end{equation}
Let's pick $g(q)>0$ and $f(r)<0$. Then
\begin{equation}
f\land g=\begin{bmatrix}
0\land bg(q) \\
0\\
f(r)
\end{bmatrix}.
\end{equation}
Since $0\land bg(q)=b0+cf(r)=cf(r)=0$ must hold, we must have $bg(q)\geq0$ and thus $b\geq0$. Similarly, if $b=0$, we must have $c\geq0$.
(2) $f(r)=0$ and $g(q)=0$. Same conclusion as in the last case.
(3) $f(q)=0$ and $g(r)\neq0$. We can add appropriate multiple of $f$ to $g$ to make $g(r)=0$. Note that $g(q)$ would not be affected since $f(q)=0$. Then the conclusion is the same as in case 1.
(4) $g(q)=0$ and $f(r)\neq0$. Same conclusion as in the last case.
(5) $f(r)=0$ and $g(q)\neq0$. Same conclusion as in case 3.
(6) $g(r)=0$ and $f(q)\neq0$. Same conclusion as in the last case.
(5) All of $f(q),f(r),g(q)$ and $g(r)$ are nonzero. We can add appropriate multiple of $f$ to $g$ to make $g(r)=0$. Same conclusion as in the last case.
Problem
Let $\mathcal{F}$ be a vector lattice of functions on a set $X$ such that $\mathcal{F}$ is a finite-dimensional vector space. Show that $\mathcal{F}$ is a direct sum (as in Lemma 1) of one-dimensional vector lattices. Hint: Use Lemma 2 and Lemma 3.
My efforts
$\mathcal{F}$ being one-dimensional is trivial case. We will assume $\mathcal{F}$ to be at least two-dimensional.
(1) $X=\{p_1,p_2,\ldots,p_k\}$ and $\mathcal{F}$ is $k$-dimensional where $k$ is a positive integer. Then each $f\in\mathcal{F}$ can be represented as a $k\times1$ vector whose $j$th entry is $f(p_j)$. We can take the $k$ functions $f_1,f_2,\ldots,f_k$ whose vector representations are $[1\;0\;\ldots\;0]^T,[0\;1\;\ldots\;0]^T$ and $[0\;0\;\ldots\;1]^T$ as bases for $\mathcal{F}$. Then $\mathcal{F}$ is a direct sum of $k$ one-dimensional vector lattices $\{cf_1(p):c\in\mathbb{R}\}$ on $\{p\}$ and $\{cf_2(q):c\in\mathbb{R}\}$ on $\{q\}$.
(3) $X=\{p,q,r\}$. Assume $\mathcal{F}$ to be two-dimensional. Each $f\in\mathcal{F}$ can be represented as $f=[f(p)\;f(q)\;f(r)]^T$. Denote the two bases for $f\in\mathcal{F}$ as $b_1=[b_{11}\;b_{12}\;b_{13}]^T$ and $b_2=[b_{21}\;b_{22}\;b_{23}]^T$. Then $f$ can be written as $f=c_1b_1+c_2b_2$ where $c_1$ and $c_2$ are real coefficients. To determine $c_1$ and $c_2$, we need to solve the equation
\begin{equation}
B[c_1\;c_2]^T=f.
\end{equation}
where $B=[b_1\;b_2]$. This non-homogeneous equation system must have a solution. The matrix $B$ has rank 2. One row, say, row 1, of the augmented matrix $[B\;f]$ must be linear combination of the other two and be canceled by row operation. Note that the row to be canceled is determined by $B$ and is thus independent of $f$. Thus for some $b$ and $c,f(p)\equiv bf(q)+cf(r)$ for all $f\in\mathcal{F}$. By lemma 3, either $\mathcal{F}$ is one-dimensional, or $b\geq0,c\geq0$, and $bc=0$. But $\mathcal{F}$ is two-dimensional by assumption. So we have $b\geq0,c\geq0$, and $bc=0$. Then $B$ has rank 1, a contradiction. Therefore a vector lattice on a set of three points must be either one-dimensional or three dimensional.
Let $\mathcal{F}$ be vector lattice on a four-point set. Is it still vector lattice if all functions are restricted on a three-point subset? Yes.
I don't know how to apply Lemma 3 to set with more than 3 points.
 A: Let the $k$ bases of $\mathcal{F}$ be $f_1,f_2,\ldots,f_k$. Each $f\in\mathcal{F}$ is completely determined by its value $f(x)$ at each $x\in X$. We can consider $f(x)$ as coordinates. This is intuitive when $X$ is countable. Even if $X$ is uncountable, we can still take this view. We can write the basis functions in the form of column vectors:
\begin{equation}
f_1 = 
\begin{bmatrix}
f_1(x_1) \\
f_1(x_2) \\
\vdots \\
f_1(x_k) \\
\vdots \\
f_1(x_\alpha) \\
f_1(x_\beta) \\
\vdots
\end{bmatrix},
f_2 = 
\begin{bmatrix}
f_2(x_1) \\
f_2(x_2) \\
\vdots \\
f_2(x_k) \\
\vdots \\
f_2(x_\alpha) \\
f_2(x_\beta) \\
\vdots
\end{bmatrix}, \cdots,
f_k = 
\begin{bmatrix}
f_k(x_1) \\
f_k(x_2) \\
\vdots \\
f_k(x_k) \\
\vdots \\
f_k(x_\alpha) \\
f_k(x_\beta) \\
\vdots
\end{bmatrix},
\end{equation}
When $X$ is uncountable, we can view the column vector as a vertical line. We can pick the $x_k$ here as in Lemma 2. Note that the $x_k$ and $x_\alpha$ are not necessarily in ascending order. Then by column operations on the matrix $[f_1\;f_2\;\cdots\;f_k]$, we can obtain the new basis functions
\begin{equation}
b_1 = 
\begin{bmatrix}
1 \\
0 \\
\vdots \\
0 \\
\vdots \\
a_{1,\alpha} \\
a_{1,\beta} \\
\vdots
\end{bmatrix},
b_2 = 
\begin{bmatrix}
0 \\
1 \\
\vdots \\
0 \\
\vdots \\
a_{2,\alpha} \\
a_{2,\beta} \\
\vdots
\end{bmatrix}, \cdots,
b_k = 
\begin{bmatrix}
0 \\
0 \\
\vdots \\
1 \\
\vdots \\
a_{k,\alpha} \\
a_{k,\beta} \\
\vdots
\end{bmatrix}.
\end{equation}
Pick any two of the basis functions $b_i$ and $b_j$.
\begin{equation}
b_i\land b_j = 
\begin{bmatrix}
0 \\
0 \\
\vdots \\
0 \\
\vdots \\
a_{i,\alpha}\land a_{j,\alpha} \\
a_{i,\beta}\land a_{j,\beta} \\
\vdots
\end{bmatrix}
\end{equation}
Since $\mathcal{F}$ is vector lattice, $b_i\land b_j\in\mathcal{F}$ and must be linear combination of $b_i$ and $b_j$. Since the first $k$ entries of $b_i\land b_j$ are all zero,
\begin{equation}
b_i\land b_j=0b_i+0b_j=0.
\end{equation}
Thus
\begin{equation}
a_{i,\alpha}\land a_{j,\alpha}=0,
\end{equation}
\begin{equation}
a_{i,\beta}\land a_{j,\beta} =0,
\end{equation}
\begin{equation}
\cdots
\end{equation}
Since the above equations hold for any $i\neq j$,
\begin{equation}
a_{1,\alpha}=a_{2,\alpha}=\ldots=a_{k,\alpha}=0,
\end{equation}
\begin{equation}
a_{1,\beta}=a_{2,\beta}=\ldots=a_{k,\beta} =0,
\end{equation}
\begin{equation}
\cdots
\end{equation}
It is simply the same as the vector space direct sum decomposition.
