linear algebra foundation of Riemann integrals Let $V$ be the vector space of real functions $f\colon [a,b]\to \mathbb R$ and let $X$ be the set of characteristic (indicatrix) functions of subintervals: $X=\{\mathbb 1_I\colon I\subset [a,b] $ interval $\}$. We define $T\colon X \to \mathbb R$ as $T(\mathbb 1_I) = |I|$ where $|I|$ is the length of the interval $I$. Notice that $X$ is not a set of independent vectors because the sum of adjacent intervals is again an interval, but in that case $T$ is defined to be additive.
So it is clear that $T$ can be extended as a linear map on the vector space generated by $X$ (which is the space of so called simple functions). 
What are the abstract properties of $X$ and $T$ (in the setting of linear algebra) which can be applied to the above example to prove that $T$ is linear on $X$?
For example $X$ and $T$ have the following property: $x,y,x+y\in X \implies T(x+y) = T(x)+T(y)$ and $x,\lambda x \in X \implies \lambda=1$. Is this enough to prove that $T$ is linear i.e. that $x,y,\lambda x +\mu y \in X \implies T(\lambda x + \mu y) = \lambda T(x) + \mu T(y)$? And is this enough to prove that $T$ has a linear extension to the span of $X$?
 A: I'm not entirely sure I've understood your question properly, but here's an attempt.
Proposition: Let $V$ be a real vector space, let $X \subset V$, let $W = \operatorname{span} X$, and let $f: X \rightarrow \mathbb{R}$ be a function.  Then:
a) There is at most one linear function $F: W \rightarrow \mathbb{R}$ such that $F|_X = f$.
b) If such a function exists, there is at least one linear function $G: V \rightarrow \mathbb{R}$ such that $G|_X = f$.
c) The following are equivalent:
(i) There is a linear extension $F$ to $W$, namely: for $x_1,\ldots,x_k \in X$ and $\alpha_1,\ldots,\alpha_k \in \mathbb{R}$, 
$F(\alpha_1 x_1 + \ldots + \alpha_k x_k) = \alpha_1 f(x_1) + \ldots + \alpha_k f(x_k)$.
(ii) For all $x_1,\ldots,x_k \in X$ and $\alpha_1,\ldots,\alpha_k \in \mathbb{R}$, if 
$\alpha_1 x_1 + \ldots + \alpha_k x_k = 0$, then $\alpha_1 f(x_1) + \ldots + \alpha_k f(x_k) = 0$.  
Proof: a) If there is such a linear extension $F$ to $W$, then it is uniquely determined by the formula of condition (i) of part c), so it is unique.  b) Every linear map on a subspace $W$ extends to a linear map on the entire space $V$: just choose a basis for the subspace, extend it to a basis for the entire space, and define the map arbitrarily on the basis vectors lying outside of $W$.  c) The implication (i) $\implies$ (ii) is immediate.  For (ii) $\implies$ (i): we would like to define the extension $F$ by the formula of (i): 
$F(\alpha_1 x_1 + \ldots + \alpha_k x_k) = \alpha_1 f(x_1) + \ldots + \alpha_k f(x_k)$.  
But we need to check that this is well-defined: suppose $x \in W$ has two expressions
$\alpha_1 x_1 + \ldots + \alpha_k x_k = x = \beta_1 x_1 + \ldots + \beta_k x_k$ 
for $\alpha_1,\ldots,\alpha_k,\beta_1,\ldots,\beta_k \in \mathbb{R}$.  Then
$(\alpha_1 - \beta_1)x_1 + \ldots + (\alpha_k - \beta_k) x_k = 0$, so using (iii) we get 
$0 = (\alpha_1-\beta_1)f(x_1) + \ldots + (\alpha_k-\beta_k)f(x_k)$,
and thus 
$\alpha_1 f(x_1) + \ldots + \alpha_k f(x_k) = \beta_1 f(x_1) + \ldots + \beta_k f(x_k)$.
Notice that condition (ii) of part c) is certainly satisfied if $X$ is linearly independent.
I invite you to verify criterion (ii) in the example you've given.  
A: I think the only thing that needs to be proved for $T$ to be extendable to a linear map on the vector space generated by $X$ is the following. Assume $f=\mathbb{1}_{[c,d)}$ and $g = \mathbb{1}_{[d,e)}$ with $f,g \in X$. Then $f+g$ is also in $X.$ (I am assuming that half-open intervals are used here.) It is then needed that $T(f+g) = T(f) + T(g)$ for all such $f$ and $g$ in $X.$ 
This is because if the above is true (which it is!), then you can show that any function that can be written as a linear combination of elements of $X$ (i.e. a simple function) maps to a unique value by the extension of $T$, regardless of how it is written as a linear combination.
