Consider $$ f^+(0):=\lim_{r\searrow 0}\frac{f(r)}{r},~~~~~~\mathcal{F}:=\left\{f\in C([0,1],\mathbb{R}): f(0)=0, f^+(0)\text{ exists}\right\}. $$ Moreover, let $\mathcal{F}^+$ be the set of all non-negative function of $\mathcal{F}$ and $\mathcal{F}'$ the vector lattice of all linear forms over $\mathcal{F}$. Finally, let $$ (\mathcal{F}')^+:=\left\{L\in\mathcal{F}': \forall f\in\mathcal{F}^+~L(f)\geq 0\right\}. $$ $L\in\mathcal{F}'$ is called $\sigma$-continous, if $$ (f_n)\in\mathcal{F}^{\mathbb{N}}\text{ with } f_n\searrow 0\implies\lim_{n\to\infty}L(f_n)=0. $$
I would like to show that
(1) $\mathcal{F}$ is a $\mathbb{R}$-vector space of real-valued functions on $[0,1]$.
(2) For each $f,g\in\mathcal{F}$, we have that $$ f\vee g\colon\ x\mapsto\max\left\{f(x),g(x)\right\},~~~f\wedge g\colon x\mapsto \min\left\{f(x),g(x)\right\} $$ are in $\mathcal{F}$.
(3) For $f\in\mathcal{F}$, it follows that $f\wedge 1_{[0,1]}\in\mathcal{F}$, where $1_{(0,1)}$ is the indicator function on $[0,1]$.
(4) $L(f):=f^+(0)\in (\mathcal{F}')^+$ and $L(f)$ is not $\sigma$-continuous
I think (1), (2) and (3) can be shown this way:
(1) should be clear. Just use $(f+g)(x):=f(x)+g(x)$ as addition, and $(\alpha f)(x):=\alpha f(x)$ as multiplication and all the vector space axioms are fulfilled since the functions in $\mathcal{F}$ are real-valued.
(2) First of all, the maximum and minimum of two continuous functions is continious. Moreover, we have $$ (f\vee g)(x)=\frac{f(x)+g(x)+\lvert f(x)-g(x)\rvert}{2},~~(f\wedge g)(x)=\frac{f(x)+g(x)-\lvert f(x)-g(x)\rvert}{2}. $$
So $(f\vee g)(0)=0$ since $f(0)=g(0)=0$ by assumption and since the absolute value function is continuous, $$ (f\vee g)^+(0)=\lim_{r\searrow 0}\left(\frac{f(r)}{2r}+\frac{g(r)}{2r}+\left\lvert\frac{f(r)-g(r)}{2r}\right\rvert\right)=\frac{1}{2}(f^+(0)+g^+(0))+\frac{1}{2}\lvert f^+(0)-g^+(0)\rvert $$ which exists by assumption.
All together, $f\vee g\in\mathcal{F}$. In the same way, $f\wedge g\in\mathcal{F}$.
(3) Let $0<\varepsilon <1$. Since $f$ is continuous in $x=0$ and $f(0)=0$, there exists some $\delta >0$ such that $$ \lvert f(x)-f(0)\rvert=\lvert f(x)\rvert\leq\varepsilon~~\forall~x\in [0,\delta]. $$ This implies $$ (f\wedge 1_{[0,1]})(x)=f(x)~\forall~x\in [0,\delta]. $$ Hence $(f\wedge\mathbb{1})^+(0)=f^+(0)$ which exists by assumption. Moreover, one can use again $$ (f\wedge 1)(x)=\frac{f(x)+1(x)-\lvert f(x)-1(x)\rvert}{2} $$ to see that $(f\wedge 1)(0)=0$ and that $f\wedge 1$ is continuous on $[0,1]$. All together, $f\wedge 1\in\mathcal{F}$.
(4) without idea so far
Thanks for hints concerning (4) and feedback for (1), (2), (3).