Dense subspace of linear functionals We know that any real measurable function can be approximated by increasing simple functions. So, integral of real valued measurable function can be written as a limit of integrals of simple functions. We can observe that the integral of simple functions is just a linear combination of projection maps. I was thinking if this procedure could be done for any linear functional. To be precise, my question is as follows :

Let $X$ be a set and let $L$ be the space of all real valued functions and equip it with uniform norm. Can any linear functional on $L$ be written as a limit of linear combination of projection maps?

 A: Note that you need to require your functions to be bounded for the norm to make sense. So $L=\ell^\infty(X)$. 
Regarding your question, let us take $X=\mathbb N$. It is well-known that there are functionals that annihilate all finitely supported functions. Concretely, you take any free ultrafilter $\omega\in\beta\mathbb N\setminus \mathbb N$ and define a functional 
$$
\varphi:f\mapsto \bar f(\omega),
$$
where $\bar f$ is the extension of $f$ to $\beta\mathbb N$. For any $g\in\ell^\infty(\mathbb N)$ with $g(n)=0$, $n\geq m$ for some $m$, we will have $\varphi(g)=\lim_{n\to\infty}g(n)=0$. 
Now let $\varphi_0$ be a "linear combinations of projections maps" as you say: 
$$
\varphi_0(f)=\sum_{j=1}^hc_jf(n_j)
$$
for some fixed $n_1,\ldots,n_h\in\mathbb N$ and $c_1,\ldots,c_h\in\mathbb C$. Now define a function
$$
f(n)=\begin{cases}0,&\mbox{ if }n\in\{n_1,\ldots,n_h\}\\ 1,&\mbox{ otherwise} \end{cases}.
$$
Then $\|f\|=1$, and
$$
|\varphi(f)-\varphi_0(f)|=|1=0|=1.
$$
In other words, $\|\varphi-\varphi_0\|\geq1$, i.e. the distance from $\varphi$ to any linear combination of projection maps is at least $1$. 
*See this answer for some more information on free-ultrafilters in this context. 
