Why the element of dual space of linfinity can be represented as sum of l1 and c0 elements?

Obviously, the OP intended to ask about this sentence "$f\in\ell_\infty^*$ is the sum of an element of $\ell_1$ and an element null on $c_0$" from the paper D. H. Fremlin and M. Talagrand: A Gaussian Measure on $l^\infty$ http://jstor.org/stable/2243023 (Which is different claim from what was in the question.) The authors refer to the book Day, M. (1973). Normed Linear Spaces. Springer, Berlin. I was not able to find the exact place in Day's book where this is shown, but I think that for this special case it is relatively easy. For $f\in\ell_\infty^*$ put $a_i=f(e^i)$. Then the sequence $a=(a_i)$ belongs to $\ell_1$. (Since $\sum\limits_{i=1}^n a_i = \sum\limits_{i=1}^n f(e^i) = f(\sum\limits_{i=1}^n \varepsilon_ie^i) \le \lVert f \rVert$, where $\varepsilon_i=\pm1$ are chosen according to the signs of $f(e^i)$.) Now, if $x_n\to 0$, then $$f(x)a^*(x)= \lim\limits_{n\to\infty} f(\sum\limits_{i=1}^n x_ie^i)\sum\limits_{i=1}^n a_ix_i=0.$$ I hope I haven't overlooked something and that someone will provide the reference to the result (probably more general) which the authors of the abovementioned paper had in mind. 


The fact stated above by Martin is a special case of the general property of a bounded functional on a von Neumann algebra  it can be always decomposed into a sum of a normal functional (in other words an image of a functional in the predual, in this case a functional represented by a sequence in $l^1$) and a singular functional (a `highly nonnormal' functional, in the special case a functional vanishing on $c_0$). One can even achieve the decomposition respecting the functional norms in a suitable sense The general result together with some discussion can be found in the first volume of Takesaki's `Theory of Operator Algebras'. 

