I am writing concerning support of functionals on C*-algebras. I feel there are two different definitions for this notion. Let $A$ be a C*-algebra and $\phi$ be a positive linear functional on $A$. Then one may also consider $\phi$ as a $w^*$-continuous linear functional on $A^{**}$.
1st (Based on the Takesaki's book Vol I- page 134)
There is a unique projection in $e \in A^{**}$ with $e\phi=e\phi e=\phi$ and $\phi$ is faithful on $eAe$, such a projection is called the support of $\phi$.
2ed (Based on the Sakai's book- page 31) There is a largest projection $q\in A^{**}$ with $\phi(q)=0$. Then $1-q$ is called the support of $\phi$.
Example. Let us consider the functional $\phi:C[0,1]\to \mathbb{C}$ given by $\phi(f)=\int_0^1 f(t)dt$. I think by the 1st approach, the support of $\phi$ is just the constant function $1$. Obviously $\phi$ vanishes on all characteristic functions $\chi_I$, where $I$ is a countable subset in $[0,1]$. These characteristic functions are all in $C[0,1]^{**}$ and so by the second approach the support of $\phi$ is not $1$.
Do you confirm that these are different too!?
The two approaches are the same: $e=1-q$.
Your question assumes that the functional extends like integration against Lebesgue measure; that's not the case (it doesn't even make sense). Below is a proof of the equivalence of the two approaches.
1 $\implies$ 2: Let $K\subset A^{**}$ be the set of projections $$K=\{p\in A^{**}:\ \phi(p)=0\}$$ and let $q$ be the supremum of $K$. So we can choose an increasing net of projections $q_j\in K$ with $q_j\nearrow q$. As every functional of $A$ becomes normal on $A^{**}$, we have $$\phi(q)=\lim\phi(q_j)=0.$$ Note that $1-e\leq q$ (because $\phi(1-e)=0$). Also, since $\phi$ is positive, $$|\phi(qe)|\leq\phi(q)^{1/2}\phi(e)^{1/2}=0.$$ So $\phi(eqe)=\phi(qe)=0$, and as $\phi$ is faithful on $eAe$, $eqe=0$. Then $qe=0$, and it follows that $q=(1-e)q$, and so $q\leq 1-e$. Then $q=1-e$ $$ \ $$ 2 $\implies$ 1: let $e=1-q$. The computation above shows that $\phi(qa)=0$ for all $a\in A^{**}$. Thus $$\phi(a)=\phi(aq)+\phi(ae)=\phi(ae).$$ Similarly, $\phi(a)=\phi(ea)$; so $$\phi(a)=\phi(eae)=\phi(ae)=\phi(ea).$$ Finally, if $0=\phi(eae)$, then $\phi(ae)=0$. Subsequently, $\phi(a^ne)=0$ for all $n$, and by linearity and w$^*$-continuity, $\phi(x)=\phi(xe)=0$ for all $x\in W^*(a)$. Let $r$ be any projection in $W^*(A)$; we have $\phi(r)=0$. Then $r\leq q=1-e$. Thus $ere=0$; as $r$ is any projection in $W^*(a)$, $exe=0$ for all $x\in W^*(a)$, and so $eae=0$ and $\phi$ is faithful on $eAe$.