Let $(X, \mathcal A, \mu)$ be a probability space, and $v\in \mathcal M^+(\mathcal A)$ be a positive measurable function. I am interested in how to show that
$$ \left| \int_X v ~ \text d\mu \right| \leq \sup\{|v(x)|~:~x\in X\} $$
I am completely stooped, and don't even know where to begin.
Since $v$ is positive, we don't need to take absolute values. We know that $$\int_X v\,d\mu$$ is given by the limit of the integrals of simple functions $s=\sum_{k= 0}^n a_k\chi_{A_k}\leq v$, where the coefficients $a_k$ are nonnegative and the sets $A_k$ are disjoint and measurable. Note the coefficients satisfy $a_k\leq\sup\{v(x): x\in X\}$ because $s\leq v$ on $A_k$. These integrals are equal to $$\sum_{k= 0}^n a_k\mu(A_k)\leq\max\{a_1,\ldots, a_n\}\sum_{k= 0}^n\mu(A_k)=\max\{a_1,\ldots, a_n\}\mu(A_1\cup\cdots\cup A_k)\\ \leq\max\{a_1,\ldots, a_n\}\mu(X)\leq\sup\{v(x): x\in X\}$$ Thus $\sup\{v(x): x\in X\}$ is an upper bound to $\int_X v\,d\mu$.