Proving that a mapping makes sense and defines a bounded linear functional. If we let $\mathfrak{M}=\{E\subset[0,1]:E$ is countable or $[0,1]-E$ is countable$\}$, $\mu$  be the counting measure on $\mathfrak{M}$, and $g(x)=x$ on $[0,1]$, how would we go about showing that the map $f\to\sum xf(x)=\int fgd\mu$ makes sense for every $f\in L^1(\mu)$ and defines a bounded linear functional on $L^1(\mu)$?

Because $f \in L_1$ we have that for some countable $E = (x_n)_{n \geq 1} \subseteq [0,1]$  $$\int_E f =  \sum_{n = 1}^{\infty} f(x_n) < +\infty$$
Then 
$$ \sum_{n = 1}^{\infty} x_n f(x_n) \leq \sum_{n = 1}^{\infty}  f(x_n) < +\infty $$
So it is well defined and if $T(f) = \int fg$,  then  $|T(f)| \leq ||f||_1 $
