I need to prove the following result:
Remark 4.30$\quad$ Let $M(X,\mathscr{A},\mathbb{R})$ be the collection of all finite signed measures on $(X,\mathscr{A})$. Let $B(X,\mathscr{A},\mathbb{R})$ be the vector space of bounded real-valued $\mathscr{A}$-measurable functions on $X$. Let $\mu\in M(X,\mathscr{A},\mathbb{R})$ and let $f\in B(X,\mathscr{A},\mathbb{R})$. Define a function $F:M(X,\mathscr{A},\mathbb{R})\to\mathbb{R}$ by letting $$ F(\mu) = \int fd\mu. $$ Then $F$ is a linear functional.
Here is my attempt:
Proof$\quad$ We first show that if $f$ is an $\mathscr{A}$-measurable characteristic function of a set $A\in\mathscr{A}$, then $F$ is a linear functional. So let $f=\chi_A$, then $\int fd\mu=\int\chi_Ad\mu=\int\chi_Ad\mu^+-\int\chi_Ad\mu^-=\mu^+(A)-\mu^-(A)=\mu(A)$ holds for each $\mu\in M(X,\mathscr{A},\mathbb{R})$ (note that $\mu^+$ and $\mu^-$ form the Jordan decomposition of $\mu$ and thus are positive measures). Then for any $\mu$ and $\nu$ in $M(X,\mathscr{A},\mathbb{R})$, \begin{align*} F(\mu+\nu) &= \int fd(\mu+\nu)\\ &= (\mu+\nu)(A)\\ &= \mu(A)+\nu(A)\quad \text{(well defined for finite $\mu$ and $\nu$)}\\ &= \int fd\mu + \int fd\nu\\ &= F(\mu)+F(\nu) \end{align*} (note that $\mu+\nu\in M(X,\mathscr{A},\mathbb{R})$ because $M(X,\mathscr{A},\mathbb{R})$ is a vector space).
Next suppose that $f$ is a simple $\mathscr{A}$-measurable function. Write $f=\sum_{i=1}^na_i\chi_{A_i}$. Then \begin{align*} \int fd\mu &= \int\left(\sum_{i=1}^na_i\chi_{A_i}\right)d\mu\\ &= \int\left(\sum_{i=1}^na_i\chi_{A_i}\right)d\mu^+-\int\left(\sum_{i=1}^na_i\chi_{A_i}\right)d\mu^-\\ &= \sum_{i=1}^na_i\mu^+(A_i)-\sum_{i=1}^na_i\mu^-(A_i)\\ &= \sum_{i=1}^na_i\left(\mu^+(A_i)-\mu^-(A_i)\right)\\ &= \sum_{i=1}^na_i\mu(A_i) \end{align*} holds for each finite signed measure. Then for any $\mu$ and $\nu$ in $M(X,\mathscr{A},\mathbb{R})$, \begin{align*} F(\mu+\nu) &= \int fd(\mu+\nu)\\ &= \sum_{i=1}^na_i(\mu+\nu)(A_i)\\ &= \sum_{i=1}^na_i(\mu(A_i)+\nu(A_i))\\ &= \sum_{i=1}^na_i\mu(A_i) + \sum_{i=1}^na_i\nu(A_i)\\ &= \int fd\mu + \int fd\nu\\ &= F(\mu)+F(\nu). \end{align*}
Finally, let $f$ be an arbitrary bounded real-valued $\mathscr{A}$-measurable function. Then there is a sequence $\{f_n\}$ of simple $\mathscr{A}$-measurable functions such that $f(x)=\lim_{n\to\infty}f_n(x)$ holds at each $x\in X$. Since $f$ is bounded, there is an $M\in\mathbb{R}$ such that $|f(x)|\leq M$ for all $x\in X$. Without loss of generality, we may assume $\{f_n\}$ is a sequence of bounded simple $\mathscr{A}$-measurable functions, and so $|f_n(x)|\leq W$ for some $W\in\mathbb{R}$ and for all $n\in\mathbb{N}$. Then be Lebesgue's Dominated Convergence Theorem, \begin{align*} \int fd\mu &= \int fd\mu^+ - \int fd\mu^-\\ &= \lim_{n\to\infty}\int f_nd\mu^+ - \lim_{n\to\infty}\int f_nd\mu^-\\ &= \lim_{n\to\infty}\left(\int f_nd\mu^+ - \int f_nd\mu^-\right)\\ &= \lim_{n\to\infty}\int f_nd\mu. \end{align*} So for any $\mu$ and $\nu$ in $M(X,\mathscr{A},\mathbb{R})$, \begin{align*} F(\mu+\nu) &= \int fd(\mu+\nu)\\ &= \lim_{n\to\infty}\int f_nd(\mu+\nu)\\ &= \lim_{n\to\infty}\left(\int f_nd\mu + \int f_nd\nu\right)\\ &= \lim_{n\to\infty}\int f_nd\mu + \lim_{n\to\infty}\int f_nd\nu\\ &= \int fd\mu + \int fd\nu\\ &= F(\mu)+F(\nu). \end{align*}
A similar argument shows $F(\alpha \mu)=\alpha F(\mu)$ for any $\alpha\in\mathbb{R}$ and any $\mu\in M(X,\mathscr{A},\mathbb{R})$.
I would really appreciate it if someone could help me check if my work is correct or not! Thanks a lot in advance!
