Let $\mathcal{A}$ be the $\sigma$-field on $[0,1]$ that consists of all subsets of $[0,1]$ that are (at most) countable or whose complement is (at most) countable. Let $\mu$ be the counting measure on $([0,1], \mathcal{A})$. (1) Verify that a function $f:[0,1] \longrightarrow \mathbb{R}$ is in $L^1([0,1], \mathcal{A}, \mu)$ if and only if the set $E_f=\{x \in[0,1]: f(x) \neq 0\}$ is at most countable, and $$ \sum_{x \in E_f}|f(x)|<\infty $$ Furthermore, we have $\|f\|_1=\sum_{x \in E_f}|f(x)|<\infty$ in that case.
(2) For every $f \in L^1([0,1], \mathcal{A}, \mu)$, set $$ \Phi(f)=\sum_{x \in E_f} x f(x) . $$ Prove that $\Phi$ is a continuous linear form on $L^1([0,1], \mathcal{A}, \mu)$.
(3) Show that there exists no function $g \in L^{\infty}([0,1], \mathcal{A}, \mu)$ such that $\Phi(f)=$ $\int f g \mathrm{~d} \mu$ for every $f \in L^1([0,1], \mathcal{A}, \mu)$.
Can someone give me idea on how to solve it, main problem of mine is how to use counting measure on real line and do manipulation with it. Thank you. In case, I start with Indicator function of $A$, we know it $f$ will be integrable if $A$ is finite and in that case, $\{ x \in [0, 1]: f(x) \neq 0 \}$ is countable.
Here are some hints:
(1) Suppose $f\in L^1([0,1], \mathcal{A}, \mu)$. Then $f$ is in particular measurable, and thus, for each $n\in \mathbb{N}$, the set $E_{f,n}=\{x\in [0,1]\mid |f(x)|\geq \frac{1}{n}\}$ is measurable. We have $0\leq \frac{1}{n}\chi_{E_{f,n}}\leq |f|$ and thus by the monotonicity of the Lebesgue integral we conclude $0\leq \frac{1}{n}\mu(E_{f,n})\leq \|f\|_{L^1(\mu)}<\infty$, which shows that $E_{f,n}$ is finite and therefore countable. Hence $E_f=\bigcup_{n\in \mathbb{N}} E_{f,n}$ is countable as a union of countable sets. We claim that $\sum_{x\in E_f} |f(x)|=\|f\|_{L^1(\mu)}$. Let $\epsilon>0$. If $E_f$ is finite, then $f$ is a step function and the identity $\sum_{x\in E_f} |f(x)|=\|f\|_{L^1(\mu)}$ follows immediately from the definition of the Lebesgue integral. Otherwise $E_f$ is countably infinite, and we can thus write $E_f=\{x_k\in [0,1]\mid k\in N\subseteq \mathbb{N}\}$ (we may assume that $x_k\neq x_j$ for $k\neq j$). Then $f=\sum_{k=1}^\infty f(x_k)\chi_{\{x_k\}}$ and by Lebesgue's theorem of dominated convergence (or monotone convergence theorem) we have $\sum_{k=1}^\infty |f(x_k)|=\|f\|_{L^1(\mu)}$. Thus, there is some $n_0=n_0(\epsilon)\in \mathbb{N}$ such that $\big|\|f\|_{L^1(\mu)}-\sum_{k=1}^n |f(x_k)|\big|<\epsilon$ for all $n\geq n_0$. In other words: setting $I=\{x_1,\dots, x_{n_0}\}\subseteq E_f$ we have for all finite $J$ with $I\subseteq J \subseteq E_f$ that $\big|\|f\|_{L^1(\mu)}-\sum_{x\in J} |f(x)|\big|<\epsilon$. This shows the unconditional convergence of the series $\sum_{x\in E_f} |f(x)|$ with limit $\|f\|_{L^1(\mu)}$, i.e., $ \sum_{x\in E_f} |f(x)|= \|f\|_{L^1(\mu)}$.
Conversely, suppose $f\colon [0,1]\to \mathbb{R}$ is a function such that $E_f$ is countable and such that $\sum_{x\in E_f} |f(x)|=:\alpha <\infty$. Then, we can write $E_f=\{x_k\in [0,1]\mid k\in \mathbb{N}\}$. Then $f=\sum_{k=1}^\infty f(x_k)\chi_{\{x_k\}}$ is a limit of measurable functions and therefore measurable. By the monotone convergence theorem, we conclude $\|f\|_{L^1(\mu)}=\sum_{k=1}^\infty |f(x_k)|=\alpha<\infty$. This shows that $f\in L^1([0,1], \mathcal(A), \mu)$.
(2) Let $f\in L^1([0,1], \mathcal{A}, \mu)$. Consider the function $g\colon [0,1]\to \mathbb{R}$, $g(x)=x$ (note that $g$ is $\textit{not}$ measurable!). By (1) we can check that $gf$ is in $L^1([0,1], \mathcal{A}, \mu)$: We have $E_{gf}\subseteq E_{f}$, thus $E_{gf}$ is countable as $E_f$ is countable by (1). Moreover,
$\sum_{x\in E_{gf}} |x| |f(x)|\leq \sum_{x\in E_{f}} |f(x)| = \|f\|_{L^1(\mu)}<\infty$.
By (1), we conclude that $gf$ is integrable. In particular, $\Phi(f)=\sum_{x\in E_f} x f(x)$ exists and $|\Phi(f)|\leq \|f\|_{L^1(\mu)}$, which shows that $\Phi$ is a continuous linear functional on $L^1([0,1], \mathcal{A}, \mu)$.
(3) Suppose there exists such a $g$. For each $x\in [0,1]$, we then must have $g(x)=\Phi(\chi_{\{x\}})=x$. But $g$ cannot be measurable since $g^{-1}([0, 1/2])=[0,1/2]$ does not lie in $\mathcal{A}$ since $[0, 1/2]$ and its compliment $(1/2,1]$ are uncountable. In particular $g\notin L^\infty([0,1], \mathcal{A}, \mu)$.