$J$ is surjective but not injective on $L^{\infty}\to (L^{1})^{*}$

Question

Let Let $X:=\{0, 1\}$ and $\mathcal{A}:=\mathcal{P}(X)$ and $\mu(\{0\})=1$ while $\mu(\{1\})=\infty$

I know that $f\in L^{1}(X,\mathcal{A}, \mu)$ iff $f\vert_{\{1\}}=0$ and $g \in L^{\infty}(X,\mathcal{A}, \mu)$ for any well-defined $g$ on $X$.

Now, show that: $J: L^{\infty} \to (L^{1})^{*}, Jg(f)=\int_{X}fgd\mu=f(0)g(0)$

I claim that $J$ is surjective but not injective.

My idea: $L^{\infty}$ can be identified as $\mathbb R^{2}$, since $L^{\infty}=\{g|(g(0),g(1))\in \mathbb R^{2}\}$ so we can now interpret $J$ as $J: \mathbb R^{2}\to (L^{1})^{*}, (x,y)\mapsto J(x,y)$ and$ J(x,y)(g)=xg(0)$.

Surjectivity: let $K\in (L^{1})^{*}$ and for any $g \in L^{1}$ it follows that $K(g)\in \mathbb R$ then choose $(x,y)\in \mathbb R^{2}$ so that $x=\frac{K(g)}{g(0)}$ so that $J(x,y)(g)=K(g)$ but that would only be for one particular $g$ so I actually have no surjectivity, any ideas?

Not Injective: Easy since if $y\neq z$ then $J(x,y)(g)=xg(0)=J(x,z)(g)$ for any $g \in L^{1}$ so $J(x,y)=J(x,z)$

Answer 1

$J$ is surjective: let $K \in (L^{1})^{*}$. Note that $L^{1}$ is one-dimensional since $f(1)=0$ for all $f \in L^{1}$. Let $f_0(1)=0,f_0(0)=1$. Define $g \in L^{\infty}$ by $g(x)=K(f_0)$ for all $x$. Then $J_g(f)=K(f)$ for all $f$: $f$ is given by $f=af_0$ for some $a$. So $K(f)=aK(f_0)$. Also $J_g(f)=f(0)g(0)=af_0(0)g(0)=ag(0)=aK(f_0)$. Thus $J_g(f)=K(f)$ for all $f \in L^{1}$.