Is my Interpretation of $L^{1}$ and $L^{\infty}$ correct in this case Let $X:=[0,1]$ and $\mathcal{A}:=\{A \subseteq [0,1]: A \operatorname{or} A^{c} \operatorname{countable}\}$ and $\mu$ be the counting measure. 
I am asked to characterize both $L^{1}(X,\mathcal{A}, \mu)$ as well as $L^{\infty}(X,\mathcal{A}, \mu)$
My ideas: 
for $L^{\infty}(X,\mathcal{A}, \mu)$ it is clear that in order for $f \in L^{\infty}(X,\mathcal{A},\mu)$: $\sup\limits_{x \in [0,1]}\vert f(x)\vert<\infty$ since every point has positive measure $1$. 
for $L^{1}(X,\mathcal{A}, \mu)$, I want to say that: 
$f \in L^{1}(X,\mathcal{A}, \mu)\iff f1_{A} \in L^{1}(A,\mathcal{A},\mu) \wedge f1_{A^{c}}=0$ for all $A \subseteq [0,1]$ that are countable
Am I missing something? 
 A: We first characterize measurable functions: A function $f:[0,1]\to\mathbb{C}$ is measurable if and only if there exists a sequence $(x_n)\subset [0,1]$ such that $f$ is constant when restricted to $[0,1]\setminus\{x_n:n\geq1\}$. I don't think one can explain this better than the accepted answer in this post.
For $L^\infty(\mu)$: as you observed, if $f\in L^\infty(\mu)$ and $x\in [0,1]$, since $\mu(\{x\})=1>0$, we must have that $|f(x)|<\|f\|_\infty$, so $\sup_{x\in[0,1]}|f(x)|<\infty$. On the other hand, any $f$ that satisfies $\sup_{x\in [0,1]}|f(x)|<\infty$ is obviously an element of $L^\infty(\mu)$.
Now for $L^1(\mu)$. Let $f:[0,1]\to\mathbb{C}$ be a measurable function, so there exists a constant $c\in\mathbb{C}$ and a sequence $(x_n)\subset[0,1]$ so that $f=c$ on $[0,1]\setminus\{x_n:n\geq1\}$. Now
$$\int_{[0,1]}|f|d\mu=\int_{[0,1]\setminus\{x_n:n\geq1\}}|f|d\mu+\int_{\{x_n:n\geq1\}}|f|d\mu=$$
$$\int_{[0,1]\setminus\{x_n:n\geq1\}}|f|d\mu+\sum_{n=1}^\infty\int_{\{x_n\}}|f|d\mu=$$ $$=\int_{[0,1]\setminus\{x_n:n\geq1\}}|f|d\mu+\sum_{n=1}^\infty|f(x_n)|=|c|\cdot\mu([0,1]\setminus\{x_n:n\geq1\})+\sum_{n=1}^\infty|f(x_n)|=$$
$$=|c|\cdot\infty+\sum_{n=1}^\infty|f(x_n)| $$
Now this quantity on the last line of the equation is finite if and only if $c=0$ and $\sum_n|f(x_n)|<\infty$, i.e. $f=0$ everywhere except a countable set of points in $[0,1]$ on which $f$ assumes values that form a sequence of $\ell^1$.
