I have two questions:
1, Give an example of a measure space such that $L^{1}(X,\mathcal{A},\mu) = \mathcal{L}^{1}(X,\mathcal{A},\mu)$.
2, State, and prove, a condition on $\mu$ which is equivalent to the identity in 1.
Attempt at solution:
1, Let $X = \mathbb{N}$, $\mathcal{A} = \mathcal{P}(\mathbb{N})$. The idea is that every $n\in \mathbb{N}$ defines a function. But I do not feel confident about this proposal.
2, Let $\mu(A) = 0$ iff $A = \emptyset$.
Edit: $\mathcal{L}^{1}(X,\mathcal{A},\mu) = \{f:X\rightarrow \mathbb{R}^{d} \text{ or } \mathbb{C}: f \text{ is measurable and }|f| \text{ is integrable.}\}$
$L^{1}(X,\mathcal{A},\mu)=\mathcal{L}^{1}(X,\mathcal{A},\mu) \diagup \mathcal{N}^{1}(X,\mathcal{A},\mu)$, where $\mathcal{N}^{1}(X,\mathcal{A},\mu) = \{ f\in\mathcal{L}^{1}(X,\mathcal{A},\mu): f \sim 0\text{ where } f\sim 0 \text{ iff } f(x) = 0 \text{ almost everywhere}\}$.
For the point 1), indeed your attempt is a correct approach. Let me clarify: We know that in the space $\mathcal{L}^p(X,\mathcal{A}, \mu)$, $\| f \|_{\mathcal{L}^p}=0$ iff $f=0 \ a.e$, then the function $\| \cdot \|_{\mathcal{L}^p}: \mathcal{L}^p\to \mathbb{C}$ is just a semi norm.
The motivation on constructing the quotient space $L^p=\mathcal{L}^p/\mathcal{N}^p$ is that $\|[f]\|_{L^P}=0$ iff $[f]=\mathcal{N}^p=0\in L^p$, then $\|\cdot\|_{L^p}: L^p\to \mathbb{C}$ is in fact a norm. Usually in measure theory we do not distinguish between $[f]$ and $f$, and we use that $\|f \|_{\mathcal{L}^p}=\|[f] \|_{L^p}$, so that $(\mathcal{L}^p, \| \|_{\mathcal{L}^p})$ can be seen as a Banach Space.
However to answer your question, we need to look for a space in which $\mathcal{N}^p$ contains only the $0$ element. Then, following your attempt we put $X=\mathbb{N}$, $\mathcal{A}=\mathcal{P}(\mathbb{N})$ and $\mu=$counting measure (here $\mathcal{L}^p(\mathbb{N}, \mathcal{P}(\mathbb{N}), \mu)$ is the classical $\ell^p(\mathbb{N})$ Banach Space). Now, if we take $f: \mathbb{N} \to \mathbb{C}$ such that $\|{f}\|_{\mathcal{L}^p}^p=0$, then $$ \|{f}\|_{\mathcal{L}^p}^p = \int_{\mathbb{N}} |f|^p d\mu = \sum_{n \in \mathbb{N}} | f(n)|^p=0, $$
it is clear that $ \ \|f\|_{\mathcal{L}^p}=0$ iff $f(n)=0 \ \forall \ n \in \mathbb{N}$ iff $f=0$, this gives $\mathcal{N}^p(\mathbb{N}, \mathcal{P}(\mathbb{N}), \mu)=\left\{ 0 \right\}$, and that is why $\mathcal{L^p}=L^p$
For the point 2), now you can guess that $\mu$ needs to be such that $\mathcal{N}^p(X, \mathcal{A}, \mu)=\left\{ 0 \right\}$ to have that $\mathcal{L^p}=L^p$. Indeed, is easy to prove that if $\mathcal{N}^p(X, \mathcal{A}, \mu)=\left\{ 0 \right\}$ then $L^p=\mathcal{L}^p/\left\{ 0 \right\} $ and $\mathcal{L}^p$ are isometrically isomorphic