I am attempting to prove that $$\lim_{\mu(A)\rightarrow 0}\int_{A}\lvert f \lvert {\rm d}\mu=0.$$
Given $(X,\mathcal{M},\mu)$ a measured space , we suppose that $$f\in L^{1}(X,\mathcal{M},\mu) \text{ prove that } \displaystyle\lim_{\mu(A)\rightarrow 0}\int_{A}\lvert f \lvert{\rm d}\mu=0\text{ with } \mu \text{ a positive measure}$$
My attempt: we put $\lvert f \lvert=g\ge 0$ then there exists $g_{n}$ an increasing sequence of positive simple function such that $\displaystyle\lim_{n\rightarrow\infty}g_{n}(x)=g(x)$ and: $$g_{n}=\sum_{i=1}^{N}\alpha_{i}^{n}\chi_{A_{i}^{n}} \text{ where }A=\bigcup_{i=1}^{N}A_{i}^{n} \text{ and } A_{i}^{n}\cap A_{j}^{n}=\emptyset \text{ if } i\neq j \implies \int_{A}g_{n}{\rm d}\mu=\sum_{i=1}^{N}\alpha_{i}^{n}\mu({A_{i}^{n}})$$ and because $\mu(A)=\displaystyle \sum_{i=1}^{N}\mu(A_{i}^{n}) \text{ then } \mu(A)\rightarrow 0 \iff \mu(A_{i}^{n})\rightarrow 0 , \forall i\in\{1,2,...,N\} $ then: $$\lim_{\mu(A)\rightarrow 0}\int_{A}g_{n}{\rm d}\mu=0$$ by the monotone convergence theorem we have $\displaystyle\lim_{n\rightarrow \infty}\int_{A}g_{n}{\rm d}\mu=\int_{A}{\rm d}\mu$ we can deduce in this case that: $$\lim_{\mu(A)\rightarrow 0}\int_{A}g{\rm d}\mu=0$$ but I am not sure of this answer.
Let us consider a large quantity $\alpha$. Now, $$\int_A |f| d\mu = \int_{|f| \leq\alpha} |f| I_A d\mu + \int_{|f| \geq\alpha} |f|I_A d\mu $$ $$ \leq \alpha \mu(A) + \int_{|f| \geq\alpha} |f|I_A d\mu $$
Now, $|f|I_{A \cap \{|f| \geq\alpha\}} < \epsilon$ for large $\alpha$. Therefore, $$ \int_{|f| \geq\alpha} |f|I_A d\mu < \epsilon \mu(A \cap \{|f| \geq\alpha\}) < \epsilon \mu(A) $$
So, $$ \int_A |f| d\mu \leq \alpha \mu(A) + \epsilon \mu(A) $$ Take $\mu(A) \rightarrow 0$ both side,you get the desired result.