$\int \limits_{E}|f|d\mu<\varepsilon$ whenever $\mu(E)<\delta$. Suppose $f\in L^1(\mu)$. Prove that to each $\varepsilon>0$ there exists a $\delta>0$ such that $\int \limits_{E}|f|d\mu<\varepsilon$ whenever $\mu(E)<\delta$.
Proof: Let $\varepsilon>0$ be given and $E$ be an arbitrary measurable set. Since $\int \limits_{E}|f|d\mu=\sup \limits_{0\leqslant s \leqslant |f|} \int \limits_{E}sd\mu$, there exists a simple measurable function $s(x)=\sum \limits_{i=1}^{n}\alpha_{i}1_{A_i}$ with $0\leqslant s \leqslant |f|$ such that: $$\left(\int \limits_{E}|f|d\mu\right)-\frac{\varepsilon}{2}\leqslant \int \limits_{E}sd\mu \Rightarrow \int \limits_{E}|f|d\mu\leqslant \left(\int \limits_{E}sd\mu\right)+\frac{\varepsilon}{2}$$
Let's wotk with last integral: $$\int \limits_{E}sd\mu=\sum \limits_{i=1}^{n}\alpha_i\mu(A_i)\leq \max\{\alpha_1,\dots, \alpha_n\}\sum \limits_{i=1}^{n}\mu(A_i)=\max\{\alpha_1,\dots, \alpha_n\}\mu(E)=L\mu(E)$$ where $L=\max\{\alpha_1,\dots, \alpha_n\}$. Taking $\delta=\frac{\varepsilon}{2L}$(if $L>0$, otherwise is obvious), for all $\mu(E)<\delta$, we get: $$\int \limits_{E}|f|d\mu<\frac{\varepsilon}{2}+L\mu(E)<\frac{\varepsilon}{2}+L\frac{\varepsilon}{2L}=\varepsilon$$
Sorry if this topic is repeated but I would like to know is my proof correct?
EDITED VERSION: Since $|f|$ is measurable on $X$ and $\int \limits_{X}|f|d\mu=\sup \limits_{0\leqslant s\leqslant |f|}\int \limits_{X}sd\mu$.
Let $\varepsilon>0$ be given then exists simple measurable function $s: 0\leqslant s\leqslant |f|$ on $X$ such that $\int \limits_{X}|f|d\mu-\dfrac{\varepsilon}{2}\leqslant \int \limits_{X}sd\mu$, where  $s=\sum \limits_{i=1}^{n}\alpha_i1_{A_i}$ where $\alpha_1, \dots, \alpha_n$ are distinct positive reals and $A_i=\{x: s(x)=\alpha_i\}$. Also $\sqcup_{i=1}^{n}A_i=X$. Note that for any $E\in \mathfrak{M}$ we have $$\int \limits_{E}(|f|-s)d\mu \leqslant \int \limits_{X}(|f|-s)d\mu\leqslant \frac{\varepsilon}{2}.$$
Taking $\delta=\dfrac{\varepsilon}{2L},$ where $L=\max\{\alpha_1,\dots,\alpha_n\}$. Note that $\delta$ in this case does not depends on $E$! Hence for $E$ with $\mu(E)<\delta$ we have: $$\int \limits_{E}|f|d\mu\leqslant \frac{\varepsilon}{2}+\int \limits_{E}sd\mu\leqslant \dfrac{\varepsilon}{2}+\sum \limits_{i=1}^{n}\alpha_i\mu(A_i\cap E)\leqslant \dfrac{\varepsilon}{2}+L\sum \limits_{i=1}^{n}\mu(A_i\cap E)$$$$\leqslant \dfrac{\varepsilon}{2}+L\mu((\sqcup_{i=1}^nA_i)\cap E)=\dfrac{\varepsilon}{2}+L\mu(E)=\varepsilon.$$
 A: I don't think your proof is correct. Your choice of $\delta$ depends on $E$ (through your choice of $s$ on $E$). In the following I will prove a theorem, of which your problem is a special case.
In case you don't know about absolutely continuous measures, here is the definition:

Let $\mu$ and $\lambda$ be positive measures on a $\sigma$-algebra $\mathfrak{M}$. $\lambda$ is said to be absolutely continuous with respect to $\mu$ iff $\forall E\in \mathfrak{M}$, $\mu(E) = 0$ indicates $\lambda(E) = 0$.

The theorem (which justifies the "absolute continuity"):

Suppose that $\mu$ and $\lambda$ are positive measures on $\mathfrak{M}$, with $\lambda$ being finite. Then $\lambda$ is absolutely continuous with respect to $\mu$ iff $\forall \epsilon >0$, $\exists \delta > 0$ such that $\forall E \in \mathfrak{M}$, $\mu(E)< \delta$ indicates $\lambda(E) < \epsilon$.

Suppose the latter holds. Then given $E\in \mathfrak{M}$ with $\mu(E) = 0$, it follows directly that $\forall \epsilon > 0$, $\lambda(E)<\epsilon$. Hence $\lambda(E) = 0$.
Suppose the latter doesn't hold. Then there exist $\epsilon > 0$ and a sequence of measurable sets $\{E_n\}$, such that $\mu(E_n) < 2^{-n}$ and $\lambda(E_n)\geq \epsilon$ for all $n$. Put $A_n = \bigcup_{i=n}^\infty E_i$ and $A = \bigcap_{n=1}^\infty A_n$. Then
$$
\mu(A) = \lim_{n\to \infty}\mu (A_n) \leq \lim_{n\to \infty}2^{-n+1} = 0,
$$
while
$$
\lambda(A) = \lim_{n\to \infty} \lambda(A_n) \geq \epsilon > 0.
$$
Hence $\lambda$ is not absolutely continuous with respect to $\mu$. This finishes the proof.
If you know about complex measures, the definition and theorem here can actually be stated with $\lambda$ being a complex measure instead, and the proof is exactly the same.
In your case, observe that given $f\in L^1(\mu)$, $d\lambda = |f|d\mu$ defines an absolutely continuous finite measure $\lambda$ with respect to $\mu$.
