For every $\epsilon>0$ there exists $\delta>0$ such that $\int_A|f(x)|\mu(dx) < \epsilon$ whenever $\mu(A) < \delta$

If $f$ is integrable there exists a simple function $0 \le \phi \le f$ with the property that $$\int f - \phi \, d\mu < \frac{\epsilon}{2}.$$ Write $$\phi = \sum_{k=1}^n c_k \chi_{C_k}.$$ Since $$ \int_A \phi \, d\mu = \sum_k c_k \mu(A \cap C_k) \le \mu(A) \sum_{k=1}^n c_k$$ you can take $$\delta = \frac{\epsilon}{2 \sum _{k=1}^n c_k}$$ to conclude $$\mu(A) < \delta \implies \int_A f \, d\mu \le \int (f-\phi) \, d\mu + \int_A \phi \, d\mu< \epsilon.$$

Hint: $$ \int_A|f|\leqslant x\mu(A)+\int_{\{|f|\gt x\}}|f|\qquad\&\qquad\lim_{x\to\infty}\int_{\{|f|\gt x\}}|f|=0$$

Define the sequence $A_n=\{|f|\geq n\}$ (the set of all the points in which the value of $|f|$ is greater or equal than $n$). Since $f$ is integrable, clearly $|f|$ is measurable and therefore $A_n$ is measurable for any $n$. $A_n$ is defined a.e. (as $f$ is); should it sound odd to you, since everything holds up to a.e. equivalence, you could replace $f$ with any representative of its equivalence class.

Obviously $A_n$ is decreasing ($A_n\supseteq A_{n+1}$ for any $n$), therefore there exists the limit $$ A_\infty\dot=\lim_{n\to\infty}A_n=\bigcap_{n=1}^\infty A_n $$ and $A_\infty$ is measurable. From the integrability of $f$ it follows that $\mu(A_\infty)=0$ (if $\mu(A_\infty)>0$ then $\int_X |f|{\rm d}\mu \geq \int_{A_\infty}|f|{\rm d}\mu = \infty \cdot \mu(A_\infty) = \infty$, so $f$ would not be integrable).

Therefore the sequence $$ f_n ~\dot=~ f\chi_{A_n} $$ converges to $0$ a.e. ($\chi_{A_n}$ is the indicator function of the set $A_n$). It is not difficult to argue that $f_n$ is measurable for any $n$.

Now apply Lebesgue's dominated convergence theorem to $f_n$ which is pointwise less or equal than $|f|$, which is integrable.

Therefore $$ \int_{A_n}f_n{\rm d}\mu ~\stackrel{n\to\infty}{\longrightarrow} 0 $$ It should be easy from here... :)