For every $\epsilon>0$ there exists $\delta>0$ such that $\int_A|f(x)|\mu(dx) < \epsilon$ whenever $\mu(A) < \delta$ Hello all mathematicians!! Again, I am struggling with solving the exercises in Lebesgue Integral for preparing the quiz. At this moment, I and my friend are handling this problem, but both of us agreed this problem is a bit tricky.
The problem is following. 
Let ($X,\mathcal{A},\mu$) be a measure space and suppose $\mu$ is $\sigma$-finite. Suppose $f$ is integrable. Prove that given $\epsilon$ there exist $\delta$ such that 
$$
\int_A |f(x)|\mu(dx) \;\;<\;\;\epsilon
$$
whenever $\mu(A) < \delta$.
Could anybody give some good idea for us? Think you very much for your suggestion in advance. 
 A: 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... :)
A: 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 $$\mu(A) < \delta \implies \int_A f \, d\mu  \le \int (f-\phi) \, d\mu  + \int_A \phi \, d\mu< \epsilon.$$
A: Hint:
$$
\int_A|f|\leqslant x\mu(A)+\int_{\{|f|\gt x\}}|f|\qquad\&\qquad\lim_{x\to\infty}\int_{\{|f|\gt x\}}|f|=0$$
