TITLE: For every $\epsilon>0$ there exists $\delta>0$ such that $\int_A|f(x)|\mu(dx) < \epsilon$ whenever $\mu(A) < \delta$
QUESTION [8 upvotes]: 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.

REPLY [17 votes]: 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 $$\mu(A) < \delta \implies \int_A f \, d\mu  \le \int (f-\phi) \, d\mu  + \int_A \phi \, d\mu< \epsilon.$$