Find $\delta >0$ such that $\int_E |f| d\mu < \epsilon$ whenever $\mu(E)<\delta$ I am studying for a qualifying exam, and I am struggling with this problem since $f$ is not necessarily integrable. 
Let $(X,\Sigma, \mu)$ be a measure space and let
$$\mathcal{L}(\mu) = \{ \text{ measurable } f \quad| \quad  \chi_Ef \in L^1(\mu) \text{ whenever } \mu(E)<\infty\}.$$
Show that for any $f\in \mathcal{L}(\mu)$ and any $\epsilon >0$ there is $\delta >0$ such that $\int_E|f| d\mu < \epsilon$ whenever $\mu(E)< \delta$.
A technique I've used in other similar problems is to define $A_n = \{ x\in X \, | \, 1/n \leq |f(x)| \leq n \}$ and let $A = \displaystyle \bigcup_{n=1}^\infty A_n$. We can also define $A_0 = \{ x\in X \,|\, f(x) = 0\}$ and $A_\infty= \{x\in X\, | \, |f(x)| = \infty\}$. The part where I'm stuck is now that 
$$\int_X|f|d\mu = \int_{A_0} |f|d\mu + \int_{A} |f| d\mu + \int_{A_\infty} |f|d\mu$$
where the first term on the right is zero, and I want the last term on the right to be zero.
Is there another way to go about this problem? Explanations are helpful to me since I'm studying and I don't want to confuse myself further. Thanks!

My try:
the best thing to do in my opinion is working by absurd. Suppose the thesis does not hold.
Then $\exists \varepsilon >0$ such that $\forall n \geq 1$ exists $E_n \in \Sigma$ such that $\mu (E_n) < 2^{-n} \varepsilon$ BUT $\int_{E_n} |f| d \mu > \varepsilon$.
Then call $E=\bigcup_{n\geq1} E_n$. We have that $\mu(E) < \varepsilon$ but
 $$\int_{E} |f| d\mu = \sum_{n\geq1} \varepsilon = +\infty$$
This contradicts the assumption on $f$.
The only problem is that this works only if $\{E_n \}_{n\geq1}$ are pairwise disjoint.
