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!

Suppose not: there is $\varepsilon_0\gt 0$ such that for each positive $\delta$, there is a measurable set $A$ such that $$\mu(A)\lt \delta\quad \mbox{ and }\quad \int_A|f|\mathrm d\mu\gt\varepsilon_0.$$
In particular, for each integer $k$ and $\delta:=2^{-k}$, there exists $A_k$ of measure smaller than $2^{-k}$ for which $\int_{A_k}|f|\mathrm d\mu\gt\varepsilon$. Define $A:=\bigcup_k A_k$ (a set of finite measure). We have 
$$\varepsilon_0\lt \int_{A_k}|f|\mathrm d\mu\leqslant n\mu(A_k)+\int \chi_{A_k}\chi_{\{|f|\gt n\}}|f|\mathrm d\mu\\\leqslant n2^{-k}+\int \chi_A\chi_{\{|f|\gt n\}}|f|\mathrm d\mu,$$
hence 
$$\varepsilon_0\leqslant\int \chi_A\chi_{\{|f|\gt n\}}|f|\mathrm d\mu,$$
and by monotone convergence, $\lim_{n\to\infty}\int \chi_A\chi_{\{|f|\gt n\}}|f|\mathrm d\mu=0$, a contradiction.
