The problem is
Let $(X,\cal M, \mu)$ be a measure space and consider $f\in L^1(X,\cal M, \mu)$. Show that for each $\epsilon > 0$ there exists $\delta > 0$ such that $\int_E {|f|d\mu } < \varepsilon $ for all $E\in \cal M$ with $\mu(E) < \delta$.
I can see if $f$ is bounded by some $M$, we can choose $\delta = \frac{\epsilon}{M} $ and the claim holds. However, $f\in L^1$ only implies $f$ being finite a.e. but not bounded. What should I do with the case when $f$ is not bounded? Thank you!
Following the hint by Robert Israel,
Without loss of generality, we can assume that $f\geq 0$. Assume the conclusion is false. Then for some $\epsilon_0>0$, we have a sequence of sets $E_n$ with $\mu(E_n)<2^{-n}$ such that $\int_{E_n} f>\epsilon_0$. Let $E_n^{*}=\cup_{k\geq n} E_k$, then $\mu(E_n^{*})\leq 2^{-n+1}$.
Since $\mu(E_1^{*})<\infty$, we have $\mu(\cap_{n=1}^{\infty}E_n^{*})=\lim_{n\rightarrow\infty}\mu(E_n^{*})=0$. Therefore, $f \cdot 1_{E_n^{*}} \rightarrow 0$ a.e.
By Dominated convergence theorem, $\int_{E_n^{*}}f d\mu = \int_X f\cdot 1_{E_n^{*}}d\mu\rightarrow 0$. However, $\int_{E_n^{*}}fd\mu\geq\int_{E_n}fd\mu\geq \epsilon_0>0$. This is a contradiction.
$\textbf{Hint:}$
Represent $f$ as sum of two functions: first one is bounded, and the second one is supported on a small set.
$$f(x) = f(x)\chi_{\{|f(t)|\le n\}}(x) + f(x)\chi_{\{|f(t)|> n\}}(x)$$
