Continuity of integration in $L^1(X)$. Let $\varepsilon>0$ and let $f\in L^1(X)$. Show that there exists $\delta>0$ such that for any $E\subseteq X$ with $m(E)<\delta$ we have 
$$
\int_E \lvert f\rvert<\varepsilon.
$$
MY THOUGHTS: Since $|f|$ is nonnegative and $f\in L^1(X)$ then I know that there exists some $g$ that is bounded on $F\subseteq X$ and $m(F)<\infty$ such that
$$
\int_F|f|-g<\varepsilon
$$
so let $\delta=m(X\sim F)$ and we are done.
PROBLEMS: My outline above assumes two things which we don't know:
1.) $m(X)<\infty$ (In order for $\delta=m(X\sim F)$ to be finite using excision);
2.) All I have shown is that there exists some set whose measure is equal to $\delta$ that has the desired property, not that any set of measure less than $\delta$ satisfies the property.
Any help would be appreciated.
 A: We use the fact that simple functions are dense in $L^1(X)$. Choose one such function, $g$, clearly a positive function, so that
$$\int_X \big ||f|-g\big|\,dm <\epsilon'$$
Then since $g$ is simple, $\sup_{x\in X} g(x) = M<\infty$. Then we have that if $m(E)<\delta$ then
$$\int_E g\,dm< M\delta$$
But then

$$\int_E |f|\,dm \le\int_E \big| |f|-g\big|\,dm+\int_E g\,dm< \epsilon'+\delta M$$

So if we choose $\epsilon$ as in the question, then we let $\epsilon', \delta$ be chosen so that $\epsilon'+\delta M\le\epsilon$, eg. $\epsilon '={\epsilon \over 2}, \delta = {\epsilon\over 2M}$.
A: Let $f_n(x) = \min\{|f(x)|,n\}$. Then $f_n$ is a sequence of measurable functions on $X$ such that $f_n \to |f|$, $|f_n| \le |f|$, and $\int_X |f| < \infty$. Hence, by the dominated convergence theorem, $\int_X f_n \to \int_X |f|$. Given $\varepsilon > 0$, there exists a positive integer $k$ such that $\int_X |f| < \int _X f_k + \epsilon/2$. Let $\delta = \varepsilon/(2k)$. If $E$ is a measurable subset of $X$ with $m(E) < \delta$, then 
\begin{align}\int_E |f| &= \int_X |f| - \int_{X\setminus E} |f| < \left(\int_X f_k + \frac{\varepsilon}{2}\right) - \int_{X\setminus E} f_k \\
&= \int_E f_k + \frac{\varepsilon}{2}< km(E) + \frac{\varepsilon}{2} < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.
\end{align}
