lebesgue integral - 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.

Answer

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. $\epsilon '={\epsilon \over 2}, \delta = {\epsilon\over 2M}$.