Proving $\epsilon$-$\delta$ convergence of Lebesgue integrals just from the definition

Question

I am pretty lost with how to do this.

The problem is to show that, in a measure space $(X,M,\mu)$ with a measurable function $f: X \to [0,\infty)$ such that $\int_X f d\mu < \infty$. Then, for all $\epsilon > 0$, there is a $\delta > 0$ such that for all $E \in M$ if $\mu(E) < \delta$ then

$$\int_E f d\mu < \epsilon$$

Apparently, I should do this from just the definition, which is

$$\int_Efd\mu = \sup_{0 \leq s \leq f} \int_E s d\mu$$ and $$\int_Esd\mu = \sum_{j=1}^n \alpha_j \mu(A_j \cap E)$$

But I have no idea how to continue

Answer 1

If $\int_X f d\mu =0$, the result is trivial. So suppose that $\int_X f d\mu >0$ and take $\epsilon >0$. We can find a simple function $0 \le s \le f$ such that $$s = \sum_{i=1}^n \alpha_i \chi_{A_i}$$ where the $A_i$ are disjoint measurable subsets and

$$0 < \int_X f d\mu - \int_X s d\mu = \int_X f d\mu - \sum_{i=1}^n \alpha_i \mu(A_i) \le \epsilon/2$$

For $E \in M$, you have $$\int_E s d\mu = \sum_{i=1}^n \alpha_i \mu(A_i \cap E) \le \mu(E) \left(\sum_{i=1}^n \alpha_i \right).$$

Now take $\delta=(\epsilon/2)/\left(\sum_{i=1}^n \alpha_i \right)$ and suppose that $\mu(E) \le \delta$.

You have

$$\int_E f d\mu = \int_E (f-s) d\mu + \int_E s d\mu \le \epsilon/2 + \epsilon/2=\epsilon$$

as desired.

Answer 2

There exists a simple function $s=\sum\limits_{i=1}^n a_i \chi_{E_i}$ satisfying $0\leq s \leq f$ and

$$\int_X f d\mu - \int_X s d\mu < \frac{\epsilon}{2}.$$

For this simple function $s$, it is easy to check that $\int_E s d\mu \leq \max\{a_i ~|~ i=1,\ldots,n \}\mu(E)$ for all $E\in M$.

Hence, there exists $\delta >0$ satisfying $$\int_E s d\mu < \frac{\epsilon}{2}$$ whenever $\mu(E) < \delta$.

Therefore, for all $E\in M$ satisfying $\mu(E)<\delta$, the inequality

$$ \int_E f d\mu = \int_E (f-s)d\mu + \int_E s d\mu < \int_X (f-s)d\mu + \frac{\epsilon}{2} < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon $$

holds.