This is the Excercise 1.12 of Rudin's Real and Complex Analysis:
Suppose $f\in L^1(\mu)$. Prove that to each $\epsilon>0$ there exists a $\delta>0$ such that $\int_{E}|f|d\mu<\epsilon$ whenever $\mu(E)<\delta$.
This problem likes the uniformly continuous property. I tried to prove it by making contradiction, but I can't figure it out. Thanks!
Fix $\varepsilon>0$. By definition of Lebesgue integral, there exists a step function $s$ such that $\int_X (|f|-s)d\mu<\varepsilon$, and $0\leq s\leq |f|$. So we are reduced to show that the result when $s$ is a step function.
Let $E$ a measurable set, and assume that $s$ has the form $\sum_{j=1}^Na_j\chi_{A_j}$, where $a_j\geq 0$ and $A_j$ are measurable sets. Then $$\int_E sd\mu=\sum_{j=1}^Na_j\mu(A_j\cap E)\leq \mu(E)\sum_{j=1}^Na_j.$$
Here is another solution.
Define $A_j=\{x:|f(x)|\geq j \},\forall j\in \mathbb N$ .
The given statement is clearly true if $|f|$ is bounded. So if we can show the integrals over the defined sets tend to zero, we are done.
Now, $ \chi_{_{A_j}} $ is a collection of monotonically decreasing sequence with limit $0$. And so is true for $ |f| \chi _{_{A_j}} $ with $|f|\chi _{_{A_1}}\in L^1(\mu)$. So apply Dominated Convergence Theorem to get $\lim_{j \to \infty}\int_{A_j}|f| d\mu=0 $.
