Absolue continuity of the integral

Question

Let $f \in L^1(X,\mu)$ Show that for every $\epsilon \gt 0$ it exists $\sigma \gt 0$ such that $\int_{A}f \lt \epsilon$ if $\mu(A) \lt \sigma$

Answer 1

If the thesis were not true, it would exist an $\varepsilon > 0$ such that for all $\sigma$ then: if $\mu(A) < \sigma$, then $|\int_{A} f| \geq \varepsilon$. If this happens, you can make the integral arbitrarily large.

Answer 2

Hint: Given $\epsilon>0,$ you can find a bounded $b$ such that $\int_X|f-b|<\epsilon.$ The result is clearly true for $b.$

Answer 3

Hint: We could assume that $f$ is positive. And there is a increasing sequence of simple functions that converge a.e. to $f$, i.e., $$0{\le\varphi}_1\le\varphi_2\le\cdots\le f,$$ $$\varphi_n\rightarrow f,a.e.,as\ n\rightarrow+\infty$$

and it's easy to prove the proposition for simple functions. And by using the convergence, we could finish the proof.