Lebesgue integral/measure issue

Question

So I have an exercise that seems trivial to me, although, I could have done the proof completely wrong; I'm worried my negation is wrong. Here is the statement:

Given $\epsilon > 0$, show that there exists a $\delta >0$ such that $\int_{E} |f| < \epsilon$ whenever $m(E) < \delta$.

Here is my attempt:

Suppose not. Then, fix $\epsilon >0$. For every positive $\delta$ we have that $\int_{E} |f| \geq \epsilon$ whenever $m(E) < \delta$. However, if $E$ is a null-set, this is impossible.

Answer 1

You have done the negation wrong. Note that your statement reads: $$ \forall \epsilon > 0 \; \exists \delta > 0 \; \forall E:\left( m(E) < \delta \Rightarrow \int_E |f| < \epsilon\right) $$ Hence, the negation is: $$ \exists \epsilon > 0\; \forall \delta > 0 \; \exists E:\left( m(E) < \delta \land \int_E |f| \ge \epsilon \right) $$ You do not have it for all $E$, as you used, there is some $E$ such that $m(E) < \delta$ and $\int_E |f| \ge \epsilon$.

Answer 2

To prove it use the following hint:

WLOG assume that $f$ is positive and define the sequence of functions $f_n(x)=\min {\{f(x),n\}}$ which tends to $f(x)$ pointwise a.e and monotonic increasingly. Then use Beppo Levi's theorem : $\int\limits_{X}{f_n}\to\int\limits_{X}{f}$ where $(X,\Sigma,\mu)$ is your measure space. So for each $\epsilon>0,\exists N_\epsilon\in\mathbb N: \int\limits_{X}{(f-f_n)}<\frac{\epsilon}{2},\forall n>N_\epsilon$. Then $$\int\limits_{E}{f}\leq \int\limits_{E}{f_n}+\int\limits_{X}{f-f_n}<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$ for each $E:\mu(E)<\frac{\epsilon}{2N_\epsilon}$

Answer 3

Hint: I'm assuming we know $ \int_X|f|\,d\mu < \epsilon.$ (You didn't say.) Show the result holds if $f$ is bounded, and then recall there are simple functions close to $f$ in the $L^1$ norm.