Lebesgue integrable function when integrated on a set is less that $\epsilon$ Let $f$ be a non-negative measurable and integrable function on measure space $(X,M,\mu)$. Show that for every $\epsilon >0 $ there exists an $\delta >0 $ such that for every $E \in M$ with $\mu(E) \le \delta$ we have$\int\limits_Efd\mu \le \epsilon$  
Have been trying to approach it via simple function approximation. However depending on different $E$ we get different simple functions. Hence a uniform bound on them is not possible. Any suggestions?
 A: For every measurable $E$ and every $x\gt0$,
$$
\int_Ef\mathrm d\mu=\int_{E\cap[f\lt x]}f\mathrm d\mu+\int_{E\cap[f\geqslant x]}f\mathrm d\mu\leqslant x\mu(E)+\int_{[f\geqslant x]}f\mathrm d\mu.
$$
Choose $\varepsilon\gt0$. Since $f$ is integrable and $f\cdot\mathbf 1_{[f\geqslant x]}\leqslant f$, Lebesgue dominated convergence theorem shows that the rightmost integral in the RHS goes to zero when $x\to\infty$. Hence, there exists $x_\varepsilon$ such that this integral is $\leqslant\varepsilon/2$. Now, $\delta=\varepsilon/(2x_\varepsilon)$ guarantees that the LHS is $\leqslant\varepsilon$ for every $E$ such that $\mu(E)\leqslant\delta$.
A: Hints:


*

*If $f$ is bounded, the claim is clearly true.

*For arbitrary $f$, write $$\left| \int_E fd\mu\right|\le \left|\int_E fd\mu-\int_E (f\wedge n)d\mu\right|+\left|\int_E(f\wedge n)d\mu\right|$$ and notice that the first term can be made arbitrary small (since Levi's lemma assures that $\int (f\wedge n)\to\int f$). Thanks to previous hint, the second term poses no problem also.

*Now, given $\epsilon\gt0$, by Levi's lemma you can find $n$ s.t. first term is less than $\ldots$, second term is obviously less than $\ldots$, thus it suffices to take $\delta=\ldots$ (dependent on $n$, which is formerly chosen and fixed), to obtain that the LHS is smaller than $\epsilon$.


Can you fill up the gaps?
