# Show that $\exists \delta >0$ so the inequality holds (Lebesgue integral)

Let $$(X, \mathcal{A}, \mu)$$ be a measure space, $$u \in \mathcal{L}^1(\mu)$$ and $$K_n = \{|u| \leq n\}$$ for every $$n\geq 1$$.

Show that there exist $$\delta>0$$ so:

$$\forall E \in \mathcal{A}: \mu(E) < \delta \implies \left| \int_E u d\mu \right| < \frac{1}{2019}$$

I can't seem to understand this questions fully. Can someone explain or give a hint to how I am suppose to approach this problem? It seems I need to show that for $$\mu(E) < \delta$$ there exist an $$\epsilon= \frac{1}{2019}$$, so the inequality hold

Can I use the following for anything? $$\Bigg|\int_E u d\mu\Bigg| \leq n \cdot \mu(E) + \int_{X\backslash K_n}{|u|}d\mu$$

$$\left|\int_E ud\mu\right| \leq \left|\int_{E\cap K_n} ud\mu\right| + \left|\int_{E\cap K_n^{c}} ud\mu\right|.$$
The first term does not exceed $$n \mu (E)$$ and the second one does not exceed $$\int_{K_n^{c}} |u|d\mu$$. By DCT this last integral tends to $$0$$ as $$n \to \infty$$. Choose $$n$$ such that it is less than $$\epsilon/2$$ and then choose $$\delta$$ such that $$n\delta <\epsilon/2$$. Take $$\epsilon =\frac 1 {2019}$$.