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 $\epsilon =\frac 1 {2019}$.


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